Question 15 Marks
The first ionization constant of $H_2S$ is $9.1 \times 10^{–8}$. Calculate the concentration of $HS^–$ ion in its $0.1M$ solution. How will this concentration be affected if the solution is $0.1M$ in HCl also ? If the second dissociation constant of $H_2S$ is $1.2 \times 10^{–13},$ calculate the concentration of $S^{2–}$ under both conditions.
Answer
- To calculate the concentration of HS– ion:
Case I (in the absence of HCl):
Let the concentration of $HS^–$ be xM.
| |
$\text{H}_2\text{S}$ |
$\leftrightarrow$ |
$\text{H}^+$ |
$+$ |
$\text{HS}^-$ |
| $\text{C}_\text{f}$ |
$0.1$ |
|
$0$ |
|
$0$ |
| $\text{C}_\text{f}$ |
$0.1-\text{x}$ |
|
$\text{x}$ |
|
$\text{x}$ |
Then, $\text{K}_{\text{a}_1}=\frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]}$
$9.1\times10^{-8}=\frac{\text{(x)(x)}}{0.1-\text{x}}$
$(9.1\times10^{-8})(0.1-\text{x})=\text{x}^2$
$\text{Taking 0.1}-\text{xM};0.1\text{M, we have }(9.1\times10^{-8})(0.1)=\text{x}^2.$
$9.1\times10^{-9}=\text{x}^2$
$\text{x}=\sqrt{9.1\times10^{-9}}$
$=9.54\times10^{-5}\text{M}$
$\Rightarrow[\text{HS}^-]=9.54\times10^{-5}\text{M}$
Case II (in the presence of HCl):
In the presence of 0.1 M of HCl, let $[HS^-]$ be yM.
Then,
| |
$\text{H}_2\text{S}$ |
$\leftrightarrow$ |
$\text{HS}^-$ |
$+$ |
$\text{H}^+$ |
| $\text{C}_\text{f}$ |
$0.1$ |
|
$0$ |
|
$0$ |
| $\text{C}_\text{f}$ |
$0.1-\text{y}$ |
|
$\text{y}$ |
|
$\text{y}$ |
|
Also,
| $\text{HCl}$ |
$\leftrightarrow$ |
$\text{H}^+$ |
$+$ |
$\text{Cl}^-$ |
| |
|
$0.1$ |
|
$0.1$ |
Now, $\text{K}_{\text{a}_1}=\frac{[\text{HS}^-][\text{H}^+]}{[\text{H}_2\text{S}]}$
$\text{K}_{\text{c}_1}=\frac{[\text{y}(0.1+\text{y})]}{0.1-\text{y}}$ $(\because\ 0.1-\text{y};0.1\text{M})$
$9.1\times10^{-8}=\frac{\text{y}\times0.1}{0.1}$
$9.1\times10^{-8}=\text{y}$ $(\text{and}\ 0.1+\text{y};0.1\text{M})$
$\Rightarrow[\text{HS}^-]=9.1\times10^{-8}$
- To calculate the concentration of $[S^{2-}]$:
Case I (in the absence of 0.1 M HCl):
$\text{HS}^-\leftrightarrow\text{H}^++\text{S}^{2-}$
$[\text{HS}^-]=9.54\times10^{-5}\text{M}$ (From first ionization, case I)
Let $[\text{S}^{2-}]$ be x.
Also, $[\text{H}^+]=9.54\times10^{-5}\text{M}$ (From first ionization, case I)
$\text{K}_{\text{a}_2}=\frac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]}$
$\text{K}_{\text{a}_2}=\frac{(9.54\times10^{-5})(\text{x})}{9.54\times10^{-5}}$
$1.2\times10^{-13}=\text{x}=[\text{S}^{2-}]$
Case II (in the presence of 0.1 M HCl):
Again, let the concentration of $HS^–$ be X'M.
$[\text{HS}^-]=9.1\times10^{-8}\text{M}$ (From first ionization, case II)
$[\text{H}^+]=0.1\text{M}$ (From HCl, case II)
$[\text{S}^{2-}]=\text{x}'$
Then, $\text{K}_{\text{a}_2}=\frac{[\text{H}^+][\text{S}^{2-}]}{[\text{HS}^-]}$
$1.2\times10^{-13}=\frac{(0.1)(\text{x}')}{9.1\times10^{-8}}$
$10.92\times10^{-21}=0.1\text{x}'$
$\frac{10.92\times10^{-21}}{0.1}=\text{x}'$
$\text{x}'=\frac{1.092\times10^{-20}}{0.1}$
$=1.092\times10^{-19}\text{M}$
$\Rightarrow\text{K}_{\text{a}_1}=1.74\times10^{-5}$ View full question & answer→Question 25 Marks
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
AnswerThe hydrogen ion concentration in the given substances can be calculated by using the given relation:
$\text{pH}=-\log[\text{H}^+]$
- pH of milk = 6.8
Since, $\text{pH}=-\log[\text{H}^+]$
$6.8=-\log[\text{H}^+]$
$\log[\text{H}^+]=-6.8$
$[\text{H}^+]=\text{anitlog}(-6.8)$
$=1.5\times10^{-7}\text{M}$
- pH of black coffee = 5.0
Since, $\text{pH}=-\log[\text{H}^+]$
$5.0=-\log[\text{H}^+]$
$\log[\text{H}^+]=-5.0$
$[\text{H}^+]=\text{anitlog}(-5.0)$
$=10^{-5}\text{M}$
- pH of tomato juice = 4.2
Since, $\text{pH}=-\log[\text{H}^+]$
$4.2=-\log[\text{H}^+]$
$\log[\text{H}^+]=-4.2$
$[\text{H}^+]=\text{anitlog}(-4.2)$
$=6.31\times10^{-5}\text{M}$
- pH of lemon juice = 2.2
Since, $\text{pH}=-\log[\text{H}^+]$
$2.2=-\log[\text{H}^+]$
$\log[\text{H}^+]=-2.2$
$[\text{H}^+]=\text{anitlog}(-2.2)$
$=6.31\times10^{-3}\text{M}$
- pH of egg white = 7.8
Since, $\text{pH}=-\log[\text{H}^+]$
$7.8=-\log[\text{H}^+]$
$\log[\text{H}^+]=-7.8$
$[\text{H}^+]=\text{anitlog}(-7.8)$
$=1.58\times10^{-8}\text{M}$ View full question & answer→Question 35 Marks
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at $298K$ from their solubility product constants given in Table $7.9$. Determine also the molarities of individual ions.
AnswerSilver chromate: $\text{Ag}_2\text{CrO}_4\rightarrow2\text{Ag}^++\text{CrO}_4^{2-}$ Then, $\text{K}_\text{sp}=[\text{Ag}^+]^2[\text{CrO}_4^{2-}]$ Let the solubility of $\text{Ag}_2\text{CrO}_4$ be s. $\Rightarrow[\text{Ag}6+]2\text{s and}[\text{CrO}_4^{2-}]=\text{s}$ Then, $\text{K}_\text{sp}=(2\text{s})^2.\text{s}=4\text{s}^3$
$\Rightarrow1.1\times10^{-12}=4\text{s}^3$
$.275\times10^{-12}=\text{s}^3$
$\text{s}=0.65\times10^{-4}\text{M}$ Molarity of $\text{Ag}^+=2\text{s}=2\times0.65\times10^{-4}=1.30\times10^{-4}\text{M}$ Molarity of $\text{CrO}_4^{2-}=\text{s}=0.65\times10^{-4}\text{M}$ Barium chromate: $\text{Ba}\text{CrO}_4\rightarrow\text{Ba}^{2+}+\text{CrO}_4^{2-}$ Then, $\text{K}_\text{sp}=[\text{Ba}^{2+}]^2[\text{CrO}_4^{2-}]$ Let s be the solubility of $\text{Ba}\text{CrO}_4.$ Thus, $[\text{Ba}^{2+}]=\text{s and}[\text{CrO}_4^{2-}]=\text{s}$
$\Rightarrow\text{K}_\text{sp}=\text{s}^2$
$\Rightarrow1.2\times10^{-10}=\text{s}^2$
$\Rightarrow\text{s}=1.09\times10^{-5}\text{M}$ Molarity of $\text{Ba}^{2+}$ = Molarity of $\text{CrO}_4^{2-}=\text{s}=1.09\times10^{-5}\text{M}$ Ferric hydroxide: $\text{Fe}\text{(OH)}_3\rightarrow\text{Fe}^{2+}+\text{3HO}^-$ Then, $\text{K}_\text{sp}=[\text{Fe}^{2+}][\text{OH}^-]$ Let s be the solubility of $\text{Fe}\text{(OH)}_3.$ Thus, $[\text{Fe}^{3+}]=\text{s and}[\text{OH}^-]=3\text{s}$
$\Rightarrow\text{K}_\text{sp}=\text{s}.(3\text{s})^3$
$=\text{s}.27\text{s}^3$
$\text{K}_\text{sp}=27\text{s}^3$$1.0\times10^{-38}=27\text{s}^4$
$.037\times10^{-38}=\text{s}^4$
$.00037\times10^{-36}=\text{s}^4\ \Rightarrow1039\times10^{-10}\text{M=S}$
Molarity of $\text{Fe}^{3+}=\text{s}=1.39\times10^{-10}\text{M}$ Molarity of $\text{OH}^-=3\text{s}=4.17\times10^{-10}\text{M}$ Lead chloride: $\text{Pb}\text{(Cl)}_2\rightarrow\text{Pb}^{2+}+\text{2Cl}^-$
$\text{K}_\text{sp}=[\text{Pb}^{2+}][\text{Cl}^-]^2$ Let $K_{sp}$ be the solubility of $\text{Pb}\text{Cl}_2.$
$[\text{PB}^{2+}]=\text{s and}[\text{Cl}^-]=2\text{s}$
$\text{Thus, }\text{K}_\text{sp}=\text{s}.(2\text{s})^2$
$=4\text{s}^3$
$\Rightarrow1.6\times10^{-5}=4\text{s}^3$$\Rightarrow0.4\times10^{-5}=\text{s}^3$
$4\times10^{-6}=\text{s}^3\Rightarrow1.58\times10^{-2}\text{M}=\text{S.1}$
Molarity of $\text{PB}^{2+}=\text{s}=1.58\times10^{-2}\text{M}$ Molarity of chloride $=2\text{s}=3.16\times10^{-2}\text{M}$ Mercurous iodide: $\text{Hg}_2\text{I}_2\rightarrow\text{Hg}^{2+}+\text{2I}^-$
$\text{K}_\text{sp}=[\text{Hg}_2^{2+}]^2[\text{I}^-]^2$ Let s be the solubility of $\text{Hg}_2\text{I}_2.$
$\Rightarrow[\text{Hg}_2^{2+}]=\text{s and}[\text{I}^-]=2\text{s}$ Thus, $\text{K}_\text{sp}=\text{s}(2\text{s})^2\Rightarrow\text{K}_\text{sp}=4\text{s}^3$
$4.5\times10^{-29}=4\text{s}^3$
$1.125\times10^{-29}=\text{s}^3$$\Rightarrow\text{s}=2.24\times10^{-10}\text{M}$
Molarity of $\text{Hg}_2^{2+}=\text{s}=2.24\times10^{-10}\text{M}$ Molarity of $\text{I}^-=2\text{s}=4.48\times10^{-10}\text{M}$
View full question & answer→Question 45 Marks
Calculate the degree of ionization of $0.05M$ acetic acid if its $pK_a$ value is $4.74$. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl?
Answer$\text{c}=0.05\text{M}$ $\text{pK}_\text{a}=4.74$ $\text{pK}_\text{a}=-\log(\text{K}_\text{a})$ $\text{K}_\text{a}=1.82\times10^{-5}$ $\text{K}_\text{a}=\text{c}\alpha^2\ \alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$ $\alpha=\sqrt{\frac{1.82\times10^{-5}}{5\times10^{-2}}}=1.908\times10^{-2}$ When HCl is added to the solution, the concentration of $H^+$ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease.Case I: When 0.01M HCl is taken.
Let x be the amount of acetic acid dissociated after the addition of HCl.
| |
$\text{CH}_3\text{COOH}$ |
$\leftrightarrow$ |
$\text{H}^+$ |
$+$ |
$\text{CH}_3\text{COO}^-$ |
| Inilial conc. |
$0.05\text{M}$ |
|
$0$ |
|
$0$ |
| After dissociation |
$0.05-\text{x}$ |
|
$0.01+\text{x}$ |
|
$\text{x}$ |
As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05-x and 0.01 + xcan be taken as 0.05 and 0.01 respectively.
$\text{K}_\text{a}=\frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}$
$\therefore\ \text{K}_\text{a}=\frac{(0.01)\text{x}}{0.05}$
$\text{x}=\frac{1.82\times10^{-5}\times0.05}{0.01}$
$\text{x}=1.82\times10^{-3}\times0.05\text{M}$
Now,
$\alpha=\frac{\text{Amount of acid dissociated}}{\text{Amount of acid taken}}$
$=\frac{1.82\times10^{-3}\times0.05}{0.05}$
$=1.82\times10^{-3}$
Case II: When 0.1M HCl is taken.Let the amount of acetic acid dissociated in this case be X. As we have done in the first case, the concentrations of various species involved in the reaction are:
$[\text{CH}_3\text{COOH}]=0.05-\text{x ; }0.05\text{M}$
$[\text{CH}_3\text{COO}^-]=\text{x}$
$[\text{H}^+]=0.1+\text{x ; }0.1\text{M}$
$\text{K}_\text{a}=\frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]}$
$\therefore\ \text{K}_\text{a}=\frac{(0.01)\text{x}}{0.05}$
$\text{x}=\frac{1.82\times10^{-5}\times0.05}{0.01}$
$\text{x}=1.82\times10^{-4}\times0.05\text{M}$
Now,
$\alpha=\frac{\text{Amount of acid dissociated}}{\text{Amount of acid taken}}$
$=\frac{1.82\times10^{-4}\times0.05}{0.05}$
$=1.82\times10^{-4}$ View full question & answer→Question 55 Marks
The ionization constant of acetic acid is $1.74 \times 10^{–5}$. Calculate the degree of dissociation of acetic acid in its $0.05M$ solution. Calculate the concentration of acetate ion in the solution and its pH.
AnswerMethod 1
- $\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+\ \text{K}_\text{a}=1.74\times10^{-5}$
- $\text{H}_2\text{O}+\text{H}_2\text{O}\leftrightarrow\text{H}_3\text{O}^++\text{OH}^-\ \text{K}_\text{w}=1.0\times10^{-14}$
Since $Ka >> K_w$:
| |
|
$$$\text{CH}_3\text{COOH}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$\text{CH}_3\text{COO}^-$ |
$+$ |
$\text{H}_3\text{O}^+$ |
| $\text{C}_\text{i}$ |
$=$ |
$0.05$ |
|
|
|
$0$ |
|
$0$ |
| |
|
$0.05-.05\alpha$ |
|
|
|
$0.05\alpha$ |
|
$0.05\alpha$ |
$\text{K}_\text{a}=\frac{(.05\alpha)(.05\alpha)}{(.05-0.05\alpha)}$
$=\frac{(.05\alpha)(0.05\alpha)}{.05(1-\alpha)}$
$=\frac{.05\alpha^2}{1-\alpha}$
$1.74\times10^{-5}=\frac{0.05\alpha^2}{1-\alpha}$
$1.74\times10^{-5}-1.74\times10^{-5}\alpha=0.05\alpha^2$
$0.05\alpha^2+1.74\times10^{-5}\alpha-1.74\times10^{-5}$
$\text{D}=\text{b}^2-4\text{ac}$
$=(1.74\times10^{-5})^2-4(.05)(1.74\times10^{-5})$
$=3.02\times10^{-25}+.348\times10^{-5}$
$\alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$
$\alpha=\sqrt{\frac{1.74\times10^{-5}}{.05}}$
$=\sqrt{\frac{34.8\times10^{-5}\times10}{10}}$
$=\sqrt{3.48\times10^{-6}}$
$=\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+$
$\alpha=1.86\times10^{-3}$
$[\text{CH}_3\text{COO}^-]=0.05\times1.86\times10^{-3}$
$=\frac{0.93\times10^{-3}}{1000}$
$=.000093$
Method 2
Degree of dissociation,
$\alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$
c = 0.05M
$K_a = 1.74 \times 10^{–5}$
Then, $\alpha=\sqrt{\frac{1.74\times10^{-5}}{.05}}$
$\alpha=\sqrt{34.8\times10^{-5}}$
$\alpha=\sqrt{3.48}\times10^{-4}$
$\alpha=1.8610^{-2}$
$\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+$
Thus, concentration of $CH_3COO– = c.\alpha$
$=.05\times1.86\times10^{-2}$
$=.093\times10^{-2}$
$=.00093\text{M}$
$\text{Since}[\text{oAc}^-]=[\text{H}^+],$
$[\text{H}^+]=.00093=.093\times10^{-2}.$
$\text{pH}=-\log[\text{H}^+]$
$=-\log(.093\times10^{-2})$
$\therefore\ \text{pH}=3.03$
Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03. View full question & answer→Question 65 Marks
The ionization constant of phenol is $1.0 \times 10^{–10}$. What is the concentration of phenolate ion in $0.05M$ solution of phenol? What will be its degree of ionization if the solution is also $0.01M$ in sodium phenolate?
AnswerIonization of phenol:
| |
$\text{C}_6\text{H}_5\text{OH}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$\text{C}_6\text{H}_5\text{O}^-$ |
$+$ |
$\text{H}_3\text{O}^+$ |
| Initial conc. |
$0.05$ |
|
|
|
$0$ |
|
$0$ |
| At equilibrium |
$0.05-\text{x}$ |
|
|
|
$\text{x}$ |
|
$\text{x}$ |
$\text{K}_\text{a}=\frac{[\text{C}_6\text{H}_5\text{O}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{OH}]}$ $\text{K}_\text{a}=\frac{\text{x}\times\text{x}}{0.05-\text{x}}$ As the value of the ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator. $\therefore\ \text{x}=\sqrt{1\times10^{-10}\times0.05}$ $=\sqrt{5\times10^{-12}}$ $=2.2\times10^{-6}\text{M}=[\text{H}_3\text{O}^+]$ Since $[\text{H}_3\text{O}^+]=[\text{C}_6\text{H}_5\text{O}^-],$ $[\text{C}_6\text{H}_5\text{O}^-]=2.2\times10^{-6}\text{M.}$ Now, let ∝ be the degree of ionization of phenol in the presence of 0.01 $M C_6H_5ONa$.
| |
$\text{C}_6\text{H}_5\text{ONa}$ |
$\rightarrow$ |
$\text{C}_6\text{H}_5\text{O}^-$ |
$+$ |
$\text{Na}^+$ |
| Conc. |
|
|
|
|
$0.01$ |
Also,
| |
$\text{C}_6\text{H}_5\text{OH}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$\text{C}_6\text{H}_5\text{O}^-$ |
$+$ |
$\text{H}_3\text{O}^+$ |
| Conc. |
$0.05-0.05\alpha$ |
|
|
|
$0.05\alpha$ |
|
$0.05\alpha$ |
$[\text{C}_6\text{H}_5\text{O}^-]=0.01+0.05\alpha;0.01\text{M}$ $[\text{H}_3\text{O}^-]=0.05\alpha$ $\text{K}_\text{a}=\frac{[\text{C}_6\text{H}_5\text{O}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{OH}]}$ $\text{K}_\text{a}=\frac{(0.01)(0.05\alpha)}{0.05}$ $1.0\times10^{-10}=.01\alpha$ $\alpha=1\times10^{-8}$ View full question & answer→Question 75 Marks
The ionization constant of nitrous acid is $4.5 \times 10^{–4}$. Calculate the pH of $0.04M$ sodium nitrite solution and also its degree of hydrolysis.
Answer$NaNO_2$_ is the salt of a strong base (NaOH) and a weak acid $(HNO_2)$. $\text{NO}_2^-+\text{H}_2\text{O}\leftrightarrow\text{HNO}_2+\text{OH}^-$
$\text{K}_\text{h}=\frac{[\text{HNO}_2][\text{OH}^-]}{[\text{NO}_2^-]}$
$\Rightarrow\frac{\text{K}_\text{w}}{\text{K}_\text{a}}=\frac{10^{-14}}{4.5\times10^{-4}}=.22\times10^{-10}$ Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be: $[\text{NO}^-_2]=.04-\text{x};0.04$
$[\text{HNO}_2]=\text{x}$
$[\text{OH}^-]=\text{x}$
$\text{K}_\text{h}=\frac{\text{x}^2}{0.04}=0.22\times10^{-10}$
$\text{x}^2=.0088\times10^{-10}$
$\text{x}=.093\times10^{-5}$
$\therefore\ [\text{OH}^-]=0.093\times10^{-5}\text{M}$
$[\text{H}_3\text{O}^+]=\frac{10^{-14}}{.093\times10^{-5}}=10.75\times10^{-9}\text{M}$$\Rightarrow\text{pH}=-\log(10.75\times10^{-9})$
$=7.96$ Therefore, degree of hydrolysis $=\frac{\text{x}}{0.04}=\frac{.093\times10^{-5}}{.04}=2.325\times10^{-5}$
View full question & answer→Question 85 Marks
The ionization constant of chloroacetic acid is $1.35 \times 10^{–3}$. What will be the pH of $0.1M$ acid and its $0.1M$ sodium salt solution?
AnswerIt is given that $K_a$ for $ClCH_2COOH$ is $1.35 \times 10^{–3}$.
$\Rightarrow\text{K}_\text{a}=\text{c}\alpha^2$
$\therefore\ \alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$
$=\sqrt{\frac{1.35\times10^{-3}}{0.1}}$ $(\therefore\ \text{concentration of acid = 0.1m})$
$\alpha=\sqrt{1.35\times10^{-2}}$
$=0.116$
$\therefore\ [\text{H}^+]=\text{c}\alpha=0.1\times0.116$
$=0.116$
$\Rightarrow\text{pH}=-\log[\text{H}^+]=1.94$
$ClCH_2COONa$ is the salt of a weak acid i.e., $ClCH_2COOH$ and a strong base i.e., NaOH.
$\text{ClCH}_2\text{COO}^-+\text{H}_2\text{O}\leftrightarrow\text{ClCH}^2\text{COOH}+\text{OH}^-$
$\text{K}_\text{h}=\frac{[\text{ClCH}^2\text{COOH}][\text{OH}^-]}{[\text{ClCH}_2\text{COO}^-]}$
$\text{K}_\text{h}=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$\text{K}_\text{h}=\frac{10^{-14}}{1.35\times10^{-3}}$
$=0.740\times10^{-11}$
Also, $\text{K}_\text{h}=\frac{\text{x}^2}{0.1}$ $(\text{where x is the concentration of OH}^-\text{and ClCH}_2\text{COOH})$
$0.740\times10^{-11}=\frac{\text{x}^2}{0.1}$
$0.074\times10^{-11}=\text{x}^2$
$\Rightarrow\text{x}^2=0.74\times10^{-12}$
$\text{x}=0.86\times10^{-6}$
$[\text{OH}^-]=0.86\times10^{-6}$
$\therefore\ [\text{H}^+]=\frac{\text{K}_\text{w}}{0.86\times10^{-6}}$
$=\frac{10^{-14}}{0.86\times10^{-6}}$
$[\text{H}^+]=1.162\times10^{-8}$
$\text{pH}=-\log[\text{H}^+]$
$=7.94$
View full question & answer→Question 95 Marks
The ionization constant of benzoic acid is $6.46 \times 10^{–5}$ and $K_{sp}$ for silver benzoate is $2.5 \times 10^{–13}$. How many times is silver benzoate more soluble in a buffer of $pH\ 3.19$ compared to its solubility in pure water?
AnswerSince pH = 3.19,
$[\text{H}_3\text{O}^+]=6.46\times10^{-4}\text{M}$
$\text{C}_6\text{H}_5\text{COOH}+\text{H}_2\text{O}\leftrightarrow\text{C}_6\text{H}_5\text{COO}^-+\text{H}_3\text{O}$
$\text{K}_\text{a}\frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{COOH}]}$
$\frac{[\text{C}_6\text{H}_5\text{COOH}]}{[\text{C}_6\text{H}_5\text{COO}^-]}=\frac{[\text{H}_3\text{O]}}{\text{K}_\text{a}}=\frac{6.46\times10^{-4}}{6.46\times10^{-5}}=10$
Let the solubility of $C_6H_5COO$Ag be xmol/L.
Then,
$[\text{Ag}^+]=\text{x}$
$[\text{C}_6\text{H}_5\text{COOH}]+[\text{C}_6\text{H}_5\text{COO}^-]=\text{x}$
$10[\text{C}_6\text{H}_5\text{COO}^-]+[\text{C}_6\text{H}_5\text{COO}^-]=\text{x}$
$[\text{C}_6\text{H}_5\text{COO}^-]=\frac{\text{x}}{11}$
$\text{K}_\text{sp}[\text{Ag}^+][\text{C}_6\text{H}_5\text{COO}^-]$
$2.5\times10^{-13}=\text{x}\Big(\frac{\text{x}}{11}\Big)$
$\text{x}1.66\times10^{-6}\text{mol/L}$
Thus, the solubility of silver benzoate in a pH 3.19 solution is $1.66 \times 10^{–6}mol/L$.
Now, let the solubility of $C_6H_5COO$ Ag be x'mol/L.
Then,
$[\text{Ag}^+]=\text{x}'\text{M and}[\text{CH}_3\text{COO}^-]=\text{x}'\text{M}.$
$\text{K}_\text{sp}=[\text{Ag}^+][\text{CH}_3\text{COO}^-]$
$\text{K}_\text{sp}=\text{(x}')^2$
$\text{x}'=\sqrt{\text{K}_\text{sp}}=\sqrt{2.5\times10^{-13}}=5\times10^{-7}\text{mol/L}$
$\therefore\ \frac{\text{x}}{\text{x}'}=\frac{1.66\times10^{-6}}{5\times10^{-7}}=3.32$
Hence, $C_6H_5COO$Ag is approximately 3.317 times more soluble in a low pH solution.
View full question & answer→Question 105 Marks
A mixture of 1.57 mol of $N_2$, 1.92 mol of $H_2$ and 8.13 mol of $NH_3$ is introduced into a $20L$ reaction vessel at $500K$. At this temperature, the equilibrium constant, $K_c$ for the reaction $\text{N}_2\text{ (g) + }3\text{H}_2\text{ (g)}\rightleftharpoons2\text{NH}_3\text{ (g) is }1.7\times10^2.$ Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
AnswerThe given reaction is:
$\text{N}_{2\text{(g)}}+3\text{H}_{2\text{(g)}}\leftrightarrow2\text{NH}_{3\text{(g)}}$
The given concentration of various species is
$[\text{N}_2]=\frac{1.57}{20}\text{mol L}^{-1}$ $[\text{H}_2]=\frac{1.92}{20}\text{mol L}^{-1}$
$[\text{NH}_3]=\frac{8.13}{20}\text{mol L}^{-1}$
Now, reaction quotient $Q_c$ is:
$\text{Q}_{\text{c}}=\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$
$=\frac{\Big(\frac{(8.13)}{20}\Big)^2}{\Big(\frac{1.57}{20}\Big)\Big(\frac{1.92}{20}\Big)^3}$
$=2.4\times10^3$
Since, $\text{Q}_{\text{c}}\neq\text{K}_{\text{c}},$ the reaction mixture is not at equilibrium.
Again, $\text{Q}_{\text{c}}>\text{K}_{\text{c}}.$ Hence, the reaction will proceed in the reverse direction.
View full question & answer→Question 115 Marks
The ionization constant of dimethylamine is $5.4 \times 10^{–4}$. Calculate its degree of ionization in its $0.02M$ solution. What percentage of dimethylamine is ionized if the solution is also $0.1M$ in NaOH?
Answer$\text{K}_\text{b}=5.4\times10^{-4}$
$\text{c}=0.02\text{M}$
$\text{Then, }\alpha=\sqrt{\frac{\text{K}_\text{b}}{\text{c}}}$
$=\sqrt{\frac{5.4\times10^{-4}}{0.02}}$
$=0.1643$
Now, if 0.1M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.
| $\text{NaOH}_\text{(aq)}$ |
$\leftrightarrow$ |
$\text{Na}^+_\text{(aq)}$ |
$+$ |
$\text{OH}^-_\text{(aq)}$ |
| |
|
$0.1\text{M}$ |
|
$0.1\text{M}$ |
And,
| $\text{(CH}_3)_2\text{NH}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$(\text{CH}_3)_2\text{NH}_2^+$ |
$+$ |
$\text{OH}^-$ |
| $(0.02-\text{x})$ |
|
|
|
$\text{x}$ |
|
$\text{x}$ |
| $;0.02\text{M}$ |
|
|
|
|
|
$;0.1\text{M}$ |
Then, $[(\text{CH}_3)_2\text{NH}_2^+]=\text{x}$
$[\text{OH}^-]=\text{x}+0.1;0.1$
$\Rightarrow\text{K}_\text{b}=\frac{[(\text{CH}_3)_2\text{NH}_2^+][\text{OH}^-]}{[(\text{CH}_3)_2\text{NH]}}$
$5.4\times10^{-4}=\frac{\text{x}\times0.1}{0.02}$
$\text{x}=0.0054$
It means that in the presence of 0.1M NaOH, 0.54% of dimethylamine will get dissociated. View full question & answer→Question 125 Marks
The equilibrium constant for the following reaction is $1.6 \times 10^5$ at 1024K $\text{H}_2\text{ (g) + Br (g)}\rightleftharpoons2\text{HBr (g)}$ Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a
sealed container at 1024K.
AnswerGiven,$\text{K}_\text{p}$ for the reaction i.e., $\text{H}_{2\text{ (g)}}\text{ + Br }_{2\text{(g)}}\leftrightarrow2\text{HBr}_\text{(g)}\text{ is }1.6\times10^5.$
Therefore, for the reaction $$$2\text{HBr}_\text{(g)}\leftrightarrow\text{H}_{2\text{ (g)}}\text{ + Br }_{2\text{(g)}},$ the equilibrium constant will be,
$\text{K}'_\text{p}=\frac{1}{\text{K}_\text{p}}$
$=\frac{1}{1.6\times10^5}$
$=6.25\times10^{-6}$
Now, let p be the pressure of both $H_2$ and $Br_2$ at equilibrium.
| |
$2\text{HBr}_\text{(g)}$ |
$\leftrightarrow$ |
$\text{H}_{2\text{ (g)}}$ |
$+$ |
$\text{Br }_{2\text{(g)}}$ |
| Initial conc. |
$10$ |
|
$0$ |
|
$0$ |
| At equilibrium |
$10-2\text{p}$ |
|
$\text{p}$ |
|
$\text{p}$ |
Now, we can write, $\frac{\text{p}_{\text{HBr}}\times\text{p}_2}{\text{p}^2_\text{HBr}}=\text{K}'_\text{p}$ $\frac{\text{p}\times\text{p}}{(10-2\text{p})^2}=6.25\times10^{-6}$ $\frac{\text{p}}{10-2\text{p}}=2.5\times10^{-3}$ $\text{p}=2.5\times10^{-2}-(5.0\times10^{-3})\text{p}$ $\text{p}+(5.0\times10^{-3})\text{p}=2.5\times10^{-2}$ $(1005\times10^{-3})\text{p}=2.5\times10^{-2}$$\text{p}=2.49\times10^{-2}\text{ bar }=2.5\times10^{-2}(\text{approximately})$
Therefore, at equilibrium, $[\text{H}_2]=[\text{Br}_2]=2.49\times10^{-2}\text{ bar}$ $[\text{HBr}]=10-2\times(2.49\times10^{-2)}\text{ bar}$ $=9.95\text{ bar}=10\text{ bar}(\text{approximately})$ View full question & answer→Question 135 Marks
One of the reaction that takes place in producing steel from iron ore is the
reduction of iron(II) oxide by carbon monoxide to give iron metal and $CO_2$.
$\text{FeO (S) + CO (g)}\rightleftharpoons\text{Fe (S) + CO}_2\text{ (g)};\text{ K}_{\text{p}}=0.265\text{atm at}1050\text{K}$
What are the equilibrium partial pressures of $CO$ and $CO_2$ at 1050 K if the initial partial pressures are: $p_{CO}= 1.4atm$ and = $0.80atm$?
AnswerFor the given reaction,
| $\text{FeO}_{\text{(g)}}$ |
$+$ |
$\text{CO}_{\text{(g)}}$ |
$\leftrightarrow$ |
$\text{Fe}_{\text{(s)}}$ |
$+$ |
$\text{CO}_{2\text{(g)}}$ |
| Initialy, |
|
$1.4\text{atm}$ |
|
|
|
$0.80\text{atm}$ |
$\text{Q}_{\text{p}}=\frac{\text{p}_{\text{co}_2}}{\text{p}_{\text{co}}}$
$=\frac{0.80}{1.4}$
$=0.571$ It is given that $\text{K}_{\text{p}}=0.265.$ Since $\text{Q}_{\text{p}}>\text{K}_{\text{p}},$ the reaction will proceed in the backward direction. Therefore, we can say that the pressure of CO will increase while the pressure of $CO_2$_ will decrease. Now, let the increase in pressure of CO = decrease in pressure of $CO_2$ be p. Then, we can write, $\text{K}_{\text{p}}=\frac{\text{p}_{\text{co}_2}}{\text{p}_{\text{co}}}$
$\Rightarrow0.265=\frac{0.80-\text{p}}{1.4+\text{p}}$
$\Rightarrow0.371+0.265\text{p}=0.80-\text{p}$
$\Rightarrow1.265\text{p}=0.429$
$\Rightarrow\text{p}=0.339\text{atm}$ Therefore, equilibrium partial of $\text{CO}_2,\text{ p}_{\text{co}_2}=0.80-0.339=0.461\text{atm.}$ And, equilibrium partial pressure of $\text{CO},\text{ p}_{\text{co}_2}=1.4-0.339=1.739\text{atm.}$ View full question & answer→Question 145 Marks
Nitric oxide reacts with $Br_2$ and gives nitrosyl bromide as per reaction given below:
$2\text{NO (g) + Br}_2\text{ (g)}\leftrightharpoons2\text{NOBr (g)}$
When $0.087$ mol of $NO$ and $0.0437$ mol of $Br _2$ are mixed in a closed container at constant temperature, $0.0518$ mol of $NOBr$ is obtained at equilibrium. Calculate equilibrium amount of $NO$ and $Br _2$.
AnswerThe given reaction is:
$\begin{matrix}2\text{NO}_{\text{(g)}}&+&\text{Br}_{2\text{(g)}}&\leftrightarrow&2\text{NOBr}_\text{(g)}\\2\text{ mol}&&\text{1 mol}&&\text{2 mol}\end{matrix}$
Now, 2mol of NOBr are formed from 2mol of NO. Therefore, 0.0518mol of NOBr are formed from 0.0518mol of NO.
Again, 2mol of NOBr are formed from 1mol of Br.
Therefore, 0.0518mol of NOBr are formed from $\frac{0.0518}{2}$mol of Br, or 0.0259mol of NO.
The amount of NO and Br present initially is as follows:
$[NO] = 0.087mol [Br_2] = 0.0437mol$
Therefore, the amount of $NO$ present at equilibrium is:
$[NO] = 0.087 – 0.0518$
$= 0.0352mol$
And, the amount of Br present at equilibrium is:
$[Br_2] = 0.0437 – 0.0259$
$= 0.0178mol$
View full question & answer→Question 155 Marks
Reaction between $N_2$ and $O_{2–}$ takes place as follows:
$\text{2N}_2\text{(g) + O}_2\text{(g)}\rightleftharpoons2\text{N}_2\text{O(g)}$
If a mixture of 0.482mol $N_2$ and 0.933mol of $O_2$ is placed in a 10L reaction vessel and allowed to form $N_2O$ at a temperature for which $\text{K}_{\text{c}}=2.0\times10^{-37},$ determine the composition of equilibrium mixture.
AnswerLet the concentration of $N_2O$ at equilibrium be x.
The given reaction is:
$\begin{matrix}&2\text{N}_{2(\text{g})}&+&\text{O}_{2\text{(g)}}&\leftrightarrow&2\text{N}_2\text{O}_{\text{(g)}}\\\text{Initial Conc.}&0.482\text{ mol}&&0.933\text{ mol}&&0\\\text{At equiluibrium}&(0.482-\text{x})\text{mol}&&(0.933-\text{x})\text{mol}&&\text{x mol} \end{matrix}$
Therefore, at equilibrium, in the 10L vessel:
$[\text{N}_2]=\frac{0.482-\text{x}}{10},[\text{O}_2]=\frac{0.933-\text{x/2}}{10},[\text{N}_2\text{O]}=\frac{\text{x}}{10}$
The value of equilibrium constant i.e., $\text{K}_{\text{c}}=2.0\times10^{-37}$ is very small. Therefore, the amount of $N_2$ and $O_2$ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of $N_2$ and $O_2$.
Then,
$[\text{N}_2]=\frac{0.482}{10}=0.0482\text{mol L}^{-1}\text{and}[\text{O}_2]=\frac{0.933}{10}=0.0933\text{mol L}^{-1}$
Now,
$\text{K}_{\text{c}}=\frac{\big[\text{N}_2\text{O}_{\text{(g)}}\big]^2}{\big[\text{N}_{2\text{(g)}}\big]^2\big[\text{O}_{2\text{(g)}}\big]}$
$\Rightarrow2.0\times10^{-37}=\frac{\big(\frac{\text{x}}{10}\big)^2}{(0.0482)^2(0.0933)}$
$\Rightarrow\frac{\text{x}^2}{100}=2.0\times10^{-37}\times(0.0482)^2\times(0.0933)$
$\Rightarrow\text{x}^2=43.35\times10^{-40}$
$\Rightarrow\text{x}=6.6\times10^{-20}$
$[\text{N}_2\text{O}]=\frac{\text{x}}{10}=\frac{6.6\times10^{-20}}{10}$
$=6.6\times10^{-21}$
View full question & answer→Question 165 Marks
At $450K, K_p= 2.0 \times 10^{10}/bar$ for the given reaction at equilibrium.
$2\text{SO}_2\text{(g) + O}_2\text{(g)}\rightleftharpoons\text{2SO}_3\text{(g)}$
What is $K_c$ at this temperature?
AnswerFor the given reaction,
$\Delta\text{n}=2-3=-1$
$\text{T}=450\text{K}$
$\text{R}=0.0831 \text{ bar L bar K}^{-1}\text{mol}^{-1}$
$\text{K}_{\text{p}}=2.0\times10^{10}\text{ bar}^{-1}$
We know that,
$\text{K}_{\text{p}}=\text{K}_{\text{c}}(\text{RT})\Delta\text{n}$
$\Rightarrow2.0\times10^{10}\text{ bar}^{-1}=\text{K}_{\text{c}}(0.0831\text{ L bar K}^{-1}\text{mol}^{-1}\times450\text{K})^{-1}$
$\Rightarrow\text{K}_{\text{c}}=\frac{2.0\times10^{-10}\text{ bar}^{-1}}{(0.0831\text{ L bar}^{-1}\text{ K}^{-1}\text{mol}^{-1}\times450\text{K})^{-1}}$
$=(2.0\times10^{10}\text{ bar}^{-1})(0.0831\text{ L bar K}^{-1}\text{mol}^{-1}\times450\text{K})$
$=74.79\times10^{10}\text{L mol}^{-1}$
$=7.48\times10^{11}\text{L mol}^{-1}$
$=7.48\times10^{11}\text{M}^{-1}$
View full question & answer→Question 175 Marks
Calculate the pH of the following solutions:
0.3g of Ca(OH)2 dissolved in water to give 500mL of solution.
AnswerFor 0.3 g of $Ca(OH)_2$ dissolved in water to give 500 mL of solution:
$\text{Ca(OH)}_2\rightarrow\text{Ca}^{2+}+2\text{OH}^-$
$[\text{Ca(OH})_2]=0.3\times\frac{1000}{500}=0.6\text{M}$
$[\text{OH}^-_\text{aq}]=2\times[\text{Ca(OH})_{2\text{aq}}]=2\times0.6$
$=1.2\text{M}$
$[\text{H}^+]=\frac{\text{K}_\text{w}}{[\text{OH}^-_\text{aq}]}$
$=\frac{10-14}{1.2}\text{M}$
$=0.833\times10^{-14}$
$\text{pH}=-\log(0.833\times10^{-14})$
$=-\log(8.33\times10^{-13})$
$=(-0.902+13)$
$=12.098$
View full question & answer→Question 185 Marks
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
$\text{CH}_4\text{ (g) + H}_2\text{O}\rightleftharpoons\text{CO (g) + 3H}_2\text{ (g)}$
- Write as expression for $K_p$ for the above reaction.
- How will the values of $K_p$ and composition of equilibrium mixture be affected by.
- increasing the pressure
- increasing the temperature
- using a catalyst?
Answer
- $\text{K}_\text{p}=\frac{\text{p}_\text{CO}\times\text{p}^3_{\text{H}_2}}{\text{p}_{\text{CH}_4}\times\text{p}_{\text{H}_2\text{O}}}$
-
- By Le Chatelier’s principle, on increasing pressure, equilibrium will shift in the backward direction where decreases number of moles.
- As the given reaction is endothermic, by Le Chatelier ‘s principle, equilibrium will shift in the forward direction with increasing temperature.
- Equilibrium composition will not be disturbed by the presence of catalyst but equilibrium will be attained quickly.
View full question & answer→Question 195 Marks
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:$2\text{BrCl (g)}\rightleftharpoons\text{Br}_2\text{ (g) + Cl}_2\text{ (g)}$for which $K_c= 32$ at $500K$. If initially pure BrCl is present at a concentration of
$3.3 \times 10^{–3} mol L^{–1}$, what is its molar concentration in the mixture at equilibrium?
Answer
| |
$2\text{BrCl}_{\text{(g)}}$ |
$\rightleftharpoons$ |
$\text{Br}_{2\text{(g)}}$ |
$+$ |
$\text{Cl}_{2\text{(g)}}$ |
| Initial |
$3.30\times10^{-3}\text{mol L}^{-1}$ |
|
$0$ |
|
$0$ |
| At eqm. |
$(3.30\times10^{-3}-\text{x})$ |
|
$\frac{\text{x}}{2}$ |
|
$\frac{\text{x}}{2}$ |
$\text{K}_{\text{c}}=\frac{(\text{x/2})(\text{x/2})}{(3.30\times10^{-3}-\text{x})^2}=32\text{ (Given)}$
$\therefore\ \frac{\text{x}^2}{4(3.30\times10^{-3}-\text{x})^2}=32$
$\text{or, }\frac{\text{x}}{2(3.30\times10^{-3}-\text{x})^2}=\sqrt{32}=5.66$
$\text{or, }\text{x=11.32(3.30}\times10^{-3}-\text{x})$
$\text{or, }12.32\text{x}=11.32\times3.30\times10^{-3}$
$\text{or, }\text{x}=3.0\times10^{-3}$
$\therefore\ \text{At eqm., [BrCl]}=(3.30\times10^{-3}-3.0\times10^{-3})$
$=0.30\times10^{-3}$
$=3.0\times10^{-4}\text{mol L}^{-1}$ View full question & answer→Question 205 Marks
Calculate the pH of the following solutions:
2g of TlOH dissolved in water to give 2litre of solution.
AnswerFor 2g of TlOH dissolved in water to give 2L of solution:
$[\text{TIOH}_\text{(aq)}]=\frac{2}{2}\text{g/L}$
$\frac{2}{2}\times\frac{1}{221}\text{M}$
$=\frac{1}{221}\text{M}$
$\text{TIOH}_\text{(aq)}\rightarrow\text{TI}_\text{(aq)}^++\text{OH}_\text{(aq)}^-$
$[\text{OH}_\text{(aq)}^-]=[\text{TIOH}_\text{(aq)}]=\frac{1}{221}\text{M}$
$\text{K}_\text{w}=[\text{H}^+][\text{OH}^-]$
$10^{-14}=[\text{H}^+]\big(\frac{1}{221}\big)$
$221\times10^{-14}=[\text{H}^+]$
$\Rightarrow\text{pH}=-\log[\text{H}^+]=-\log(221\times10^{-14})$
$=-\log(2.21\times10^{-12})$
$=11.65$
View full question & answer→Question 215 Marks
The ionization constant of propanoic acid is $1.32 \times 10^{–5}$. Calculate the degree of ionization of the acid in its $0.05M$ solution and also its pH. What will be its degree of ionization if the solution is $0.01M$ in HCl also?
AnswerLet the degree of ionization of propanoic acid be α.
Then, representing propionic acid as HA, we have:
| $\text{HA}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$\text{H}_3\text{O}^+$ |
$+$ |
$\text{A}^-$ |
| $(.05-0.0\alpha)\approx.05$ |
$.05\alpha$ |
|
$.05\alpha$ |
$\text{K}_\text{a}=\frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA]}}$
$=\frac{(.05\alpha)(.05\alpha)}{0.05}=.05\alpha^2$
$\alpha=\sqrt{\frac{\text{K}_\text{a}}{.05}}=1.63\times10^{-2}$
Then, $[\text{H}_3\text{O}^+]=.05\alpha=0.5\times1.63\times10^{-2}=\text{K}_\text{b}.15\times10^{-4}\text{M}$
$\therefore\ \text{pH}=3.09$
In the presence of 0.1M of HCl, let α´ be the degree of ionization.
$\text{Then, }[\text{H}_3\text{O}^+]=0.01$
$[\text{A}^-]=005\alpha'$
$[\text{HA]}=.05$
$\text{K}_\text{a}=\frac{0.01\times0.5\alpha'}{.05}$
$1.32\times10^{-5}=.01\times\alpha'$
$\alpha'=1.32\times10^{-3}$ View full question & answer→Question 225 Marks
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: $\text{CH}_3\text{COOH (1) + C}_2\text{H}_5\text{OH (1)}\rightleftharpoons\text{CH}_3\text{COOC}_2\text{H}_5 \text{ (1) + H}_2\text{O (1)}$Starting with 0.5mol of ethanol and 1.0mol of acetic acid and maintaining it at 293K, 0.214mol of ethyl acetate is found after sometime. Has equilibrium been reached?
AnswerLet the volume of the reaction mixture be V.
| |
$\text{CH}_3\text{COOH}_{\text{(l)}}$ |
$+$ |
$\text{C}_2\text{H}_5\text{OH}_{\text{(l)}}$ |
$\leftrightarrow$ |
$\text{CH}_3\text{COOC}_2\text{H}_{5\text{(l)}}$ |
$+$ |
$\text {H}_2\text{O}$ |
| Initial conc. |
$\frac{1.0}{\text{V}}\text{M}$ |
|
$\frac{0.5}{\text{V}}\text{M}$ |
|
$0$ |
|
$0$ |
| After some time |
$\frac{10-0.214}{\text{V}}$ |
|
$\frac{0.5-0.214}{\text{V}}$ |
|
$\frac{0.214}{\text{V}}\text{M}$ |
|
$\frac{0.214}{\text{V}}\text{M}$ |
| |
$=\frac{0.786}{\text{V}}\text{M}$ |
|
$=\frac{0.286}{\text{V}}\text{M}$ |
|
|
|
|
Therefore, the reaction quotient is,$\text{ Q}_{\text{c}}=\frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O]}}{[\text{CH}_3\text{COOH][}\text{C}_2\text{H}_5\text{OH}] }$
$=\frac{\frac{0.214}{\text{V}}\times\frac{0.214}{\text{V}}}{\frac{0.786}{\text{V}}\times\frac{0.286}{\text{V}}}$ $=0.2037$$=0204$ (approximately)
Since $\text{Q}_{\text{c}}<\text{K}_{\text{c}},$ equilibrium has not been reached. View full question & answer→Question 235 Marks
$K_p = 0.04$ atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of $C_2H_6$ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
$\text{C}_2\text{H}_6\text{ (g)}\rightleftharpoons\text{C}_2\text{H}_4\text{ (g) + H}_2\text{ (g)}$
AnswerLet p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium. Now, according to the reaction,
| |
$\text{C}_2\text{H}_{6\text{(g)}}$ |
$\leftrightarrow$ |
$\text{C}_2\text{H}_{4\text{(g)}}$ |
$+$ |
$\text{H}_{2\text{(g)}}$ |
| Initial conc. |
$4.0\text{ atm}$ |
|
$0$ |
|
$0$ |
| At equilibrium |
$4.0-\text{p}$ |
|
$\text{p}$ |
|
$\text{p}$ |
We can write
$\frac{\text{p}_{\text{C}_2\text{H}_4}\times\text{p}_{\text{H}_2}}{\text{p}_{\text{C}_2\text{H}_6}}=\text{K}_{\text{p}}$
$\Rightarrow\frac{\text{p}\times\text{p}}{4.0-\text{p}}=0.04$
$\Rightarrow\text{p}^2+0.16-0.04\text{ p}$
$\Rightarrow\text{p}^2+0.04\text{p}-0.16=0$
Now, $\text{p}\frac{-0.04\pm\sqrt{(0.04)^2-4\times1\times(-0.16)}}{2\times1}$
$=\frac{-0.04\pm0.80}{2}$
$=\frac{0.76}{2}$ (Taking positive value)
$=0.38$
Hence, at equilibrium,
$[\text{C}_2\text{H}_6]-4-\text{p}=4-0.38$
$=3.62\text{ atm}$ View full question & answer→Question 245 Marks
It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its $pK_a$.
AnswerLet the organic acid be HA.
$\Rightarrow\text{HA}\leftrightarrow\text{H}^++\text{A}^-$
Concentration of HA = 0.01M
pH = 4.15
$-\log[\text{H}^+]=4.15$
$[\text{H}^+]=7.08\times10^{-5}$
$\text{K}_\text{a}=\frac{[\text{H}^+][\text{A}^-]}{[\text{HA]}}$
Now,
$\text{[H}^+]=[\text{A}^-]=7.08\times10^{-5}$
$[\text{HA}]=0.01$
Then,
$\text{K}_\text{a}=\frac{(7.08\times10^{-5})(7.08\times10^{-5})}{0.01}$
$\text{K}_\text{a}=5.01\times10^{-7}$
$\text{pK}_\text{a}=-\log\text{K}_\text{a}$
$=\log(5.01\times10^{-7})$
$\text{pK}_\text{a}=6.3001$
View full question & answer→Question 255 Marks
The concentration of sulphide ion in $0.1$ M HCl solution saturated with hydrogen sulphide is $1.0 \times 10^{-19} \mathrm{M}$. If $10$ mL of this is added to $5$ mL of $0.04$ M solution of the following: $\mathrm{FeSO}_4, \mathrm{MnCl}_2, \mathrm{ZnCl}_2$ and $\mathrm{CdCl}_2$. in which of these solutions precipitation will take place?
AnswerFor precipitation to take place, it is required that the calculated ionic product exceeds the $K_{sp}$ value. Before mixing:
| $[\text{S}^{2-}]=1.0\times10^{-19}\text{M}$ |
$[\text{M}^{2+}]=0.04\text{M}$ |
| $\text{volume=10mL}$ |
$\text{volume=5mL}$ |
After mixing:
| $$$[\text{S}^{2-}]=?$ |
$[\text{M}^{2+}]=?$ |
| $\text{volume=(10+5)=15mL}$ |
$\text{volume=15mL}$ |
$[\text{S}^{2-}]=\frac{1.0\times10^{-19}\times10}{15}=6.67\times10^{-20}\text{M}$ $[\text{M}^{2+}]=\frac{0.04\times5}{154}=1.33\times10^{-2}\text{M}$ $\text{Ionic product}=[\text{M}^{2+}][\text{S}^{2-}]$ $(1.33\times10^{-2})(6.67\times10^{-20})$ $=8.87\times10^{-22}$ This ionic product exceeds the $K_{sp}$ of Zns and CdS. Therefore, precipitation will occur in $CdCl_2$ and $ZnCl_2$ solutions. View full question & answer→Question 265 Marks
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78M?
$2\text{ICI (g)}\rightleftharpoons\text{I}_2\text{ (g) + Cl}_2\text{ (g)};\text{ K}_{\text{c}}=0.14$
Answer
| |
$2\text{ICl}_{\text{(g)}}$ |
$\rightleftharpoons$ |
$\text{I}_{2\text{(g)}}$ |
$+$ |
$\text{Cl}_{2\text{(g)}}$ |
$;$ |
$\text{K}_{\text{c}}$ |
$=$ |
$0.14$ |
| Initial molar conc. |
$0.78$ |
|
$0$ |
|
$0$ |
|
|
|
|
| Eqm. molar conc. |
$0.78-2\text{x}$ |
|
$\text{x}$ |
|
$\text{x}$ |
|
|
|
|
Applying law of chemical equilibrium,
$\text{K}_{\text{c}}=\frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2}\Rightarrow0.14=\frac{\text{x.x}}{(0.78-2\text{x})^2}$
$\text{x}^2=0.14(0.78-2\text{x})^2$
$>$
$\text{or }\frac{\text{x}}{0.78-2\text{x}}=\sqrt{0.14}=0.374$
$\text{or }\text{x}=0.292-0.748\text{x}$
$\text{or }1.748\text{x}=0.292$
$\text{or }\text{x}=0.167$
Hence at equilibrium, $[\text{I}_2]=[\text{Cl}_2]=0.167\text{ M}$
$[\text{ICl}]=0.78-2\times0.167=0.446\text{ M}$ View full question & answer→Question 275 Marks
At a certain temperature and total pressure of $10^5 ~Pa$, iodine vapour contains 40% by volume of I atoms $\text{I}_2\text{(g)}\rightleftharpoons2\text{I}\text{(g)}$ Calculate Kp for the equilibrium.
AnswerPartial pressure of I atoms,$\text{p}_1=\frac{40}{100}\times\text{p}_{\text{total}}$
$=\frac{40}{100}\times10^5$
$=4\times10^4\text{ Pa}$
Partial pressure of $I_2$ molecules,
$\text{p}_{\text{I}_2}=\frac{60}{100}\times\text{p}_{\text{total}}$
$=\frac{60}{100}\times10^5$
$=6\times10^4\text{ Pa}$
Now, for the given reaction,
$\text{K}_{\text{p}}=\frac{(\text{pI})^2}{\text{p}_{\text{I}_2}}$
$=\frac{(4\times10^4)^2\text{ Pa}^2}{6\times10^4\text{ Pa}}$
$=2.67\times10^4\text{ Pa}$
View full question & answer→Question 285 Marks
Equal volumes of $0.002M$ solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate $K_{sp} = 7.4 \times 10^{–8}$).
AnswerWhen equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001M. Then,
| $\text{Nal0}_3$ |
$\rightarrow$ |
$\text{Na}^+$ |
$+$ |
$\text{l0}_3^-$ |
| $0.001\text{M}$ |
|
|
$0.001\text{M}$ |
| $\text{Cu(ClO}_3)_2$ |
$\rightarrow$ |
$\text{Cu}^{2+}$ |
$+$ |
$2\text{ClO}_3^-$ |
| $0.001\text{M}$ |
|
|
$0.001\text{M}$ |
Now, the solubility equilibrium for copper iodate can be written as: $\text{Cu(l0}_3)_2\rightarrow\text{Cu}^{2+}_\text{(aq)}+\text{2l0}^-_\text{3(aq)}$Ionic product of copper iodate:
$=[\text{Cu}^{2+}][10^-_3]^2$ $=(0.001)(0.001)^2$ $=1\times10^{-9}$Since the ionic product $(1 \times 10^{–9})$ is less than $K_{sp} (7.4 \times 10^{–8})$, precipitation will not occur. View full question & answer→Question 295 Marks
What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298K? (For calcium sulphate, $K_{sp}$ is $9.1 \times 10^{–6}$).
Answer$\text{CaSO}_\text{4(s)}\leftrightarrow\text{Ca}^{2+}_\text{(aq)}+\text{SO}^{2-}_\text{4(aq)}$
$\text{K}_\text{sp}=[\text{Ca}^{2+}][\text{SO}^{2-}_4]$
Let the solubility of $\mathrm{CaSO}_4$ be s .
Then,
$\mathrm{K}_{\mathrm{sp}}=\mathrm{s}^2$
$9.1 \times 10^{-6}=\mathrm{s}^2$
$\mathrm{~s}=3.02 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$
Molecular mass of $\mathrm{CaSO}_4=136 \mathrm{~g} / \mathrm{mol}$
Solubility of $\mathrm{CaSO}_4$ in gram $/ \mathrm{L}=3.02 \times 10^{-3} \times 136=0.41 \mathrm{~g} / \mathrm{L}$
This means that we need 1 L of water to dissolve 0.41 g of $\mathrm{CaSO}_4$
Therefore, to dissolve 1 g of $\mathrm{CaSO}_4$ we require $=\frac{1}{0.41} \mathrm{~L}=2.44 \mathrm{~L}$ of water.
View full question & answer→Question 305 Marks
The ionization constant of $HF , HCOOH$ and HCN at 298 K are $6.8 \times 10^{-4}, 1.8 \times 10-4$ and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.
AnswerIt is known that,
$\text{K}_\text{b}=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
Given,
$K_a of HF = 6.8 \times 10^{–4}$
Hence, $K_b $of its conjugate base $F^–$
$=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$=\frac{10^{-14}}{6.8\times10^{-4}}$
$=1.5\times10^{-11}$
Given,
$K_a of HCOOH = 1.8 \times 10^{–4}$
Hence, $K_b $of its conjugate base $HCOO $
$=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$=\frac{10^{-14}}{1.8\times10^{-4}}$
$=5.6\times10^{-11}$
Given,
$K_a of HCN = 4.8 \times 10^{–9}$
Hence,$ K_b$ of its conjugate base $CN^–$
$=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$=\frac{10^{-14}}{4.8\times10^{-9}}$
$=2.08\times10^{-6}$
View full question & answer→Question 315 Marks
What is the pH of $0.001M$ aniline solution? The ionization constant of aniline can be taken from Table $7.7$. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer$K_b = 4.27 \times 10^{–10} c = 0.001M pH =? \alpha =$? $\text{K}_\text{b}=\text{c}\alpha^2$ $4.27\times10^{-10}=0.001\times\alpha^2$ $4270\times10^{-10}=\alpha^2$ $65.34\times10^{-5}=\alpha=6.53\times10^{-5}$ $\text{Then [anion]}=\text{c}\alpha=.001\times65.34\times10^{-5}$ $=.065\times10^{-5}$ $\text{pOH}=-\log(.065\times10^{-5})$ $=6.187$ $\text{pH}=7.813$
Now, $\text{K}_\text{a}\times\text{K}_\text{b}=\text{K}_\text{w}$
$\therefore\ 4.27\times10^{-10}\times\text{K}_\text{a}=\text{K}_\text{w}$ $\text{K}_\text{a}=\frac{10^{-14}}{4.27\times10^{-10}}$ $=2.34\times10^{-5}$
Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times 10^{–5}$.
View full question & answer→Question 325 Marks
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and $H_2$. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
$\text{CO (g) + H}_2\text{O (g)}\rightleftharpoons\text{CO}_2\text{ (g) + H}_2\text{ (g)}$
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that $\text{p}_\text{co}=\text{p}_{\text{H}_2\text{O}}=4.0$bar, what will be the partial pressure of $H_2$ at equilibrium?$\text{K}_\text{p}=10.1\text{ at }400^\circ\text{C}$
AnswerLet the partial pressure of both carbon dioxide and hydrogen gas be p. The given reaction is:
| |
$\text{CO}_\text{(g)}$ |
$+$ |
$\text{H}_2\text{O}_\text{(g)}$ |
$\rightleftharpoons$ |
$\text{CO}_{2\text{(g)}}$ |
$+$ |
$\text{H}_{2\text{(g)}}$ |
| Initial conc. |
$4.0\text{ bar}$ |
|
$4.0\text{ bar}$ |
|
$0$ |
|
$0$ |
| At equilibrium |
$4.0-\text{p}$ |
|
$4.0-\text{p}$ |
|
$\text{p}$ |
|
$\text{p}$ |
It is given that $\text{K}_\text{p}=10.1.$
Now,
$\frac{\text{p}_{\text{CO}_2}\times\text{p}_{\text{H}_2}}{\text{p}_\text{COH}\times\text{p}_{\text{H}_2\text{O}}}=\text{K}\text{p}$
$\Rightarrow\frac{\text{p}\times\text{p}}{(4.0-\text{p})(4.0-\text{p})}=10.1$
$\Rightarrow\frac{\text{p}}{4.0-\text{p}}=3.178$
$\Rightarrow\text{p}=12.712-3.178\text{p}$
$\Rightarrow4.178\text{p}=12.712$
$\Rightarrow\text{p}=3.04$
Hence, at equilibrium, the partial pressure of $H_2$ will be 3.04 bar. View full question & answer→Question 335 Marks
At 1127K and 1 atm pressure, a gaseous mixture of CO and $CO_2$ in equilibrium with soild carbon has 90.55% CO by mass
$\text{C (S) + CO}_2\text{ (g)}\rightleftharpoons2\text{CO (g)}$
Calculate $K_c$ for this reaction at the above temperature.
AnswerLet the total mass of the gaseous mixture be 100g.
Mass of CO = 90.55g
And, mass of $CO_2$ = (100 – 90.55) = 9.45g
Now, number of moles of CO, $\text{n}{_{\text{co}}}=\frac{90.55}{28}=3.234\text{mol}$
Number of moles of $CO_2$, $\text{n}_{\text{co}_2}=\frac{9.45}{44}=0.215\text{mol}$ Partial pressure of CO,
$\text{p}_{\text{co}}=\frac{\text{n}_{\text{co}}}{\text{n}_{\text{co}}+\text{n}_{\text{co}_2}}\times\text{p}_{\text{total}}$
$=\frac{3.234}{3.234+0.215}\times1$
$=0.938\text{atm}$
Partial pressure of $CO_2$,
$\text{p}_{\text{co}_2}=\frac{\text{n}_{\text{co}_2}}{\text{n}_{\text{co}}+\text{n}_{\text{co}_2}}\times\text{p}_{\text{total}}$
$=\frac{0.215}{3.234+0.215}\times1$
$=0.062\text{atm}$
$\text{Therefore, K}_{\text{p}}=\frac{[\text{CO}]^2}{[\text{CO}_2]}$
$=\frac{(0.938)^2}{0.062}$
$=14.19$
For the given reaction,
$\Delta\text{n}=2-1=1$
We know that,
View full question & answer→Question 345 Marks
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, $K_{sp} = 6.3 \times 10^{–18}$).
AnswerLet the maximum concentration of each solution be xmol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e.,$\frac{\text{x}}{2}.$ $\therefore[\text{FeSO}_4]=[\text{Na}_2\text{S}]=\frac{\text{x}}{2}\text{M}$ $\text{Then, [Fe}^{2+}]=\text{[FeSO}_4]=\frac{\text{x}}{2}\text{M}$ $\text{Also, [S}^{2-}]=[\text{Na}_2\text{S}]=\frac{\text{x}}{2}\text{M}$ $\text{FeS}_\text{(x)}\leftrightarrow\text{Fe}^{2+}_\text{(aq)}+\text{S}^{2-}_\text{(aq)}$ $\text{K}_\text{sp}=[\text{Fe}^{2+}][{\text{S}^{2-}}]$ $6.3\times10^{-18}=\Big(\frac{\text{x}}{2}\Big)\Big(\frac{\text{x}}{2}\Big)$ $\frac{\text{x}^2}{4}=6.3\times10^{-18}$ $\Rightarrow\text{x}=5.02\times10^{-9}$If the concentrations of both solutions are equal to or less than $5.02 \times 10^{–9}M$, then there will be no precipitation of iron sulphide.
View full question & answer→Question 355 Marks
The solubility product constant of $\mathrm{Ag}_2 \mathrm{CrO}_4$ and AgBr are $1.1 \times 10^{-12}$ and $5.0 \times 10^{-13}$ respectively. Calculate the ratio of the molarities of their saturated solutions.
AnswerLet s be the solubility of $Ag_2CrO_4$.Then,
$\text{Ag}_2\text{CrO}_4\rightarrow2\text{Ag}^++\text{CrO}_4^{-}$
$\text{K}_\text{sp}=(2\text{s})^2.\text{s}=4\text{s}^3$
$1.1\times10^{-12}=4\text{s}^3$
$\text{s}=6.5\times10^{-5}\text{M}$
$.275\times10^{-12}=\text{s}^3$
$\text{s}=0.65\times10^{-4}\text{M}$
Let s' be the solubility of $\text{AgBr}.$
$\text{AgBr}_\text{(s)}\leftrightarrow\text{Ag}^++\text{Br}^-$
$\text{K}_\text{sp}=\text{s}'^2=5.0\times10^{-13}$
$\therefore\ \text{s}'=7.07\times10^{-7}\text{M}$
Therefore, the ratio of the molarities of their saturated solution is
$\frac{\text{s}}{\text{s}'}=\frac{6.5\times10^{-5}\text{M}}{7.07\times10^{-7}\text{M}}=91.9.$
View full question & answer→Question 365 Marks
One mole of $H_2O$ and one mole of CO are taken in 10L vessel and heated to 725K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,
$\text{H}_2\text{O (g) + CO (g)}\rightleftharpoons\text{H}_2\text{ (g) + CO}_2\text{ (g)}$
Calculate the equilibrium constant for the reaction.
Answer
| From the above reaction, |
| |
$\text{H}_2\text{O}_{\text{(g)}}$ |
$+$ |
$\text{CO}_{\text{(g)}}$ |
$\rightleftharpoons$ |
$\text{H}_{2\text{(g)}}$ |
$+$ |
$\text{CO}_{2\text{(g)}}$ |
| Initia no. of moles |
$1$ |
|
$1$ |
|
$0$ |
|
$0$ |
| At equilibrium |
$0.6$ |
|
$0.6$ |
|
$0.4$ |
|
$0.4$ |
$\therefore\ [\text{H}_2\text{O}]=\frac{0.6}{10}\text{mol L}^{-1}=0.06\text{mol L}^{-1}$
$[\text{CO}]=\frac{0.6}{10}\text{mol L}^{-1}=0.06\text{mol L}^{-1}$
$[\text{H}_2]=\frac{0.4}{10}=0.04\text{mol L}^{-1}$ and $\text{[CO}_2]=\frac{0.4}{10}$
$=0.04\text{mol L}^{-1}$
$\text{K}_{\text{c}}=\frac{[\text{H}_2][\text{CO}_2]}{[\text{H}_2\text{O}][\text{CO]}}=\frac{\frac{0.4}{10}\times\frac{0.4}{10}}{\frac{0.6}{10}\times\frac{0.6}{10}}=0.44$ View full question & answer→Question 375 Marks
Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:
- $OH^–$
- $F^–$
- $H^+$
- $BCl_3$
Answer
- $OH^–$ ions can demate an electron pair and act as Lewis base.
- $F^–$ ions can donate an electron pair and act’ as Lewis base.
- $H^+$ ions can accept an electron pair and act as Lewis acid.
- $BCl_3$ can accept an electron pair since Boron atom is electron deficient. It is a Lewis acid.
View full question & answer→Question 385 Marks
Calculate (a) $\Delta \text{G}^\ominus$ (b) the equilibrium constant for the formation of $NO_2$ from NO and $O_2$ at 298K
$\text{NO (g) + }1/2\text{ O}_2\text{ (g)}\rightleftharpoons\text{NO}_2\text{ (g)}$
where
$\Delta_\text{r}\text{G}^\ominus\text{ (NO}_2)=52.0\text{ kJ/mol}$
$\Delta_\text{r}\text{G}^\ominus\text{ (NO})=87.0\text{ kJ/mol}$
$\Delta_\text{r}\text{G}^\ominus\text{ (O}_2)=0\text{ kJ/mol}$
Answer$\text{Step I. Calculation of }\Delta\text{G}^\ominus$
$\Delta\text{G}^\ominus=\Delta\text{G}^\ominus(\text{NO}_2)-[\Delta_{\text{f}}\text{G}^\ominus(1/2\text{O}_2)]$
$=52.0-(87+0)=-35\text{ kJmol}^{-1}$
$\text{Step II. Calculation of K}_\text{c}$
$\Delta\text{G}^\ominus=-2.303\text{ RT }\log\text{ K}_\text{c}$
$\log\text{ K}_\text{c}=-\frac{\Delta\text{G}^\ominus}{2.303\text{ RT}}$
$=\frac{(-35\times10^3\text{Jmol}^{-1})^3}{2.303\times(8.314\text{kJmol}^{-1})\times(298\text{K})}=6.134$
$\text{K}_\text{c}=\text{Antilog 6.314}=1.36\times10^6.$
$\text{K}_\text{p}=\text{K}_\text{c}\text{(RT)}^{\Delta\text{n}}$
$\Rightarrow14.19=\text{K}_\text{c}(0.082\times1127)^1$
$\Rightarrow\text{K}_\text{c}=\frac{14.19}{0.082\times1127}$
$=0.154\text{(approximately)}$
View full question & answer→Question 395 Marks
At 700K, equilibrium constant for the reaction:
$\text{H}_2\text{ (g) + I}_2\text{ (g)}\rightleftharpoons2\text{HI (g)}$ is 54.8. If $0.5 ~mol L^{–1}$ of $HI(g)$ is present at equilibrium at 700K, what are the concentration of $H^2(g)$ and $I^2(g)$ assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K?
AnswerIt is given that equilibrium constant $K_c$ for the reaction
$\text{H}_{2\text{(g)}}\text{ + I}_{2\text{(g)}}\leftrightarrow2\text{HI}_{\text{(g)}} \text{ is }54.8.$
Therefore, at equilibrium, the equilibrium constant $K_c$ for the reaction
$2\text{HI}_{\text{(g)}}\leftrightarrow\text{H}_{2\text{(g)}}\text{ + I}_{2\text{(g)}}\text{ Will be }\frac{1}{54.8}$
$[\text{HI]}=0.5\text{mol L}^{-1}$
Let the concentrations of hydrogen and iodine at equilibrium be x mol $L^{–1}$
$[\text{H}_2]=[\text{I}_2]=\text{xmol L}^{-1}$
Therefore, $\frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}=\text{K}'_{\text{c}}$
$\Rightarrow\frac{\text{x}\times\text{x}}{(0.5)^2}=\frac{1}{54.8}$
$\Rightarrow\text{x}^2=\frac{0.25}{54.8}$
$\Rightarrow\text{x}=0.06754$
$\text{x}=0.068\text{mol L}^{-1}\text{(approximately)}$
Hence, at equilibrium, $[\text{H}_2]=[\text{I}_2]=\text{0.068mol L}^{-1}$
View full question & answer→Question 405 Marks
Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as:
$\text{CH}_3\text{COOH (1) + C}_2\text{H}_5\text{OH (1)}\rightleftharpoons\text{CH}_3\text{COOC}_2\text{H}_5 \text{ (1) + H}_2\text{O (1)}$
At 293K, if one starts with 1.00mol of acetic acid and 0.18mol of ethanol, there is 0.171mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
AnswerLet the volume of the reaction mixture be V. Also, here we will consider that water is a solvent and is present in excess. The given reaction is:
| |
$\text{CH}_3\text{COOH}_{\text{(l)}}$ |
$+$ |
$\text{C}_2\text{H}_5\text{OH}_{\text{(l)}}$ |
$\leftrightarrow$ |
$\text{CH}_3\text{COOC}_2\text{H}_{5\text{(l)}}$ |
$+$ |
$\text {H}_2\text{O}$ |
| Initial conc. |
$\frac{1}{\text{V}}\text{M}$ |
|
$\frac{0.18}{\text{V}}\text{M}$ |
|
$0$ |
|
$0$ |
| At equilibrium |
$\frac{1-0.171}{\text{V}}$ |
|
$\frac{0.18-0.171}{\text{V}}$ |
|
$\frac{0.171}{\text{V}}\text{M}$ |
|
$\frac{0.171}{\text{V}}\text{M}$ |
| |
$=\frac{0.829}{\text{V}}\text{M}$ |
|
$=\frac{0.009}{\text{V}}\text{M}$ |
|
|
|
|
Therefore, equilibrium constant for the given reaction is:$\text{K}_{\text{c}}=\frac{[\text{CH}_3\text{COOC}_2\text{H}_5][\text{H}_2\text{O]}}{[\text{CH}_3\text{COOH][}\text{C}_2\text{H}_5\text{OH}] }$
$=\frac{\frac{0.171}{\text{V}}\times\frac{0.171}{\text{V}}}{\frac{0.829}{\text{V}}\times\frac{0.009}{\text{V}}}=3.919$$=3.92$ (approximately) View full question & answer→Question 415 Marks
The pH of 0.005M codeine ($C_{18}H_{21}NO_3$) solution is 9.95. Calculate its ionization constant and $pK_b$.
Answerc = 0.005
pH = 9.95
pOH = 4.05
pH = – log (4.105)
$4.05=-\log[\text{OH}^-]$
$[\text{OH}^-]=8.91\times10^{-5}$
$\text{c}\alpha=8.91\times10^{-5}$
$\alpha=\frac{8.91\times10^{-5}}{5\times10^{-3}}=1.782\times10^{-2}$
$\text{Thus, K}_\text{b}=\text{c}\alpha^2$
$=0.005\times(1.782)^2\times10^{-4}$
$=0.005\times3.1755\times10^{-4}$
$=0.0158\times10^{-4}$
$\text{K}_\text{b}=1.58\times10^{-6}$
$\text{Pk}_\text{b}=-\log\text{K}_\text{b}$
$=-\log(1.58\times10^{-6})$
$=5.80$
View full question & answer→Question 425 Marks
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Answerc = 0.1M
pH = 2.34
$-\log[\text{H}^+]=\text{pH}$
$-\log[\text{H}^+]=2.34$
$[\text{H}^+]=4.5\times10^{-3}$
Also,
$[\text{H}^+]=\text{c}\alpha$
$4.5\times10^{-3}=0.1\times\alpha$
$\frac{4.5\times10^{-3}}{0.1}=\alpha$
$\alpha=45\times10^{-3}=.045$
Then,
$\text{K}_\text{a}=\text{c}\alpha^2$
$=0.1\times(45\times10^{-3})^2$
$=202.5\times10^{-6}$
$=2.02\times10^{-4}$
View full question & answer→Question 435 Marks
Four moles of $\mathrm{PCl}_5$ are heated in a closed $4 \mathrm{dm}^3(\mathrm{~L})$ container to reach equilibrium at 400 K . At equilibrium $50 \%$ of $\mathrm{PCl}_5$ is dissociated. What is the value of $\mathrm{K}_{\mathrm{c}}$ for the dissociation of $\mathrm{PCl}_5$ into $\mathrm{PCl}_3$ and $\mathrm{Cl}_2$ at 400 K .
Answer$\begin{matrix}&\text{PCl}_5(\text{g})&\rightleftharpoons&\text{PCl}_3(\text{g})&+&\text{Cl}_2(\text{g})\\\text{Initial Conc. }&4&&0&&0\\\text{Final Conc.at equilibrium}&4-\frac{4\times50}{100}&&2&&2\end{matrix}$
$\begin{matrix}\text{Final conc.in mol L}^{-1}&\frac{2}{4}&\frac{2}{4}&\frac{2}{4}\end{matrix}$
$\text{K}_{\text{c}}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$
$=\frac{\frac{1}{2}+\frac{1}{2}}{\frac{1}{2}}=0.5$
View full question & answer→Question 445 Marks
Match the following species with the corresponding conjugate acid.
| Species |
Conjugate acid |
| i. |
$\text{NH}_3$ |
a. |
$\text{CO}_3^{2-}$ |
| ii. |
$\text{HCO}_3^-$ |
b. |
$\text{NH}_4^+$ |
| iii. |
$\text{H}_2\text{O}$ |
c. |
$\text{H}_3\text{O}^+$ |
| iv. |
$\text{HSO}_4^-$ |
d. |
$\text{H}_2\text{SO}_4$ |
| |
|
e. |
$\text{H}_2\text{CO}_3$ |
Answer
| Species |
Conjugate acid |
| i. |
$\text{NH}_3$ |
b. |
$\text{NH}_4^+$ |
| ii. |
$\text{HCO}_3^-$ |
e. |
$\text{H}_2\text{CO}_3$ |
| iii. |
$\text{H}_2\text{O}$ |
c. |
$\text{H}_3\text{O}^+$ |
| iv. |
$\text{HSO}_4^-$ |
d. |
$\text{H}_2\text{SO}_4$ |
View full question & answer→Question 455 Marks
Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.
AnswerpH of solution $\mathrm{A}=6$
$\left[\mathrm{H}^{+}\right]=10^{-6} \mathrm{~mol} \mathrm{~L}^1$
pH of solution $\mathrm{B}=4$
$\left[\mathrm{H}^{+}\right]=10^{-4} \mathrm{molL}^{-1}$
On mixing one litre of each solution Total volume $=1 \mathrm{~L}+1 \mathrm{~L}=2 \mathrm{~L}$
Total amount of $\mathrm{H}^{+}$in 2 L solution formed by mixing solutions A and $\mathrm{B}=10^{-6}+10^{-4} \mathrm{~mol}$
$\text{Total [H}^+]=\frac{10^{-4}(1+0.01)}{2}=\frac{1.01\times10^{-4}}{2}$
$=5\times10^{-5}\text{mol L}^{-1}$
$\text{pH}=-\log[\text{H}^+]=-\log(5\times10^{-5})$
$=-\log5-(-5\log10)=-\log5+5$
$=5-\log5-=5-0.6990=4.3010=4.3$
Thus, the pH of resulting solution is 4.3.
View full question & answer→Question 465 Marks
$2\text{NO}(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g});\Delta\text{H}=-17\text{KJ}$
- Predict the effect of an increase in concentration of $\mathrm{NO}$ on the equilibrium concentration of $\mathrm{NO}_2$.
- Predict the effect of pressure decrease as a result of increased volume on the equilibrium concentration of $\mathrm{NO}_2$.
Answer$2\text{NO}(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g});\Delta\text{H}=-17\text{KJ}$
- If we increase the concentration of $\mathrm{NO}$, the rate of forward reaction will increase, i.e. more $\mathrm{NO}_2$ will be formed.
- Decrease in pressure will favour backward reaction, i.e. less $\mathrm{NO}_2$ will be formed.
View full question & answer→Question 475 Marks
$50.0g$ of $CaCO_3$ are heated to 1073K in a $5L$ vessel. What per cent of the $CaCO_3$ would decompose at equilibrium? $K_p$ for the reaction.
$\text{CaCO}_3(\text{s})\rightleftharpoons\text{CaO(s)}+\text{CO}_2(\text{g})$ is 1.15a/ m at 1073K.
Answer$\text{CaCO}_3(\text{s})\rightleftharpoons\text{CaO(s)}+\text{CO}_2(\text{g})$ $\text{K}_{\text{p}}=\text{p}_{\text{CO}}=1.15\text{atm, pV}=\text{nRT}$ $\text{N}_{\text{CO}_2}=\frac{\text{p}_{\text{CO}\text{V}}}{\text{RT}}$ $=\frac{1.115\times5}{0.082\times1073}=0.065\text{mol.}$1 mole of $CO_2$ is obtained by decomposition of 1 mole $CaCO_3$. Therefore, moles of $CaCO_3$ decomposed is equal to the moles of $CO_2 = 0.065mol$.
Moles of $CaCO_3$ initially present $=\frac{50}{100}=0.5\text{mol}$ [molecular mass of $CaCo_3 = 100$] Percent of $CaCO_3$ decomposed $=\frac{0.065}{0.5}\times100=-13\%$
View full question & answer→Question 485 Marks
A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Qsp) becomes greater than its solubility product. If the solubility of $BaSO_4$ in water is $8 \times 10^{–4}mol\ dm^{–3}$. Calculate its solubility in $0.01mol\ dm^{–3}$ of $H_2SO_4$.
Answer$\text{BaSO}_4\rightleftharpoons\text{Ba}^{2+}+\text{SO}_4^{2-}$
|
At t = 0
|
1
|
0
|
0
|
|
At equilibrium in water
|
1 - S
|
S
|
S
|
|
At equilibrium in sulphuric acid
|
1 -S
|
S
|
(S + 0.01)
|
$K_{sp}$ for $BaSO_4$ in water $=\text{[Ba}^{2+}][\text{SO}_4^{2-}]$
$= S \times S = S^2$
$K_{sp} = (8 \times 10^{-4})^2 = 64 \times 10^{-8}$ ...(i)
In presence of $H_2SO_4$,
$K_{sp} = (S)(S + 0.01)$
$K_{sp}$ begin constant.
$(S)(S + 0.01) = 64 \times 10^{-8}$
$S^2+ 0.01 S = 64 \times 10^{-8}$
$S^2 + 0.01 S - 64 \times 10^{-8} = 0$
$\Rightarrow\text{S}=\frac{-0.01\pm\sqrt{(0.01)^2+(4\times64\times10^{-8})}}{2}$
$=\frac{-0.01\pm\sqrt{10^{-4}+(256\times10^{-8})}}{2}$
$=\frac{-10^{-2}+(1.012\times10^{-2})}{2}=-\frac{(-1+1.012)\times10^{-2}}{2}$
$=6\times10^{-5}\text{mol dm}^{-3}$ View full question & answer→Question 495 Marks
In a system compressing of A, B, C
$\text{A}(\text{s})\rightleftharpoons2\text{B}(\text{g})+3\text{C}(\text{g})$
It conc. of 'C' is increased by factor of 2, what will be the equilibrium concentration of 'B' with respect to its original value.
Answer$\text{K}=\frac{[\text{B}]^2[\text{C}]^3}{[\text{A}]}$ in first case ...(1)
$\text{K}=\frac{[\text{B}']^2[2\text{C}]^3}{[\text{A}]}$ in second case ...(2)
From (1) and (2)
$\frac{[\text{B}]^2[\text{C}]^3}{[\text{A}]}=\frac{[\text{B}']^2[2\text{C}]^3}{[\text{A}]}$
$\Rightarrow\Big[\frac{\text{B}}{\text{B}'}\Big]^2=\frac{8\text{C}^3}{\text{C}^3}=8$
$\Rightarrow\frac{[\text{B}]}{[\text{B}']}=\sqrt{8}=2\sqrt{2}$
$\Rightarrow[\text{B}']=\frac{1}{2\sqrt{2}}[\text{B}]$
The concentration of 'B' will become $\frac{1}{2\sqrt{2}}$ times of original value.
View full question & answer→Question 505 Marks
Match Column I with Column II.
| S. No |
Column I (Reaction) |
S. No |
Column II (Equilibrium constant) |
| 1. |
$2\text{N}_2(\text{g})+6\text{H}_2(\text{g})\rightleftharpoons4\text{NH}_3(\text{g})$ |
(i) |
$2\text{K}_{\text{c}}$ |
| 2. |
$2\text{NH}_3(\text{g})\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$ |
(ii) |
$\text{K}_{\text{c}}^{\frac{1}{2}}$ |
| 3. |
$\frac{1}{2}\text{N}_2\text{(g)}+\frac{3}{2}(\text{g})\rightleftharpoons\text{NH}_3(\text{g})$ |
(iii) |
$\frac{1}{\text{K}}$ |
| 4. |
|
(iv) |
$\text{K}^2_{\text{c}}$ |
Answer
| S. No |
Column I (Reaction) |
S. No |
Column II (Equilibrium constant) |
| 1. |
$2\text{N}_2(\text{g})+6\text{H}_2(\text{g})\rightleftharpoons4\text{NH}_3(\text{g})$ |
(iv) |
$\text{K}^2_{\text{c}}$ |
| 2. |
$2\text{NH}_3(\text{g})\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$ |
(iii) |
$\frac{1}{\text{K}}$ |
| 3. |
$\frac{1}{2}\text{N}_2\text{(g)}+\frac{3}{2}(\text{g})\rightleftharpoons\text{NH}_3(\text{g})$ |
(ii) |
$\text{K}_{\text{c}}^{\frac{1}{2}}$ |
View full question & answer→Question 515 Marks
The ionization constant of propanoic acid is $1.32 \times 10^{–5}$. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
AnswerLet the degree of ionization of propanoic acid be α.
Then, representing propionic acid as HA, we have:
| $\text{HA}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$\text{H}_3\text{O}^+$ |
$+$ |
$\text{A}^-$ |
| $(.05-0.0\alpha)\approx.05$ |
$.05\alpha$ |
|
$.05\alpha$ |
$\text{K}_\text{a}=\frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA]}}$
$=\frac{(.05\alpha)(.05\alpha)}{0.05}=.05\alpha^2$
$\alpha=\sqrt{\frac{\text{K}_\text{a}}{.05}}=1.63\times10^{-2}$
Then, $[\text{H}_3\text{O}^+]=.05\alpha=0.5\times1.63\times10^{-2}=\text{K}_\text{b}.15\times10^{-4}\text{M}$
$\therefore\ \text{pH}=3.09$
In the presence of 0.1M of HCl, let α´ be the degree of ionization.
$\text{Then, }[\text{H}_3\text{O}^+]=0.01$
$[\text{A}^-]=005\alpha'$
$[\text{HA]}=.05$
$\text{K}_\text{a}=\frac{0.01\times0.5\alpha'}{.05}$
$1.32\times10^{-5}=.01\times\alpha'$
$\alpha'=1.32\times10^{-3}$ View full question & answer→Question 525 Marks
What do you understand by following?
- Dissociation constant of an acid.
- Buffer solution.
- Solubility product.
Answer
- Dissociation Constant of an Acid $K_a$: It measures the ability of an acid to lose [$H^+$] experimentally.
$\text{CH}_3\text{COOH}(\text{l})+\text{H}_2\text{O}(\text{l})\\\rightleftharpoons\text{CH}_3\text{COO}^-(\text{aq})+\text{H}_3\text{O}^+(\text{aq})$
$\text{K}_{\text{a}}=\frac{[\text{CH}_3\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{COOH}]}$
- Buffer Solution: The solution whose pH does not change by adding small amount of H+ or $OH^+$ is called buffer solution. e.g. mixture of $CH_3COOH$ and $CH_3COONa$ is a buffer solution.
- Solubility Product: It is a product of molar concentration of ions formed in a saturated solution at a given temperature raised to the power equal to the number of each ions formed by 1 mole of sparingly soluble compound.
$\text{e.g. AgCl}(\text{s})\rightleftharpoons\text{Ag}^+(\text{aq})+\text{Cl}^-\text{(aq)};$
$\text{K}_{\text{sq}}=[\text{Ag}^+][\text{Cl}^-]$ View full question & answer→Question 535 Marks
An equilibrium mixture at 300K contains $N_2O_4$ and $NO_2$ at 0.28 and 1.1atm pressure respectively. If the volume of the container is doubled, calculate the new equilibrium pressure of two gases.
Answer$\begin{matrix}&\text{N}_2\text{O}_4(\text{g})&\rightleftharpoons&2\text{NO}_2(\text{g})\\\text{perssure at equilibrium }&0.28&&1.1\end{matrix}$
$\text{K}_{\text{p}}=\frac{\text{p}(\text{NO}_2)^2}{\text{p}(\text{N}_2\text{O}_4)}$
$=\frac{(1.1)^2}{(0.28)}=4.32\text{atm}$
If volume of the container is doubled, the pressure will be reduced to half.
$\begin{matrix}&\text{N}_2\text{O}_4(\text{g})&\rightleftharpoons&2\text{NO}_2(\text{g})\\\text{New perssure }&\Big(\frac{0.28}{2}-\text{p}\Big)&&\Big(\frac{1.1}{2}+2\text{p}\Big)\end{matrix}$
$\text{K}_{\text{p}}=\frac{\Big(\frac{1.1}{2}+2\text{p}\Big)^2}{\Big(\frac{0.28}{2}-\text{p}\Big)}=4.32$
$\text { On solving } p=0.045$
$\therefore p\left(N_2 \mathrm{O}_4\right)=0.14-0.045=0.095 \mathrm{~atm}$
$p\left(\mathrm{~N}_2\right)=0.55+0.045=0.64 \mathrm{~atm}$.
View full question & answer→Question 545 Marks
i. The equilibrium constant of a reaction is $2 \times 10^{-3}$ at $25^{\circ} \mathrm{C}$ and $2 \times 10^{-2}$ at $50^{\circ} \mathrm{C}$ Is the reaction endothermic or exothermic?
ii. The solubility of $\mathrm{CaF}_2$ in water at 298 K is $1.7 \times 10^{-3}$ gram per 100 ml of the solution.
Calculate solubility product of $\mathrm{CaF}_2$.
Answer
- The reaction is endothermic because equilibrium constant is increasing with increase in temperature.
- $\text{s}=1.7\times10^{-3}\text{g/ 100ml}$
$=\frac{1.7\times10^{-3}}{100}\times1000$
$=1.7\times10^{-2}\text{g L}^{-1}$
$=\frac{1.7\times10^{-2}\text{g L}^{-1}}{(40+38)\text{g mol}^{-1}}$
$=\frac{1.7\times10^{-2}\text{g L}^{-1}}{78\text{g mol}^{-1}}$
$=2.18\times10^{-4}\text{mol L}^{-1}$
$\text{CaF}_2\rightleftharpoons\text{Ca}^{2+}+\text{2F}^-$
$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{F}^-]^2$
$=(2.18\times10^{-4})(2\times2.18\times10^{-4})^2$
$=41.44\times10^{-12}$
$=4.144\times10^{-11}$ View full question & answer→Question 555 Marks
Match the following equilibria with the corresponding condition.
|
Column I
|
Column II
|
|
i.
|
Liquid ⇌ Vapour
|
a.
|
Saturated solution
|
|
ii.
|
Solid ⇌ Liquid
|
b.
|
Boiling point
|
|
iii.
|
Solid ⇌ Vapour
|
c.
|
Sublimation point
|
|
iv.
|
Solute(s) ⇌ Solute (solution)
|
d.
|
Melting point
|
|
|
|
e.
|
Unsaturated solution
|
Answer
|
Column I
|
Column II
|
|
i.
|
Liquid ⇌ Vapour
|
b.
|
Boiling point
|
|
ii.
|
Solid ⇌ Liquid
|
d.
|
Melting point
|
|
iii.
|
Solid ⇌ Vapour
|
c.
|
Sublimation point
|
|
iv.
|
Solute(s) ⇌ Solute (solution)
|
a.
|
Saturated solution
|
View full question & answer→Question 565 Marks
The solubility product of $Al ( OH )_3$ is $2.7 \times 10^{-11}$. Calculate its solubility in $gL ^{-1}$ and also find out pH of this solution. (Atomic mass of $Al =27 u$ ).
AnswerLet S be the solubility of $Al(OH)_3$
| |
$\text{Al(OH})_3$ |
$\rightleftharpoons$ |
$\text{Al}^{3+}\text{(aq)}$ |
$+$ |
$3\text{OH}^-\text{(aq)}$ |
| Concentration of species at t = 0 |
$1$ |
|
$0$ |
|
$0$ |
| Concentration of various species at equilibrium |
$1-\text{S}$ |
|
$\text{S}$ |
|
$3\text{S}$ |
$\text{K}_\text{sp}=[\text{Al}^{3+}][\text{OH}^-]^3=\text{(S)(3S)}^3=27\text{S}^4$
$\text{S}^4=\frac{\text{K}_\text{sp}}{27}=\frac{27\times10^{-11}}{27\times10}=1\times10^{-12}$
$\text{S}=1\times10^{-3}\text{mol L}^{-1}$
Solubility of $Al(OH)_3$
Molar mass of $Al(OH)_3 is 78g$. Therefore,
Solubility of $Al(OH)_3 in g L^{-1} = 1 \times 10^{-3} \times 78g L^{-1}$
$= 78 \times 10^{-3}g L^{-1}$
$= 7.8 \times 10^{-2}g L^{-1}$
pH of the solution
$S = 1 \times 10^{-3}mol L^{-1}$
$[OH^-] = 3S = 3 \times 1 \times 10^{-3}= 3 \times 10^{-3}$
$pOH = 3 -\log 3$
$pH = 14 -pOH = 11 + \log 3 = 11.4771$ View full question & answer→Question 575 Marks
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78M?
$2\text{ICI (g)}\rightleftharpoons\text{I}_2\text{ (g) + Cl}_2\text{ (g)};\text{ K}_{\text{c}}=0.14$
Answer
| |
$2\text{ICl}_{\text{(g)}}$ |
$\rightleftharpoons$ |
$\text{I}_{2\text{(g)}}$ |
$+$ |
$\text{Cl}_{2\text{(g)}}$ |
$;$ |
$\text{K}_{\text{c}}$ |
$=$ |
$0.14$ |
| Initial molar conc. |
$0.78$ |
|
$0$ |
|
$0$ |
|
|
|
|
| Eqm. molar conc. |
$0.78-2\text{x}$ |
|
$\text{x}$ |
|
$\text{x}$ |
|
|
|
|
Applying law of chemical equilibrium,
$\text{K}_{\text{c}}=\frac{[\text{I}_2][\text{Cl}_2]}{[\text{ICl}]^2}\Rightarrow0.14=\frac{\text{x.x}}{(0.78-2\text{x})^2}$
$\text{x}^2=0.14(0.78-2\text{x})^2$
$>$
$\text{or }\frac{\text{x}}{0.78-2\text{x}}=\sqrt{0.14}=0.374$
$\text{or }\text{x}=0.292-0.748\text{x}$
$\text{or }1.748\text{x}=0.292$
$\text{or }\text{x}=0.167$
Hence at equilibrium, $[\text{I}_2]=[\text{Cl}_2]=0.167\text{ M}$
$[\text{ICl}]=0.78-2\times0.167=0.446\text{ M}$ View full question & answer→Question 585 Marks
What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298K? (For calcium sulphate, $K_{sp}$ is $9.1 \times 10^{–6}$).
Answer$\text{CaSO}_\text{4(s)}\leftrightarrow\text{Ca}^{2+}_\text{(aq)}+\text{SO}^{2-}_\text{4(aq)}$
$\text{K}_\text{sp}=[\text{Ca}^{2+}][\text{SO}^{2-}_4]$
Let the solubility of $CaSO_4$ be s.
Then,
$\text{K}_\text{sp}=\text{s}^2$
$9.1\times10^{-6}=\text{s}^2$
$\text{s}=3.02\times10^{-3}\text{mol/L}$
Molecular mass of $CaSO_4$ = 136 g/mol
Solubility of $CaSO_4$ in gram/L = $3.02 \times 10^{–3} \times 136= 0.41g/L$
This means that we need 1L of water to dissolve 0.41g of $CaSO_4$
Therefore, to dissolve 1g of $CaSO_4$ we require $=\frac{1}{0.41}\text{L}=2.44\text{L}$ of water.
View full question & answer→Question 595 Marks
The ionization constant of $HF , HCOOH$ and HCN at 298 K are $6.8 \times 10^{-4}, 1.8 \times 10-4$ and $4.8 \times 10^{-9}$ respectively. Calculate the ionization constants of the corresponding conjugate base.
AnswerIt is known that,
$\text{K}_\text{b}=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
Given,
$K_a$ of $HF = 6.8 \times 10^{–4}$
Hence, $K_b$_ of its conjugate base $F^–$
$=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$=\frac{10^{-14}}{6.8\times10^{-4}}$
$=1.5\times10^{-11}$
Given,
$K_a $ of $ HCOOH = 1.8 \times 10^{–4}$
Hence,$ K_b $of its conjugate base $HCOO^–$
$=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$=\frac{10^{-14}}{1.8\times10^{-4}}$
$=5.6\times10^{-11}$
Given,
$K_a$ of $ HCN = 4.8 \times 10^{–9}$
Hence, $K_b$ of its conjugate base $CN^–$
$=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$=\frac{10^{-14}}{4.8\times10^{-9}}$
$=2.08\times10^{-6}$
View full question & answer→Question 605 Marks
What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer$K_b = 4.27 \times 10^{–10} c$ = 0.001M pH =? α =? $\text{K}_\text{b}=\text{c}\alpha^2$ $4.27\times10^{-10}=0.001\times\alpha^2$ $4270\times10^{-10}=\alpha^2$ $65.34\times10^{-5}=\alpha=6.53\times10^{-5}$ $\text{Then [anion]}=\text{c}\alpha=.001\times65.34\times10^{-5}$ $=.065\times10^{-5}$ $\text{pOH}=-\log(.065\times10^{-5})$ $=6.187$ $\text{pH}=7.813$Now,
$\text{K}_\text{a}\times\text{K}_\text{b}=\text{K}_\text{w}$ $\therefore\ 4.27\times10^{-10}\times\text{K}_\text{a}=\text{K}_\text{w}$ $\text{K}_\text{a}=\frac{10^{-14}}{4.27\times10^{-10}}$ $=2.34\times10^{-5}$Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times 10^{–5}$.
View full question & answer→Question 615 Marks
Calculate the volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution.
($K_{sp}$ of $PbCl_2 = 3.2 \times 10^{–8}$ , atomic mass of $Pb = 207u$).
Answer
| |
$\text{PbCl}_2$ |
$\rightleftharpoons$ |
$\text{Pb}^{2+}$ |
$+$ |
$\text{2Cl}^-$ |
| Conc. at t = 0 |
$1$ |
|
$0$ |
|
$0$ |
| Conc. at equilibrium |
$1-\text{S}$ |
|
$\text{S}$ |
|
$2\text{S}$ |
$\text{K}_\text{sp}=\text{[Pb}^{2+}][\text{Cl}^-]^2=\text{(S)(2S)}^2=4\text{S}^3$
$\text{S}^3=\frac{\text{K}_\text{sp}}{4}=\frac{3.2\times10^{-8}}{4}8\times10^{-9}\text{mol L}^{-1}$
$\text{S}=\sqrt[3]{8\times10^{-9}}=2\times10^{-3}\text{mol L}^{-1}$
Solubility of $\text{PbCl}_4=2\times10^{-3}\times278$ (molar mass of $PbCl_2) = 556 \times 10^{-3}g L^{-1}$
$= 0.556 g L^{-1}$
To get saturated solution, $0.556g\ PbCl_2$ is dissolved in 1L water.
0.1g of $PbCl_2$, i disobed in $\frac{0.1}{0.556}=0.1798\text{L}$ water.
To make a saturated solution, 0.1g $PbCl_2$ is dissolved in $0.1798\approx0.2\text{L}$ water. View full question & answer→Question 625 Marks
A reaction between ammonia and boron trifluoride is given below:
: $NH_3 + BF_3 → H_3N: BF_3$
Identify the acid and base in this reaction. Which theory explains it? What is the hybridisation of B and N in the reactants?
AnswerAlthough $\mathrm{BF}_3$ does not have a proton but acts as Lewis acid as it is an electron deficient compound. It reacts with $\mathrm{NH}_3$ by accepting the lone pair of electrons from $\mathrm{NH}_3$ and completes its octet. The reaction can be represented by.
$\mathrm{BF}_3+: \mathrm{NH}_3 \rightarrow \mathrm{BF}_3 \leftarrow: \mathrm{NH}_3$
Lewis electronic theory of acids and bases can explain it. Boron in $\mathrm{BF}_3$ is $\mathrm{sp}^2$ hybridised, whereas N in $\mathrm{NH}_3$ is $\mathrm{sp}^3$ hybridised.
View full question & answer→Question 635 Marks
For the reaction: $\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$at 400K, $K_p = 41$. Find the value of $K_p$ for each of the following reactions at the same temperature.
- $\text{2NH}_3(\text{g})\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$
- $\frac{1}{2}\text{Na}(\text{g})+\frac{3}{2}\text{H}_2(\text{g})\rightleftharpoons\text{NH}_3(\text{g})$
- $2\text{N}_2(\text{g})+6\text{H}_2(\text{g})\rightleftharpoons4\text{NH}_3(\text{g})$
Answer
- $\text{K}_{\text{p}}=\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}=41$
$2\text{NH}_3\text{(g)}\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2(\text{g})$
$\text{K}'_{\text{p}}=\frac{\text{[N}_2][\text{H}_2]^3}{[\text{NH}_3]^2}$
$=\frac{1}{\text{K}_{\text{p}}}=\frac{1}{41}=0.024$
- $\frac{1}{2}\text{N}_2(\text{g})+\frac{3}{2}\text{H}_2\rightleftharpoons\text{NH}_3(\text{g})$
$\text{K}''_{\text{p}}=\frac{[\text{NH}_3]}{[\text{NH}_2]^{\frac{1}{2}}[\text{H}_2]^{\frac{3}{4}}}$
$=\sqrt{\text{K}_{\text{p}}}=\sqrt{41}=\sqrt{6.4}$
- $2\text{N}_2\text{(g)}+6\text{H}_2(\text{g})\rightleftharpoons4\text{NH}_3(\text{g})$
$\text{K}'''_{\text{p}}=\frac{[\text{NH}_3]^4}{[\text{N}_2]^2[\text{H}_2]^6}$
$=(\text{K}_{\text{p}})^2=(41)^2=1681$
$=1.68\times10^3$ View full question & answer→Question 645 Marks
The solubility of silver chloride (AgCl) in water at $25^\circ C is 1.06 \times 10^{-5}mol L^{-1}.$ Calculate the solubility product of AgCl at this temperature.
AnswerThe solubility equilibrium of AgCl is
$AgCl(s) \rightleftharpoons Ag^{+}(aq)+Cl^{-}(aq)$
One mole of AgCl in solution gives 1 mole of $Ag ^{+}$ions and 1 mole of $Cl ^{-}$ions.
Since, the solubility of AgCl is $1.06 \times 10^{-5} mol L ^{-1}$, it will gives $1.06 \times 15^{-5} mol L ^{-1}$ of $Ag ^{+}$ions and $1.06 \times 10^{-5} mol L ^{-}$, of $Cl ^{-}$ions. Therefore,
${\left[Ag^{+}\right]=1.06 \times 10^{-5} mol L^{-1},\left[Cl^{-}\right]=1.06 \times 10^{-5} mol L^{-1}}$
$\text { Now, } K_{sp}=\left[Ag^{+}\right]\left[Cl^{-}\right]=\left(1.06 \times 10^{-5}\right) \times\left(1.06 \times 10^{-5}\right)=1.12 \times 10^{-10}$
View full question & answer→Question 655 Marks
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, $K_{sp} = 6.3 \times 10^{–18}$).
AnswerLet the maximum concentration of each solution be xmol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e.,$\frac{\text{x}}{2}.$ $\therefore[\text{FeSO}_4]=[\text{Na}_2\text{S}]=\frac{\text{x}}{2}\text{M}$ $\text{Then, [Fe}^{2+}]=\text{[FeSO}_4]=\frac{\text{x}}{2}\text{M}$ $\text{Also, [S}^{2-}]=[\text{Na}_2\text{S}]=\frac{\text{x}}{2}\text{M}$ $\text{FeS}_\text{(x)}\leftrightarrow\text{Fe}^{2+}_\text{(aq)}+\text{S}^{2-}_\text{(aq)}$ $\text{K}_\text{sp}=[\text{Fe}^{2+}][{\text{S}^{2-}}]$ $6.3\times10^{-18}=\Big(\frac{\text{x}}{2}\Big)\Big(\frac{\text{x}}{2}\Big)$ $\frac{\text{x}^2}{4}=6.3\times10^{-18}$ $\Rightarrow\text{x}=5.02\times10^{-9}$If the concentrations of both solutions are equal to or less than $5.02 \times 10^{–9}M$, then there will be no precipitation of iron sulphide.
View full question & answer→Question 665 Marks
The solubility product of AgCl is $1.5 \times 10^{-10}$. Predict whether there will be any precipitation by mixing $50 mL$ of $0.01 M NaCl$ and $50 mL$ of $0.01 M AgNO _3$ solution.
AnswerOn mixing 50mL of $0.01M NaCl$ and 50mL pf $0.01M AgNO_3$, the total volume becomes 100mL.
Therefore,
Conc. of $NaCl$ in $100\text{mL}=\frac{0.01\times50}{100}=0.005\text{M}$
Conc. of$ AgNO_3 $in $100\text{mL}=\frac{0.01\times50}{100}=0.005\text{M}$
Now $NaCl(aq) = Na^+(aq) + Cl^-(aq)$and $\text{AgNO}_3 \text{aq}= \text{Ag}+ \text{(aq}) +\text{NO}_3^- \text{(aq)}$
$[Cl^-] = [NaCl] = 0.005N$
$[Ag^+] = [AgNO_3] = 0.005M$
$\therefore$ Ionic product of $[Ag^+][Cl^-] = 0.005 \times 0.005 = 2.5 \times 10^{-5}$
Since ionic product is greater than its solubility product, precipitation will occur.
View full question & answer→Question 675 Marks
The solubility product of $Fe ( OH )_3$ is $1 \times 10^{-36}$. What is the minimum concentration of $OH ^{-}$ions required to precipitate $Fe ( OH )_3$ from a 0.001 M solution of $FeCl _3$ ?
Answer$K _{\text {sp }}$ for $Fe ( OH )^3$ is $K _{ sp }=\left[ Fe ^{3+}\right]\left[ OH ^{-}\right]^3$
Precipitation will occur when ionic product,
$\left[ Fe ^{3+}\right]\left[ OH ^{-}\right]^3$ becomes greater than $K _{ sp }$
$\left[ Fe ^{3+}\right]=\left[ FeCl _3\right]=0.001 M$
The concentration of $OH ^{-}$ions required to start the precipitation is
$[\text{OH}^-]^3=\frac{\text{K}_{\text{sp}}}{[\text{Fe}^{3+}]}$
$=\frac{1\times10^{-6}}{0.001}=1\times10^{-33}$
$\therefore[\text{OH}^-]=(1\times10^{-33})^{\frac{1}{3}}$
$=1\times10^{-11}\text{mol L}^{-1}$
Thus, concentration of $OH^-$ required to start precipitation of $Fe(OH)_3 = 1 \times 10^{-11}mol L^{-1}$
View full question & answer→Question 685 Marks
Which of the following reactions involve homogeneous equilibria and which involve heterogeneous equilibria?
- $2\text{N}_2\text{O(g)}\rightleftharpoons2\text{N}_2(\text{g})+\text{O}_2\text{(g)}$
- $2\text{NH}_3\text{(g)}\rightleftharpoons\text{N}_2(\text{g})+3\text{H}_2\text{(g)}$
- $2\text{Cu}(\text{NO}_3)_2(\text{s})\rightleftharpoons2\text{CuO(s)}+4\text{NO}_2(\text{g})+\text{O}_2(\text{g})$
- $\text{CH}_3\text{COOC}_2\text{H}_5(\text{aq})+\text{H}_2\text{O}\text{l}\\\rightleftharpoons\text{CH}_3\text{COOH(aq)}+\text{C}_2\text{H}_5\text{OH(aq)}$
- $\text{Fe}^{3+}(\text{aq})+\text{3OH}^-\text{(aq)}\rightleftharpoons\text{Fe(OH)}_3(\text{s})$
Answer
- Homogeneous equilibriales.
- Homogeneous equilibria.
- Heterogeneous equilibria bond.
- Homogeneous equilibria.
- Heterogeneous equilibriato.
View full question & answer→Question 695 Marks
Calculate pH when $9.8g$ of $H_2SO_4$ is dissolved in $2L$ solution.
Answer$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1}{\text{Volume of solution in Litres}}$
$W_B = 9.8g$ = mass of solute
MB = Molar mass of $H_2SO_4 = 98g mol^{-1}$
Volume of solution = 2L
$\text{M}=\frac{9.8}{98}\times\frac{1}{2}=0.05\text{mol L}^{-1}$
$\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }2\text{H}^++\text{SO}^{2-}_4$
$[H^+] = 2 \times$ Molarity of $H_2SO_4$
$= 2 \times 0.05 = 0.1mol L^{-1}$
$= 10^{-1}mol L^{-1}$
$\text{pH}=-\log(\text{H}^+)=-\log10^{-1}$
$=+1\log10=1\times1=1$
$[\because\log10=1]$
View full question & answer→Question 705 Marks
$3.2mol$ of HI were taken in a sealed bulb at $440°C$ till the equilibrium state was reached. Its degree of dissociation was found to be $20\%$. Calculate the number of moles of HI. $H_2$ and $I_2$ present at equilibrium point and also determine the equilibrium constant.
Answer$\begin{matrix}&\text{2HI}(\text{g})&\rightleftharpoons&\text{H}_2(\text{g})&+&\text{l}_2(\text{g})\\\text{Initial Conc. }&3-2\text{moles}&&0&&0\\\text{Final Conc.at equilibrium}&3.2-\frac{2\times20}{100}&&\frac{20}{100}&&\frac{20}{100}\\&2.8\text{mol}&&0.2\text{mol}&&0.2\text{mol}\end{matrix}$
$\text{K}_{\text{c}}=\frac{[\text{H}_2][\text{I}_2]}{[\text{HI}^2]}$
$[\text{HI}]=2.8\text{moles},$
$[\text{H}_2]=[\text{I}_2]=0.2\text{mol.}$
$\text{K}_{\text{c}}=\frac{0.2\times0.2}{(2.8)^2}=\frac{4}{100}\times\frac{100}{28\times28}$
$\text{K}_{\text{c}}=\frac{1000}{196}\times10^{-3}$
$=5.1\times10^{-3}$
View full question & answer→Question 715 Marks
Write a relation between $\Delta\text{G}$ and Q and define the meaning of each term and answer the following:
- Why a reaction proceeds forward when Q < K and no net reaction occurs when Q = K?
- Explain the effect of increase in pressure in terms of reaction quotient Q.
For the reaction, AnswerThe relation between $\Delta\text{G}$ and Q is
$\Delta\text{G}=\Delta\text{G}^{\ominus}+\text{RT InQ}$
$\Delta\text{G}=$ change in free energy as the reaction proceeds
$\Delta\text{G}^{\ominus}$ standard free energy
Q = reaction quotienten
R= gas constant
T = absolute temperature in Kual
- Since, $\Delta\text{G}^{\ominus}=-\text{RT In K}$
$\therefore\Delta\text{G}=-\text{RT In K}+\text{RT In Q;}$
$\Delta\text{G}=\text{RT In}\frac{\text{Q}}{\text{K}}$
will be negative and the reaction proceeds in the forward direction. $\text{Q}=\text{K},\Delta\text{G}=0$ If reaction is in equilibrium and there is no net reaction.
- $\text{CO}\text{(g)}+3\text{H}_2(\text{g})\rightleftharpoons\text{CH}_4\text{(g)}+\text{H}_2\text{O}(\text{g})$
$\text{K}_{\text{c}}=\frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}$
On increasing pressure, volume decreases. If we doubled the pressure, volume will be halved but the molar concentrations will be doubled. Then,
$\text{Q}_{\text{c}}=\frac{2[\text{CH}_4].2[\text{H}_2\text{O}]}{2[\text{CO}]\{2[\text{H}_2]\}^3}$
$\frac{1}{4}\frac{[\text{CH}_4][\text{H}_2\text{O}]}{[\text{CO}][\text{H}_2]^3}=\frac{1}{4}\text{K}_{\text{c}}$
Therefore, $Q_c$ is less than $K_C$ so $Q_c$ will tend to increase to re-establish equilibrium and the reaction will go in forward direction.
$\text{CO}\text{(g)}+3\text{H}_2(\text{g})\rightleftharpoons\text{CH}_4\text{(g)}+\text{H}_2\text{O}(\text{g})$ View full question & answer→Question 725 Marks
The pH of $0.005M$ codeine $(C_{18}H_{21}NO_3)$ solution is $9.95$. Calculate its ionization constant and $pK_b$.
Answerc = 0.005
pH = 9.95
pOH = 4.05
pH = – log (4.105)
$4.05=-\log[\text{OH}^-]$
$[\text{OH}^-]=8.91\times10^{-5}$
$\text{c}\alpha=8.91\times10^{-5}$
$\alpha=\frac{8.91\times10^{-5}}{5\times10^{-3}}=1.782\times10^{-2}$
$\text{Thus, K}_\text{b}=\text{c}\alpha^2$
$=0.005\times(1.782)^2\times10^{-4}$
$=0.005\times3.1755\times10^{-4}$
$=0.0158\times10^{-4}$
$\text{K}_\text{b}=1.58\times10^{-6}$
$\text{Pk}_\text{b}=-\log\text{K}_\text{b}$
$=-\log(1.58\times10^{-6})$
$=5.80$
View full question & answer→Question 735 Marks
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Answerc = 0.1M
pH = 2.34
$-\log[\text{H}^+]=\text{pH}$
$-\log[\text{H}^+]=2.34$
$[\text{H}^+]=4.5\times10^{-3}$
Also,
$[\text{H}^+]=\text{c}\alpha$
$4.5\times10^{-3}=0.1\times\alpha$
$\frac{4.5\times10^{-3}}{0.1}=\alpha$
$\alpha=45\times10^{-3}=.045$
Then,
$\text{K}_\text{a}=\text{c}\alpha^2$
$=0.1\times(45\times10^{-3})^2$
$=202.5\times10^{-6}$
$=2.02\times10^{-4}$
View full question & answer→Question 745 Marks
- Write the conjugate acid for Bronsted base of $HCOO^-$.
- Calculate the pH of a $1.0 \times 10^{-8}$ M solution of HCl.
- Calculate the solubility of $A_2X_3$ in pure water, assuming that neither kind of ion reacts with water. [The solubility product of $A_2X_3, K_{sp} = 1.1 \times 10^{-23}$]
Answer
- $HCOOH$ is conjugate acid of $HCOO^-$.
- $\text{H}_2\text{O}+\text{H}_2\text{O}\rightleftharpoons\text{H}_3\text{O}^++\text{OH}^-$
$\text{K}_\text{w}=1.0\times10^{-14}=[\text{H}_3\text{O}^+][\text{OH}^-]$ $(\because[\text{H}_3\text{O}^+]=[\text{OH}^-])$
$=1.0\times10^{-14}=[\text{H}_3\text{O}^+]^2$
$[\text{H}_3\text{O}^+]=\sqrt{1.0\times10^{-14}}=10^{-7}\text{mol L}^{-1}$
$\text{H}_2\text{O}+\text{HCl}\xrightarrow{\ \ \ \ }\text{H}_3\text{O}s^++\text{Cl}^-$
$10^{-8}$
Total $[\text{H}_3\text{O}^+]=10^{-8}+10^{-7}=10^{-7}(1+10^{-1})$
$=1.1\times10^{-7}\text{mol L}^{-1}$
$\text{pH}=-\log[\text{H}_3\text{O}^+]$
$=-\log1.1\times10^{-7}$
$=-0.04139+7.00=6.958$
- $\text{A}_2\text{X}_3\rightleftharpoons2\text{A}^{3+}+3\text{X}^{2-}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2\text{s}\ \ \ \ \ \ \ \ \ \ \ 3\text{s}$
where solubility is 's' mol $L^{-1}$
$\text{K}_\text{sp}=[\text{A}^{3+}]^2[\text{X}^{2-}]^3$
$1.1\times10^{-23}=(2\text{s})^2(3\text{s})^3=108\text{s}^5$
$\Rightarrow \sqrt{\frac{110}{108}\times10^{-25}}=1\times10^{-5}\text{mol L}^{-1}$
$\text{s}\simeq1\times10^{-5}\text{mol L}^{-1}$ View full question & answer→Question 755 Marks
- At 473K, equilibrium constant, K for decomposition of $PCl_5$ is $8.3 \times 10^{-3}$. If decomposition is depicted as
$\text{PCl}_5(\text{s})\rightleftharpoons\text{PCl}_3(\text{s})+\text{Cl}_2(\text{g});$
$\Delta_\text{r}\text{H}^\text{o}=124.0\text{kJ mol}^{-1}$
- Write an expression for $K_c$ for the reaction.
- What is the value of $K_c$ for the reverse reaction at same temperature.
- What would be the effect on $K_c$ if:
- The pressure is increased?
- The temperature is increased?
- Write equilibrium constant for the following reactions:
- $\text{BaCO}_3(\text{s})\rightleftharpoons\text{BaO(s)}+\text{CO}_2(\text{g})$
- $\text{CH}_4(\text{g})+2\text{O}_2(\text{g})\rightleftharpoons\text{CO}_2(\text{g})+2\text{H}_2\text{O(g)}$
Answer
- $\text{PCl}_5(\text{s})\rightleftharpoons\text{PCl}_3(\text{s})+\text{Cl}_2(\text{g});$
$\Delta _\text{r}\text{H}^\text{o}=124.0\text{kJ mol}^{-1}$
- $\text{K}_\text{c}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}=8.3\times10^{-3}$
- $K_c$ for the reverse reaction $=\frac{1}{\text{K}_\text{c}\text{for the forward reaction}}$
$=\frac{1}{8.3\times10^{-3}}=120.48$
-
- When pressure increases $K_c$ reamains unchanged.
- $K_c$ increses with increases in temprature because the reaction is endothermic.
-
- $\text{K}=\frac{[\text{BaO(s)}][\text{CO}_2(\text{g})]}{[\text{BaCO}_3(\text{g})]}$
Since $[\text{BaCO}_3(\text{s})]=[\text{BaO(s)}]=1$
$\text{K}=[\text{CO}_2(\text{g})]$
- $\text{K}=\frac{[\text{CO}_2(\text{g})][\text{H}_2\text{O(g)}]^2}{[\text{CH}_2(\text{g})][\text{O}_2(\text{g})]^2}$
View full question & answer→Question 765 Marks
- Which of the following are Lewis acids?
$\text{H}_2\text{O},\text{BF}_3,\text{H}^+$ and $\text{NH}^+_4$
- The pH of a sample of vinegar is $3.76$. Calculate the concentration of hydrogen ion in it.
- What is common ion effect? Explain its application in qualitative analysis of II group radicals.
Answer
- Lewis acids are electron deficient species. So, Lewis acids are $\text{BF}_3,\text{H}^+$ and $\text{NH}^+_4.$
- $\text{pH}=-\log[\text{H}^+]$
$\Rightarrow\log[\text{H}^+]=-\text{pH}=-3.76$
$\Rightarrow \log[\text{H}^+]=-3.76+1-1$
$[\text{H}^+]=\bar{4}.24\text{ Antilog}$
$[\text{H}^+]=1.738\times10^{-4}$
$=1.74\times10^{-4}\text{M}$
- Common ion effect is the suppression of the dissociation of weak electrolyte in the presence of a strong electrolyte having a common ion. Sulphides of II group radicals are precipitated by passing $H_2S$ gas in presence of HCl. $H_2S$ being weak electrolyte ionises slightly and HCl which is a strong electrolyte is completely ionised.
$\text{H}_2\text{S}\rightleftharpoons2\text{H}^++\text{S}^{2-}$
$\text{HCl}\xrightarrow{\ \ \ \ \ \ \ }\text{H}^++\text{Cl}^-$
Due to common ion effect the degree of dissociation of $H_2S$ decreases and concentration of $S^{2-}$ ions in solution becomes small enough to precipitate only II group cations and not group IV cations. The second group cations have lower solubility product than IV group cations. View full question & answer→Question 775 Marks
$\text{K}_{\text{c}}\text{ for }\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g})+\text{Cl}_2(\text{g})\text{ is }0.04\text{ at }25^\circ\text{C}.$
How much moles of $PCl_5$ must be added to 3L flask to obtain a chlorine concentration of $0.15M$.
Answer$\begin{matrix}&\text{PCl}_5(\text{g})&\rightleftharpoons&\text{PCl}_3(\text{g})&+&\text{Cl}_2(\text{g})\\\text{Initial Conc. }&\text{x}&&0&&0\\\text{Final Conc.}&\text{x}-0.15&&0.15&&0.15\end{matrix}$
$\text{K}_{\text{c}}=\frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}$
$0.04=\frac{0.15\times0.15}{\text{x}-0.15}$
$\text{x}-0.15=\frac{0.15\times0.15}{0.04}$
$=\frac{225}{1000}\times\frac{100}{4}$
$=\frac{225}{400}=\frac{9}{16}$
$\text{x}=\frac{9}{16}+\frac{15}{100}=\frac{9}{16}+\frac{3}{20}$
$=\frac{180+48}{320}=\frac{228}{320}=0.7\text{mol L}^{-1}$
Number of moles of $PCl_5$ per litre = 0.7mol
Number of moles of $PCl_5$ in 3L flask = 0.7 × 3 = 2.1mol.
View full question & answer→Question 785 Marks
- Consider the following endothermic reaction:
$\text{CH}_4(\text{g})+\text{H}_2\text{O}(\text{g})\rightleftharpoons\text{CO(g)}+3\text{H}_2(\text{g})$
- Write expression for $K_p$ for the above reaction.
- How will the values of $K_p$ and composition of equilibrium mixture be affected by.
- increasing the pressure.
- increasing the temperature.
- using a catalyst?
- Calculate the pH of the resultant mixture of $10$ml of $0.1$M $H_2SO_4 + 10$ ml of $0.1$M KOH.
Answer
- $\text{CH}_4(\text{g})+\text{H}_2\text{O}(\text{g})\rightleftharpoons\text{CO(g)}+3\text{H}_2(\text{g})$
- $\text{K}_{\text{p}}=\frac{(\text{p}_{\text{CO}})(\text{p}_{\text{H}_2})^3}{(\text{p}_{\text{CH}_4})(\text{p}_{\text{H}_2\text{O}})}$
-
- On increasing pressure, the reaction equilibria will shift in the backward direction.
- As the given reaction is endothermic, on increasing temperature the given equilibrium will shift in forward direction.
- There is no effect of catalyst in equilibrium composition, however the equilibrium will be attained faster.
- $10ml$ of $0.1M H_2SO_4$ is mixed with $10ml$ of $0.1$M KOH.
The reaction is
$2\text{KOH}+\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{K}_2\text{SO}_4+2\text{H}_2\text{O}$
$10$ ml of $0.1 \mathrm{M} \mathrm{~H}_2 \mathrm{SO}_4=0.1 \times 10=1$ millimole
$10$ ml of $0.1 \mathrm{M} \mathrm{~KOH}=0.1 \times 10=1$ millimole
$1$ millimole of $KOH$ will react with $0.5$ millimole of $\mathrm{H}_2 \mathrm{SO}_4$
$\left[\because 0.5\right.$ millimole will produce 1 millimole of $\left.\mathrm{H}^{+}\right]$
$\therefore \mathrm{H}_2 \mathrm{SO}_4$ left $=1-0.5=0.5$ millimole
Volume of reaction mixtrue
$=10+10=20 \mathrm{ml}$
$\therefore$ Molarity of $x$ in the mixtrue.
$\therefore$ Molarity of × in the mixtrue.
$=\frac{0.5}{20}=2.5\times10^{-2}\text{M}$
$[\text{H}^+]=2\times2.5 \times10^{-2}$
$=5\times10^{-2}\text{M}$
$\text{pH}=-\log(5\times10^{-2})$
$=-\log5-\log10^{-2}$
$=-0.6990+2.0000=1.30$ View full question & answer→Question 795 Marks
The ionization constant of dimethylamine is $5.4 \times 10^{–4}$. Calculate its degree of ionization in its $0.02M$ solution. What percentage of dimethylamine is ionized if the solution is also $0.1M$ in NaOH?
Answer$\text{K}_\text{b}=5.4\times10^{-4}$
$\text{c}=0.02\text{M}$
$\text{Then, }\alpha=\sqrt{\frac{\text{K}_\text{b}}{\text{c}}}$
$=\sqrt{\frac{5.4\times10^{-4}}{0.02}}$
$=0.1643$
Now, if 0.1M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.
| $\text{NaOH}_\text{(aq)}$ |
$\leftrightarrow$ |
$\text{Na}^+_\text{(aq)}$ |
$+$ |
$\text{OH}^-_\text{(aq)}$ |
| |
|
$0.1\text{M}$ |
|
$0.1\text{M}$ |
And,
| $\text{(CH}_3)_2\text{NH}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$(\text{CH}_3)_2\text{NH}_2^+$ |
$+$ |
$\text{OH}^-$ |
| $(0.02-\text{x})$ |
|
|
|
$\text{x}$ |
|
$\text{x}$ |
| $;0.02\text{M}$ |
|
|
|
|
|
$;0.1\text{M}$ |
Then, $[(\text{CH}_3)_2\text{NH}_2^+]=\text{x}$
$[\text{OH}^-]=\text{x}+0.1;0.1$
$\Rightarrow\text{K}_\text{b}=\frac{[(\text{CH}_3)_2\text{NH}_2^+][\text{OH}^-]}{[(\text{CH}_3)_2\text{NH]}}$
$5.4\times10^{-4}=\frac{\text{x}\times0.1}{0.02}$
$\text{x}=0.0054$
It means that in the presence of 0.1M NaOH, 0.54% of dimethylamine will get dissociated. View full question & answer→Question 805 Marks
Consider the following equilibrium at 2773K
$\text{H}_2(\text{g})+\text{Cl}_2(\text{g})\rightleftharpoons2\text{HCl}(\text{g})$
Initially $0.25M H_2$ and $0.25MCl_2$ are introduced into a reaction vessel and the system in allowed to attain equilibrium. At equilibrium the concentrations of $H_2(g)$ and $Cl_2(g)$ became $0.0314M$. Calculate $K_c$ and $K_p$.
Answer
- $\begin{matrix}&\text{H}_2(\text{g})&+&\text{Cl}_2(\text{g})\rightleftharpoons&\text{2Hcl}(\text{g})\\\text{Initial Conc. }&0.25&&0.25\text{M}&0\\\text{Final Conc.at equilibrium}&0.0314\text{M}&&0.0314\text{M}&0\end{matrix}$
$2(0.25-0.0314)=0.2186\text{M}\times2$
$=0.219\text{M}\times2=0.438$
$\text{K}_{\text{c}}=\frac{[\text{HCl}]^2}{[\text{H}_2][\text{Cl}_2]}$
$=\frac{(0.438)^2}{0.0314\times0.314}=195$
- $\text{K}_{\text{p}}=\text{K}_{\text{c}}(\text{RT})^{\Delta\text{n}}$
$\Delta\text{n}=0$
$\text{K}_{\text{p}}=\text{K}_{\text{c}}=195$ View full question & answer→Question 815 Marks
$\mathrm{K}_p=0.04 \mathrm{~atm}$ at 899 K for the equilibrium shown below. What is the equilibrium concentration of $\mathrm{C}_2 \mathrm{H}_6$ when it is placed in a flask at 4.0atm pressure and allowed to come to equilibrium?
$\text{C}_2\text{H}_6(\text{g})\rightleftharpoons\text{C}_2\text{H}_4(\text{g})+\text{H}_2(\text{g})$
Answer$\begin{matrix}&\text{C}_2\text{H}_6(\text{g})&\rightleftharpoons&\text{C}_2\text{H}_4(\text{g})&+&\text{H}_2(\text{g})\\\text{Intial pressure}&4.0\text{atm}&&0&&0\\\text{Equli pressure}&(4.0-\text{p})\text{atm}&&\text{p}&&\text{P}\end{matrix}$$\text{K}_{\text{p}}=\frac{\text{p}_{\text{C}_2\text{H}_4}.\text{p}_{\text{H}_2}}{\text{p}_{\text{C}_2\text{H}_6}}$
$=\frac{\text{p}\times\text{p}}{4.0-\text{p}}$
$0.04=\frac{\text{p}^2}{4.0-\text{p}}$
$\text{or }0.16-0.04\text{p}=\text{p}^2$
$\text{p}=-0.04\pm\frac{\sqrt{0.0016-4(-0.16)}}{2}$
$\text{p}=\frac{-0.04\pm0.80}{2}$
$\Rightarrow\text{p}=0.38$
(by taking positive value)
Hence, $\text{p}_{\text{C}_2\text{H}_6}=4.0-0.38=3.62\text{atm}.$
View full question & answer→Question 825 Marks
Calculate the pH of $1 \times 10^{-8}M$ solution of $HCl.$
AnswerIf we use the relation, $\text{pH} = -\log [\text{H}_3\text{O}^+]$ we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7. It may be noted that in very dilute acidic solution, when $H^+$ concentrations from acid and water are comparable, the concentration of$ H^+ $from water cannot be neglected. Therefore,
$[H^+]_{total}= [H^+]_{acid} + [H^+]_{water}$
water Since HCl is strong acid and is completely ionized
$[H^+]_{HCl}= 1.0 \times 10^{-8}$
The concentration of $H^+$^ from ionisation is equal to the$ [OH^-]$ from water,
$[\text{H}^+]_{\text{H}_2\text{O}}=[\text{OH}^-]_{\text{H}_3\text{O}}=\text{x say}$
$[H^+]_{total} = 1.0 \times 10^{-8} + x$
$But [H^+] [OH^-] = 1.0 \times 10^{-14}$
$\therefore$ $(1.0 \times 10^{-8}+ x) (x) = 1.0 \times 10^{-14}$
$\Rightarrow x^2 + 10^{-8} x - 10^{-14} = 0$
On solving for x, we get $x = 9.5 \times 10^{-8}$
$\therefore$ $[H^+] = 1.0 \times 10^{-8} + 9.5 \times 10^{-8}$
$= 10.5 \times 10^{-8}$
$= 1.05 \times 10^{-7}$
$\text{pH}=-\log[\text{H}^+]$
$=-\log(1.05\times10^{-7})=6.98$
View full question & answer→Question 835 Marks
The solubility product of lead bromide is $8 \times 10^{-5}$ at 298 K . If the salt is $80 \%$ dissociated in saturated solution, calculate the solubility of the salt (in $g / L$ ).
AnswerThe solubility equilibrium is,
$\text{PbBr}_2(\text{s})\rightleftharpoons\text{Pb}^{2+}(\text{aq})+2\text{Br}^-(\text{aq})$
Suppose the solubility of $PbBr_2$ is S moles per litre. Then, the concentrations of various species at equilibrium are
$[Pb^{2+}] = S, [Br^-] = 2S$
$Now, K_{sp} = [Pb^{2+}][Br^-]^2$
$8 \times 10^{-5} = (S) \times (2S)^2$
$8 \times 10^{-5} = (S) \times (2S)^2$
$\text{or }\text{S}^3=\frac{8\times10^{-5}}{4}=2\times10^{-5}$
$\text{S} = (2.0 \times 10-5)^{\frac{1}{3}}$
$=2.714\times10^{-2}\text{mol L}^{-1}$
This is the solubility of PbBr, if it is 100% dissociated. Solubility of PbBr, if it is 80% dissociated
$=\frac{2.714\times10^{-2}\times80}{100}$
$=2.192\times10^{-2}\text{ mol L}^{-1}$
Molecular weight of $PbBr _2=207+2 \times 80=367$
Solubility of $PbBr _2=2.192 \times 10^{-2} \times 367=8.04 g L ^{-1}$.
View full question & answer→Question 845 Marks
- For the reaction:
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g}),$
The value of $K_p$ is $3.6 \times 10^{-2}$ at $500K$.
Colculate the value of $K_c$ for the reaction at the same temperature R = 0.083L bar $K^{-1}mol^{-1}$.
- What is the effect of increasing pressure in the reactions? Give reason.
$\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g})+\text{Cl}_2\text{(g)}$
$\text{N}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO(g)}$Answer
- The reaction is
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$
Given: $\text{K}_\text{p}=3.6\times10^{-2}$ at 500K
The reation between $K_p$ and $K_c$ is
$\text{K}_\text{p}=\text{K}_\text{c}(\text{RT})^{\Delta\text{n}}$
For the above reaction
$\Delta \text{n}=2-4=-2$
$\text{K}_\text{c}=\frac{\text{K}_\text{p}}{(\text{RT})^{\Delta\text{n}}}$ (R = 0.083bar L $K^{-1}mol^{-1}$)
$=\frac{3.6\times10^{-2}}{(0.083\times500)^{-2}}$
$=3.6\times10^{-2}\times(0.083\times500)^2$
$=62$
-
- The equilibrium will shift in backward reaction because number of moles of products are more than reactants $\Delta \text{n}>0.$
- No effect because number of moles of reactants and products are equal, i.e., $\Delta \text{n}=0.$
View full question & answer→Question 855 Marks
- The reaction quotient of a reversible reaction is $Q_C$ and the equilibrium constant is $K_c$. What do you conclude for the reaction if $Q_c < K_c$?
- State Le Chatelier's principle.
- In qualitative analysis, $NH_4Cl$ is added before adding $NH_4OH$ solution for testing of III group radicals [$Fe^{3+}, Cr^{3+}$ and $Al^{3+}$]. Explain by using concept of common ion effect.
Answer
- If $Q_c < K_c$; the reaction tends towards forward direction to attain equilibrium.
- Le Chatelier's Principle: If a system in equilibrium is subjected to a change in concentration, temperature or pressure, the equilibrium shifts in a direction that tends to undo the effect of the change.
- Hydroxides of group III are precipitated by adding $NH_4OH$ in presence of $NH_4Cl$. The role of $NH_4Cl$ is to produce common ion effect.
$\text{NH}_4\text{OH}\rightleftharpoons\text{NH}^+_4+\text{OH}^-$
$\text{NH}_4\text{Cl}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{NH}_4^++\text{Cl}^-$
Due to common ion effect, the degree of dissociation of $NH_4OH$ is suppressed and less $OH^‑$ are formed. This less concentration of $OH^-$ is sufficient to precipitate group III cations but not the cations of higher groups since the $K_{sp}$ of group III < subsequent groups. View full question & answer→Question 865 Marks
The values of K of two sparingly soluble salts $Ni ( OH )_2$ and AgCN are $2.0 \times 10^{-15}$ and $6 \times 10^{-17}$ respectively. Which salt is more soluble? Explain.
AnswerGiven: $K _{\text {sp }}$ of $Ni ( OH )_2=2 \times 10^{-15}$
$K_{sp} of AgCN = 6 \times 10^{-17}$
$K_{sp} = [Ni^2+] [OH-]^2 = 2 \times 10^{-15}$
$\text{AgCN}\rightleftharpoons\text{Ag}^++\text{CN}^-$
$K_{sp} = [Ag^+] [CN^-] = 6 \times 10^{-17}$
$Let [Ag^+] [CN^-] = s_1$
and $[Ni^{2+}] = s_2$
Hence $[OH^-] = 2s_2$
Since $\text{s}_1^2=6\times10^{-17}$
$\Rightarrow\text{s}_1=\sqrt{60\times10^{-18}}$
$\text{s}_1=7.8\times10^{-9}\text{M}$
Since $\text{s}_2\times(2\text{s}_2)^2=2\times10^{-15}$
$\Rightarrow4\text{s}^3_2=2\times10^{-15}$
$\Rightarrow\text{s}_2^3=0.5\times10^{-15}$
$\Rightarrow\text{s}_2^3=5\times10^{-16}$
$\Rightarrow\text{s}_2^3=500\times10^{-18}$
$\Rightarrow\text{s}_2^3=3\sqrt{500\times10^{-18}}$
$=7.9\times10^{-6}\text{M}$
$\text{so }\text{s}_2=7.9\times10^{-6}\text{M}$
Since $s_2 > s_1$ therefore $Ni(OH)_2$ is more soluble than AgCN.
View full question & answer→Question 875 Marks
- Hydrolysis of sucrose give
$\text{Sucrose}+\text{H}_2\text{O}\rightleftharpoons\text{Glucose}+\text{Fructose}$
Equilibrium constant $K_c$ for the reaction is $2 \times 10^{13}$ at $300K$. Calculate $\Delta\text{G}^\circ\text{ at }300\text{K.}(\log 2=0.3010)$
- The concentration of hydrogen in two sample of soft drinks A and B $4.0 \times 10^{-7}$ and $3.2 \times 10^{-6}$ respectively. Which of these two soft drinks has higher pH?
Answer
- $\text{Sucrose}+\text{H}_2\text{O}\rightleftharpoons\text{Glucose}+\text{Fructose}$
$\text{K}_{\text{c}}2\times10^{13},\text{T}=300\text{K},\Delta\text{G}^\circ=?$
$\Delta\text{G}^\circ=-2.30\text{ RT }\log\text{K}_{\text{c}}$
$=-2.303\times8.314\times300\log2\times10^{13}$
$=-19.147\times300(\log2+\log^{13})$
$=\frac{-19.147\times300\times13.3010\text{J}}{1000}$
$=-76.402\text{kJ mol}^{-1}$
- For A,
$\text{pH}=-\log[\text{H}_3\text{O}^+]$
$=-\log4.0\times10^{-7}$
$=-\log4.0-\log10^{-7}$
$=-0.6021+7.000=6.3979$
- For B,
$\text{pH}=-\log[\text{H}_3\text{O}^+]$
$=-\log3.2\times10^{-6}$
$=-\log3.2-\log10^{-6}$
$=-0.5050+6.000=5.4950$
pH of 'A' is higher than 'B'. View full question & answer→Question 885 Marks
Calculate the pH of 0.4g of NaOH dissolved in water to give 200ml of solution.
Answer$\text{M}=\frac{\text{W}_{\text{B}}}{\text{M}_{\text{B}}}\times\frac{1000}{\text{Volume of solution in ml}}$
where, Mass of solute, $\text{W}_{\text{B}}$ = 0.4
Volume of solution = 200mL
$=\frac{0.4}{40}\times\frac{1000}{200}$
$=\frac{2}{40}=\frac{1}{20}=0.05\text{M}$
$\text{pOH}=-\log[\text{OH}^-]$
$=-\log(5\times10^{-2})$
$=-\log5-\log(10^{-2})$
$=-0.6990+2.0000$
$=1.3010$
$\text{pH}=14-1.3010$
$=12.399$
View full question & answer→Question 895 Marks
The buffers of $X$ and $Y$ of pH $4.0$ and $6.0$ respectively are prepared from acid $HA$ and the salt $Na A$.
Both the buffers are $0.50M$ in $HA$. What would be the $pH$ of the solution obtained by mixing equal volumes of the two buffers?
$(K_{HA}= 1.0 \times 10^{-5})$
Answer$K_{HA}= 1.0 \times 10^{-5}$
$\therefore \text{pK}_{\text{a}}=-\log(1.0\times10^{-5})=5.0$ Determination of conc. of saltin buffer X. $\text{pH}=\text{pK}_{\text{a}}+\log\frac{[\text{Salt}]}{[\text{Acid}]};$
$4.0=5.0+\log\frac{[\text{Salt}]}{0.5}$
$\log\frac{[\text{Salt}]}{0.5}=-1$
$\Rightarrow\frac{[\text{Salt}]}{0.5}=10^{-1}$ [salt] $= 0.5 \times 10^{-1} = 0.05M$ Determination of conc. of saltin buffer Y. $6.0=5.0+\log\frac{[\text{Salt}]}{0.5}$
$\log\frac{[\text{Salt}]}{0.5}=1.0$
$\Rightarrow\frac{[\text{Salt}]}{0.5}=10$[salt] $= 10 \times 0.5 = 5.0$
Conc. of salt in the mixture
$[\text{Salt}]_{\text{mix}}=\frac{0.5+5}{2}=2.75\text{M}$
$\text{pH}=\text{pK}_{\text{a}}+\log\frac{[\text{Salt}]}{[\text{Acid}]}$
$=5.0+\log\frac{2.75}{0.5}$
$=5.0+0.74=5.74$
View full question & answer→Question 905 Marks
An aqueous solution contains an unknown concentration of $Ba ^{2+}$. When 50 mL of 1 M solution of $Na _2 SO _4$ is added, $BaSO _4$ first begin to precipitate. The final volume is $500 mL$ . The solubility product of $BaSO _4$ is $1.0 \times 10^{-10}$. What is original concentration of $Ba ^{2+}$ ?
Answer$\text{BaSO}_4\text{(s)}\rightleftharpoons\text{Ba}^{2+}+\text{SO}^{2-}_4\text{(aq)}$
For $Na_2SO_4$
$M_1V_1= M_2V_2$
$1M \times 50mL = M_2 \times 500mL$
$M_2 = 0.1M$
$[\text{SO}^{2-}_4]=0.1$
$\text{K}_{\text{sp}}=[\text{Ba}^{2+}][\text{SO}^{2-}_4]$
$\Rightarrow[\text{Ba}^{2+}]=\frac{\text{K}_{\text{sp}}}{[\text{SO}^{2-}_{4}]}$
$=\frac{1\times10^{-10}}{0.1}=1.0\times10^{-9}\text{M}$
$[Ba^{2+}] = 1.0 \times 10^{-9}M$ in $500mL$ of solution
$[Ba^{2+}] in 500mL = 1.0 \times 10^{-9}M$
$[\text{Ba}^{2+}]\text{in }1000\text{mL}=\frac{1.0\times10^{-9}}{500}\times1000$
$=2\times10^{-9}\text{mol}.$
View full question & answer→Question 915 Marks
- In qualitative analysis, on what basis cations are grouped?
- The value of $K_c$ in the reaction:
$2\text{A}\rightleftharpoons\text{B}+\text{C}$ is $2 \times 10^{-3}$. At a given time, the composition of reaction mixture is $[A] = [B] = [C] = 3 \times 10^{-4}M$. In which direction the reaction will proceed?
- The solubility of $Sr(OH)_2$ at 298K is 19.23g/L of solution. Calculate the concentration of strontium and hydroxyl ions and the pH of the solution.
Answer
- Cations are grouped on the basis of their solubility product $(K_{sp})$. i.e. value of $K_{sp}$ is closed to each other, e.g., $K_{sp}$ of sulphides of groups II cations are close to each other.
- For the reaction, the reaction quotient $Q_c$ is given by $\text{Q}_\text{c}=\frac{[\text{B}][\text{C}]}{[\text{A}]^2}$
as $[\text{A}]=[\text{B}]=[\text{C}]=3\times10^{-4}\text{M}$
$\text{Q}_\text{c}=\frac{(3\times10^{-4})(3\times10^{-4})}{(3\times10^{-4})^2}=1$
as $\text{Q}\text{c}>\text{K}_\text{c}$
So, the reaction will proceed in the reverse direction.
- Molar mass of $Sr(OH)_2$ = 87.6 + 34
$= 121.6g mol^{-1}$
Solubility of $Sr(OH)_2$ in mol $L^{-1}$
$=\frac{19.23\text{g L}^{-1}}{121.6\text{g mol}^{-1}}$
$\text{Sr(OH)}_2\xrightarrow{\ \ \ \ \ }\text{Sr}^{2+}+2\text{OH}^-$
$[\text{Sr}^{2+}]=0.1581\text{M},[\text{OH}^-]$
$=2\times0.1581=0.3162\text{M}$
$\text{p}[\text{OH}]=-\log0.3162=0.5$
$\Rightarrow \text{pH}=14-0.5=13.5$ View full question & answer→Question 925 Marks
How much volume of $0.1 \mathrm{M} \mathrm{~CH}_2\mathrm{COOH}$ should be added to 50 mL of $0.2 \mathrm{M} \mathrm{~CH}_3 \mathrm{COONa}$ solution to prepare a buffer solution of pH 4.91 . ( $\mathrm{pK}_{\mathrm{a}}$ of ACH is 4.76 ).
AnswerAccording to Henderson's equation
$\text{pH}=\text{pK}_{\text{a}}+\log\frac{[\text{Salt}]}{[\text{Acid}]};$
$\text{pH}=4.91,\text{pK}_{\text{a}}=4.76$
$4.91=4.76+\log\frac{[\text{Salt}]}{[\text{Acid}]}$
$\log\frac{[\text{NaAcl}]}{[\text{AcH}]}=4.91-4.76=0.15$
$\frac{[\text{NaAc}]}{[\text{AcH}]}=\text{antilog}(0.15)=1.41$
If V is the volume of 0.1M AcH required
$\frac{\text{NaAc}}{[\text{AcH}]}=\frac{\frac{0.2\times50}{1000}}{\frac{0.1\times\text{V}}{1000}}=1.41$
$=\frac{0.2\times50}{0.15\times\text{V}}=1.41$
$\text{or }\text{V}=\frac{0.2\times50}{0.1\times1.41}=70.92\text{mL}$
Volume of 0.1M acetic acid required = 70.92mL.
View full question & answer→Question 935 Marks
- Predict the acidic, basic or neutral nature of the following salts:
$NaCN, KBr, NaNO_2, NH_4NO_3$
- What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298K? (For calcium sulphate $K_{sp} is 9.1 \times 10^{-6}$).
- At $450K ; K_p = 2.0 \times 10^{10}bar^{-1}$ for the reaction at equilibrium:
$2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g})$
What is $K_c$ at this temperature?Answer$NaCN, NaNO_2$ - solutions are basic as they are salts of strong base and weak acid. (HCN and $HNO_2$ are weak acids and NaOH is strong base).
KBr - this solution is neutral as it is salt of strong acid HBr and strong base KOH.
$NH_4NO_3$ - its solution is acidic as it is a salt of strong acid $(HNO_3)$ and weak base $(NH_4OH)$.
- $\text{CaSO}_4(\text{s})\rightleftharpoons\text{Ca}^{2+}(\text{aq})+\text{SO}^{2-}_4(\text{aq});$
(mol. mass of $CaSO_4 = 136g/ mol$)
Let the solubility of $CaSO_4$ in mol/L is x.
$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{SO}^{2-}_4]=\text{x}^2$
$\text{x}=\sqrt{\text{K}_\text{sp}}=\sqrt{9.1\times10^{-6}}$
$=3.02\times10^{-3}\text{mol/L}$
$=3.02\times10^{-3}\times136\text{g/L}$
$=0.411\text{g/L};$
For dissolving $0.411g$ of $CaSO_4$, water required is 1L
For dissolving 1g of $CaSO_4$, water required $=\frac{1}{0.411}\text{L}=2.43\text{L}$
- For the reaction:
$2\text{SO}_2(\text{g})+\text{O}_2\text{(g)}\rightleftharpoons2\text{SO}_3(\text{g})$
$\Delta\text{n}_\text{g}=2-3=-1$
$\text{K}_\text{p}=\text{K}_\text{c}(\text{RT})^{\Delta\text{n}}$
$\text{K}_\text{c}=\text{K}_\text{p}(\text{RT})^{-\Delta\text{n}}$
For this reaction
$\text{K}_\text{c}=\text{K}_\text{p}\text{RT}=(2\times10^{10}\text{bar}^{-1}) $
$(0.083\text{L bar K}^{-1}\text{mol}^{-1})\times450\text{K}$
$=7.48\times10^{11}\text{L mol}^{-1}$ View full question & answer→Question 945 Marks
- Define solubility product. Write solubility product expression for $Zr_3(PO_4)_4$.
- Calculate the pH of $0.01 M CH_3COOH$ solution. $[K_a(CH_2COOH) = 1.74 \times 10^{-51}$
- Explain why NaCl is precipitated when HCl(g) is passed through the saturated solution of NaCl.
Answer
- Solubility Product: It is defined as the product of molar concentrations of the ions (formed in the saturated solution at a given temperature) raised to the power equal to the number of times each ion occurs in the equation for solubility equilibrium,
- $=-\log\text{C}\alpha$
$=-\log\sqrt{\text{K}_\text{a}\times\text{C}}$
$=-\log\sqrt{1.74\times10^{-5}\times0.01}$
$=-\log\sqrt{1.74\times10^{-7}}$
$=-\log\sqrt{17.4\times10^{-8}}$
$=-\log4.17\times10^{-4}$
$=-\log4.17-\log10^{-4}$
$=-0.6217+4.000$
$=3.3783$
- It is due to common ion, $CI^-$ increase, therefore rate of backward reaction increases, solubility of NaCl decreases.
- $\text{Zr}_3(\text{PO}_4)_4\rightleftharpoons3\text{Zr}^{4+}+4\text{PO}^{3-}_4$
$\text{K}_{\text{sp}}=[\text{Zr}^{4+}][\text{PO}^{3-}_4]^4$
- $\text{pH}=-\log[\text{H}_3\text{O}^+]$
View full question & answer→Question 955 Marks
- Write the conjugate acid of $NH_3$.
- Assign reason for the following:
- A solution of $NH_4Cl$ in water shows pH less than 7.
- In qualitative analysis $NH_4Cl$ is added before adding $NH_4OH$ for testing $Fe^{3+}$ or $AP^{3+}$ ions.
- Consider the reaction:
$\text{N}_2(\text{g})+3\text{H}_2\text{(g)}\rightleftharpoons2\text{NH}_3+\text{Heat}$
Indicate the direction in which the equilibrium will shift when:
- Temperature is increased.
- Pressure is increased.
Answer
- $\text{NH}^+_4$ is conjugate acid of $NH_3$.
-
- $NH_4Cl$ is salt of weak base $NH_4OH$ and strong acid HCl, therefore $H^+$ ions are more than $OH^-$ ions thus, pH is less than 7.
- It is done to decrease $[OH^-]$ due to common ion effect, so that only group III radicals $Fe^{3+}$ or $Al^{3+}$ get precipitated and higher group radicals do not.
-
- When temperature is increased equilibrium will shift to backward direction as reaction is exothermic.
- When pressure is increased rate of forward reaction will increase as there is decrease in number of moles from reactants to products.
View full question & answer→Question 965 Marks
- State Henry's Law.
- Assign reason for the following:
- A solution of $NH_4Cl$ in water shows pH less than 7.
- In qualitative analysis $NH_4Cl$ is added before adding $NH_4OH$ for testing $Fe^{3+}$ or $AP^{3+}$ ions.
- Consider the reaction:
$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3+\text{Heat}$
Indicate the direction in which the equilibrium will shift when:
- Temperature is increased.
- Pressure is increased.
Answer
- The mass of gas dissolved in given mass of a solvent at any temperature in proportional to the pressure of the gas above the solvent.
-
- $\mathrm{NH}_4 \mathrm{Cl}$ is salt of weak base $\mathrm{NH}_4 \mathrm{OH}$ and strong acid HCl , therefore $\mathrm{H}^{+}$ions are more than $\mathrm{OH}^{-}$ions thus, pH is less than 7.
- It is done to decrease $\left[\mathrm{OH}^{-}\right]$due to common ion effect, so that only group III radicals $\mathrm{Fe}^{3+}$ or $\mathrm{Al}^{3+}$ get precipitated and higher group radicals do not.
-
- When temperature is increased equilibrium will shift to backward direction as reaction is exothermic.
- When pressure is increased rate of forward reaction will increase as there is decrease in number of moles from reactants to products.
View full question & answer→Question 975 Marks
- Define solubility product. Write solubility product expression in terms of molar solubility for $FeCl_3$
- What is the effect of temperature on solubility of gases in liquids?
- Equilibrium constant for the reaction is $4.0$. What will be the equilibrium constant for the reverse reaction.
- Calculate the pH of $10^{-8}M$ HCl solution.
Answer
- Solubility product is defined as the product of molar concentration of ions raised to the power the number of ions formed per formula of the compound.
$\text{FeCl}_3(\text{s})\rightleftharpoons\text{Fe}^{3+}+3\text{Cl}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ '\text{s}' \ \ \ \ \ \ \ \ \ \ 3\text{s}$
$\text{K}_{\text{sp}}=[\text{Fe}^{3+}][\text{Cl}^-]^3$
$\text{K}_\text{sp}=(\text{s})(3\text{s})^3,$
where 's' $mol L^{-1}$ is solubility.
$\Rightarrow\text{K}_\text{sp}= 27\text{s}^4$
$\Rightarrow\text{s}=4\sqrt{\frac{\text{K}_\text{sp}}{27}}$
- Solubility of gases in liquids decreases with increase in temperature because force of attraction between gas and liquid decreases at high temperature.
- K for the reaction = 4
$\therefore \text{K}'$ for reverse reaction $=\frac{1}{4}=0.25$ [$\because \text{K}'=\frac{1}{ \text{K}}$ for reverse reaction]
- pH of $10^{-8}M$ HCl solution.
$\text{HCl}\xrightarrow{\ \ \ \ \ \ }\text{H}^++\text{Cl}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ^{10^{-8}\text{M}}$
$\text{H}_2\text{O}\rightleftharpoons\text{H}^++\text{OH}^-$
$\text{K}_\text{w}=1\times10^{-14}$
$\Rightarrow [\text{H}^+][\text{OH}^-]=10^{-14}$
$\because[\text{H}^+]=[\text{OH}]$
$\therefore [\text{H}^+]^2=10^{-14}$
$\Rightarrow [\text{H}^+]=10^{-7}\text{mol L}^{-1}$
Total concentration of
$[\text{H}^+]=(10^{-8}+10^{-7})$
$=10^{-7}(1+0.1)$
$=1.1\times10^{-7}$
$\therefore \text{pH}=-\log[\text{H}^+]$
$=-\log1.1\times10^{-7}$
$=-\log1.1-\log10^{-7}$
$\Rightarrow \text{pH}=-0.0454+7.000$
$=6.9546$ View full question & answer→Question 985 Marks
Show that the degree of dissociation $(\alpha)$ for the dissociation of $PCl_5$ into PCl3 $\alpha=\Big(\frac{\text{K}_{\text{p}}}{\text{p}+\text{K}_{\text{p}}}\Big)^{\frac{1}{2}}$ and $Cl_2$ in pressure p is given by,
Answer$\begin{matrix}&\text{PCl}_5&\rightleftharpoons&\text{PCl}_3&+&\text{Cl}_2\\\text{Initial moles }&1&&0&&0\\\text{Moles after diss.}&1-\alpha&&\alpha&&\alpha&(\text{Total }= 1+\alpha)\end{matrix}$
$\therefore\text{p}_{\text{PCl}}=\frac{1-\alpha}{1+\alpha}\times\text{p},\text{p}_{\text{PCl}_3}$ $=\frac{\alpha}{1+\alpha}\times\text{p},\text{p}_{\text{Cl}_2}=\frac{\alpha}{1+\alpha}\times\text{p},$ $\text{K}_{\text{p}}=\frac{\text{p}_{\text{Cl}_3}\times\text{p}_{\text{Cl}_2}}{\text{p}_{\text{PCl}_5}}$$=\frac{\Big(\frac{\alpha}{1+\alpha}\text{p}\Big)\Big(\frac{\alpha}{1+\alpha}\Big)}{\Big(\frac{1-\alpha}{1+\alpha}\Big)\text{p}}$
$=\frac{\alpha^2\text{p}}{1-\alpha^2}$ $\text{or }(1-\alpha^2)\text{K}_{\text{p}}=\alpha^2\text{p}$ $\text{or }(\text{p}+\text{K}_{\text{p}})\alpha^2=\text{K}_{\text{p}}$ $\text{or }\alpha=\Big(\frac{\text{K}_{\text{p}}}{\text{p}+\text{K}_{\text{p}}}\Big)^{\frac{1}{2}}$
View full question & answer→Question 995 Marks
Calculate the equilibrium constant for the following equilibrium system at 1120K. $\text{C}(\text{s})+\text{CO}_2(\text{g})+2\text{Cl}_2\stackrel{\text{K}_\text{p}}{\rightleftharpoons}2\text{COCl}_2(\text{g})$ Given the following equations and equilibrium constants: $\text{CO}(\text{s})+\text{Cl}_2(\text{g})\rightleftharpoons\text{COCl}_2(\text{g})$ $\text{Kp}_1 = 6.0 \times 10-3 \dots(1)$ $\text{CO}(\text{s})+\text{Cl}_2(\text{g})\rightleftharpoons\text{CO}_2(\text{g})$$\text{Kp}_2 = 1.3 \times 1014 \dots(2)$
Answer$\text{Multiply Eq. (1) by 2}$
$2\text{CO}(\text{g})+2\text{Cl}_2(\text{g})\rightleftharpoons2\text{COCl}_2(\text{g})$
$\text{K}'_{\text{p}1}=\left(6.0 \times 10^{-3}\right)^2=36 \times 10^{-6}$
$\text{K}'_{\text{p}1}=\frac{[\text{COCl}_2]^2}{[\text{CO}]^2[\text{Cl}_2]^2}=36\times10^{-6}\dots(3)$
$\text{K}'_{\text{p}2}=\frac{[\text{CO}]^2}{[\text{CO}_2]}=36\times10^{14}\dots(4)$
$\text{Multiply K'p}_1 \text{ and } \text{K'p}_2$
$\text{K}'_{\text{p}}=\frac{[\text{COCl}_2]^2}{[\text{CO}_2][\text{Cl}_2]^2}$
$=46.8\times10^8=4.68\times10^9$
View full question & answer→Question 1005 Marks
The average concentration of $\mathrm{SO}_2$ in atmosphere over a city on a certain day is 10 ppm , when the average temperature is 298 K . Given that the solubility of $\mathrm{SO}_2$ in water at 298 K is $1.3653 \mathrm{~mol} / \mathrm{L}$ and the $\mathrm{pK}_{\mathrm{a}}$ of $\mathrm{H}_2 \mathrm{SO}_3$ is 1.92 , estimate the pH of acid rain on that day.
AnswerAmount of $SO_2$ in atmospheres $= 10\text{ ppm}=\frac{10}{10^6}10^{-5}$
Molar conc. of $SO_2$ in pressure of water = amount of $SO_2 \times$ solubility of $SO_2$ in water
$H_2SO_3$ dissociates as = $1.3653 \times 10^{-5}$
$\begin{matrix}&\text{H}_2\text{SO}_3&\rightleftharpoons&\text{H}^+&+&\text{HSO}^-_3\\\text{Intial conc.}&1.3653\times10^{-5}&&0&&0\\\text{Molar cons. of equiv.}&(1.3653\times10^{-5})&&\text{x}&&\text{x}\end{matrix}$
$\text{K}_{\text{a}}=\frac{\text{x}^2}{(1.3653\times10^{-5}-\text{x})}$
$\because\text{PK}_{\text{a}}=1.92$
$\therefore-\log\text{K}_{\text{a}}=1.92$
$\text{K}_{\text{a}}=12\times10^{-2}$
Substituting $1.2\times10^{-2}=\frac{\text{x}^2}{(1.3653\times10^{-5}-\text{x})}$
$\text{or }\text{x}^2=1.2\times10^{-2}(1.3653\times10^{-5}-\text{x})$
On solving we, get $x = 1.3664 \times 10^{-5}$
$\therefore\text{pH}=-\log(1.364\times10^{-5})$
$=4.865$
View full question & answer→Question 1015 Marks
Calculate the $pH$ of a buffer which is $0.1 M$ in acetic acid and $0.15 M$ in sodium acetate. Given that the ionisation constants of acetic acid is $1.75 \times 10^{-5}$. Also calculate the change in $pH$ of the buffer if to $1 L$ of the buffer (i) $1$ cc of $1 M NaOH$ are added. (ii) $1$ cc of $1 M HCl$ are added. Assume that the charge in volume is negligible. (iii) What will be the buffer index of the above buffer?
Answer$\text{pH}=\text{pK}_{\text{a}}+\log\frac{[\text{Salt}]}{[\text{Acid}]}$
$=-\log(1.75\times10^{-5})+\log\frac{0.15}{0.10}$
$= (5 - 0.2430) + 0.1761$
$= 4.757 + 0.1761$
$= 4.933$
- 1cc of 1M $NaOH$ contains $NaOH = 10^{-3}mol.$
This will convent $10^{-3}mol$ of acetic acid into the salt so that salt formed $= 10^{-3}mol.$
Now, [Acid] $= 0.10 - 0.001 = 0.099M$
[Salt] $= 0.15 + 0.001 = 0.151$
$\text{pH}=4.757+\log\frac{0.151}{0.099}$
$\therefore$ Increase in $pH = 4.940 - 4.933 = 0.007$ which is negligible.
- $1cc$ of $1M HCl$ contains $HCl= 10^{-3}mol.$
This will convert $10^{-3}mol$
$CH_3COONa$ into $CH_3COOH$
$\therefore$ Now, [Acid] $= 0.10 + 0.001 = 0.101M$
[Salt] $= 0.15 - 0.001 = 0.149M$
$\text{pH}=4.757+\log\frac{0.149}{0.101}$
$= 4.757 + 0.169 = 4.925$
$\therefore$ Decrease in $pH = 4.933 = 0.007$ which is again negligible.
- Caculation of buffer index
No. of moles of HCl or NaOH added = 0.001mol Change in $pH = 0.007$
Hence, buffer index $=\frac{\text{dn}}{\text{dpH}}=\frac{0.001}{0.007}$
$=\frac{1}{7}=0.143$ View full question & answer→Question 1025 Marks
The ionization constant of phenol is $1.0 \times 10^{–10}$. What is the concentration of phenolate ion in $0.05M$ solution of phenol? What will be its degree of ionization if the solution is also $0.01M$ in sodium phenolate?
AnswerIonization of phenol:
| |
$\text{C}_6\text{H}_5\text{OH}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$\text{C}_6\text{H}_5\text{O}^-$ |
$+$ |
$\text{H}_3\text{O}^+$ |
| Initial conc. |
$0.05$ |
|
|
|
$0$ |
|
$0$ |
| At equilibrium |
$0.05-\text{x}$ |
|
|
|
$\text{x}$ |
|
$\text{x}$ |
$\text{K}_\text{a}=\frac{[\text{C}_6\text{H}_5\text{O}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{OH}]}$ $\text{K}_\text{a}=\frac{\text{x}\times\text{x}}{0.05-\text{x}}$ As the value of the ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator. $\therefore\ \text{x}=\sqrt{1\times10^{-10}\times0.05}$ $=\sqrt{5\times10^{-12}}$ $=2.2\times10^{-6}\text{M}=[\text{H}_3\text{O}^+]$ Since $[\text{H}_3\text{O}^+]=[\text{C}_6\text{H}_5\text{O}^-],$ $[\text{C}_6\text{H}_5\text{O}^-]=2.2\times10^{-6}\text{M.}$ Now, let ∝ be the degree of ionization of phenol in the presence of 0.01 $M C_6H_5ONa$.
| |
$\text{C}_6\text{H}_5\text{ONa}$ |
$\rightarrow$ |
$\text{C}_6\text{H}_5\text{O}^-$ |
$+$ |
$\text{Na}^+$ |
| Conc. |
|
|
|
|
$0.01$ |
Also,
| |
$\text{C}_6\text{H}_5\text{OH}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$\text{C}_6\text{H}_5\text{O}^-$ |
$+$ |
$\text{H}_3\text{O}^+$ |
| Conc. |
$0.05-0.05\alpha$ |
|
|
|
$0.05\alpha$ |
|
$0.05\alpha$ |
$[\text{C}_6\text{H}_5\text{O}^-]=0.01+0.05\alpha;0.01\text{M}$ $[\text{H}_3\text{O}^-]=0.05\alpha$ $\text{K}_\text{a}=\frac{[\text{C}_6\text{H}_5\text{O}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{OH}]}$ $\text{K}_\text{a}=\frac{(0.01)(0.05\alpha)}{0.05}$ $1.0\times10^{-10}=.01\alpha$ $\alpha=1\times10^{-8}$ View full question & answer→Question 1035 Marks
The ionization constant of nitrous acid is $4.5 \times 10^{–4}$. Calculate the pH of $0.04M$ sodium nitrite solution and also its degree of hydrolysis.
Answer$NaNO_2$ is the salt of a strong base $(NaOH)$ and a weak acid $(HNO_2)$. $\text{NO}_2^-+\text{H}_2\text{O}\leftrightarrow\text{HNO}_2+\text{OH}^-$
$\text{K}_\text{h}=\frac{[\text{HNO}_2][\text{OH}^-]}{[\text{NO}_2^-]}$
$\Rightarrow\frac{\text{K}_\text{w}}{\text{K}_\text{a}}=\frac{10^{-14}}{4.5\times10^{-4}}=.22\times10^{-10}$ Now, If x moles of the salt undergo hydrolysis, then the concentration of various species present in the solution will be: $[\text{NO}^-_2]=.04-\text{x};0.04$
$[\text{HNO}_2]=\text{x}$
$[\text{OH}^-]=\text{x}$
$\text{K}_\text{h}=\frac{\text{x}^2}{0.04}=0.22\times10^{-10}$
$\text{x}^2=.0088\times10^{-10}$
$\text{x}=.093\times10^{-5}$
$\therefore\ [\text{OH}^-]=0.093\times10^{-5}\text{M}$
$[\text{H}_3\text{O}^+]=\frac{10^{-14}}{.093\times10^{-5}}=10.75\times10^{-9}\text{M}$$\Rightarrow\text{pH}=-\log(10.75\times10^{-9})$
$=7.96$ Therefore, degree of hydrolysis $=\frac{\text{x}}{0.04}=\frac{.093\times10^{-5}}{.04}=2.325\times10^{-5}$
View full question & answer→Question 1045 Marks
The ionization constant of chloroacetic acid is $1.35 \times 10^{–3}$. What will be the pH of $0.1M$ acid and its $0.1M$ sodium salt solution?
AnswerIt is given that $K_a$ for $ClCH_2COOH$ is $1.35 \times 10^{–3}$.
$\Rightarrow\text{K}_\text{a}=\text{c}\alpha^2$
$\therefore\ \alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$
$=\sqrt{\frac{1.35\times10^{-3}}{0.1}}$ $(\therefore\ \text{concentration of acid = 0.1m})$
$\alpha=\sqrt{1.35\times10^{-2}}$
$=0.116$
$\therefore\ [\text{H}^+]=\text{c}\alpha=0.1\times0.116$
$=0.116$
$\Rightarrow\text{pH}=-\log[\text{H}^+]=1.94$
$ClCH_2COONa$ is the salt of a weak acid i.e., $ClCH_2COOH$ and a strong base i.e., NaOH.
$\text{ClCH}_2\text{COO}^-+\text{H}_2\text{O}\leftrightarrow\text{ClCH}^2\text{COOH}+\text{OH}^-$
$\text{K}_\text{h}=\frac{[\text{ClCH}^2\text{COOH}][\text{OH}^-]}{[\text{ClCH}_2\text{COO}^-]}$
$\text{K}_\text{h}=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$\text{K}_\text{h}=\frac{10^{-14}}{1.35\times10^{-3}}$
$=0.740\times10^{-11}$
Also, $\text{K}_\text{h}=\frac{\text{x}^2}{0.1}$ $(\text{where x is the concentration of OH}^-\text{and ClCH}_2\text{COOH})$
$0.740\times10^{-11}=\frac{\text{x}^2}{0.1}$
$0.074\times10^{-11}=\text{x}^2$
$\Rightarrow\text{x}^2=0.74\times10^{-12}$
$\text{x}=0.86\times10^{-6}$
$[\text{OH}^-]=0.86\times10^{-6}$
$\therefore\ [\text{H}^+]=\frac{\text{K}_\text{w}}{0.86\times10^{-6}}$
$=\frac{10^{-14}}{0.86\times10^{-6}}$
$[\text{H}^+]=1.162\times10^{-8}$
$\text{pH}=-\log[\text{H}^+]$
$=7.94$
View full question & answer→Question 1055 Marks
The ionization constant of benzoic acid is $6.46 \times 10^{–5}$ and $K_{sp}$ for silver benzoate is $2.5 \times 10^{–13}$. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
AnswerSince pH = 3.19,
$[\text{H}_3\text{O}^+]=6.46\times10^{-4}\text{M}$
$\text{C}_6\text{H}_5\text{COOH}+\text{H}_2\text{O}\leftrightarrow\text{C}_6\text{H}_5\text{COO}^-+\text{H}_3\text{O}$
$\text{K}_\text{a}\frac{[\text{C}_6\text{H}_5\text{COO}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{COOH}]}$
$\frac{[\text{C}_6\text{H}_5\text{COOH}]}{[\text{C}_6\text{H}_5\text{COO}^-]}=\frac{[\text{H}_3\text{O]}}{\text{K}_\text{a}}=\frac{6.46\times10^{-4}}{6.46\times10^{-5}}=10$
Let the solubility of $C_6H_5COOAg$ be xmol/L.
Then,
$[\text{Ag}^+]=\text{x}$
$[\text{C}_6\text{H}_5\text{COOH}]+[\text{C}_6\text{H}_5\text{COO}^-]=\text{x}$
$10[\text{C}_6\text{H}_5\text{COO}^-]+[\text{C}_6\text{H}_5\text{COO}^-]=\text{x}$
$[\text{C}_6\text{H}_5\text{COO}^-]=\frac{\text{x}}{11}$
$\text{K}_\text{sp}[\text{Ag}^+][\text{C}_6\text{H}_5\text{COO}^-]$
$2.5\times10^{-13}=\text{x}\Big(\frac{\text{x}}{11}\Big)$
$\text{x}1.66\times10^{-6}\text{mol/L}$
Thus, the solubility of silver benzoate in a pH 3.19 solution is $1.66 \times 10^{–6}mol/L$.
Now, let the solubility of $C_6H_5COO$ Ag be x'mol/L.
Then,
$[\text{Ag}^+]=\text{x}'\text{M and}[\text{CH}_3\text{COO}^-]=\text{x}'\text{M}.$
$\text{K}_\text{sp}=[\text{Ag}^+][\text{CH}_3\text{COO}^-]$
$\text{K}_\text{sp}=\text{(x}')^2$
$\text{x}'=\sqrt{\text{K}_\text{sp}}=\sqrt{2.5\times10^{-13}}=5\times10^{-7}\text{mol/L}$
$\therefore\ \frac{\text{x}}{\text{x}'}=\frac{1.66\times10^{-6}}{5\times10^{-7}}=3.32$
Hence, $C_6H_5COOAg$ is approximately 3.317 times more soluble in a low pH solution.
View full question & answer→Question 1065 Marks
The ionization constant of acetic acid is $1.74 \times 10^{–5}$. Calculate the degree of dissociation of acetic acid in its $0.05M$ solution. Calculate the concentration of acetate ion in the solution and its pH.
AnswerMethod 1
- $\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+\ \text{K}_\text{a}=1.74\times10^{-5}$
- $\text{H}_2\text{O}+\text{H}_2\text{O}\leftrightarrow\text{H}_3\text{O}^++\text{OH}^-\ \text{K}_\text{w}=1.0\times10^{-14}$
Since $Ka >> K_w$:
| |
|
$$$\text{CH}_3\text{COOH}$ |
$+$ |
$\text{H}_2\text{O}$ |
$\leftrightarrow$ |
$\text{CH}_3\text{COO}^-$ |
$+$ |
$\text{H}_3\text{O}^+$ |
| $\text{C}_\text{i}$ |
$=$ |
$0.05$ |
|
|
|
$0$ |
|
$0$ |
| |
|
$0.05-.05\alpha$ |
|
|
|
$0.05\alpha$ |
|
$0.05\alpha$ |
$\text{K}_\text{a}=\frac{(.05\alpha)(.05\alpha)}{(.05-0.05\alpha)}$
$=\frac{(.05\alpha)(0.05\alpha)}{.05(1-\alpha)}$
$=\frac{.05\alpha^2}{1-\alpha}$
$1.74\times10^{-5}=\frac{0.05\alpha^2}{1-\alpha}$
$1.74\times10^{-5}-1.74\times10^{-5}\alpha=0.05\alpha^2$
$0.05\alpha^2+1.74\times10^{-5}\alpha-1.74\times10^{-5}$
$\text{D}=\text{b}^2-4\text{ac}$
$=(1.74\times10^{-5})^2-4(.05)(1.74\times10^{-5})$
$=3.02\times10^{-25}+.348\times10^{-5}$
$\alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$
$\alpha=\sqrt{\frac{1.74\times10^{-5}}{.05}}$
$=\sqrt{\frac{34.8\times10^{-5}\times10}{10}}$
$=\sqrt{3.48\times10^{-6}}$
$=\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+$
$\alpha=1.86\times10^{-3}$
$[\text{CH}_3\text{COO}^-]=0.05\times1.86\times10^{-3}$
$=\frac{0.93\times10^{-3}}{1000}$
$=.000093$
Method 2
Degree of dissociation,
$\alpha=\sqrt{\frac{\text{K}_\text{a}}{\text{c}}}$
c = 0.05M
$K_a = 1.74 \times 10^{–5}$
Then,$\alpha=\sqrt{\frac{1.74\times10^{-5}}{.05}}$
$\alpha=\sqrt{34.8\times10^{-5}}$
$\alpha=\sqrt{3.48}\times10^{-4}$
$\alpha=1.8610^{-2}$
$\text{CH}_3\text{COOH}\leftrightarrow\text{CH}_3\text{COO}^-+\text{H}^+$
Thus, concentration of $CH_3COO– = c.\alpha$
$=.05\times1.86\times10^{-2}$
$=.093\times10^{-2}$
$=.00093\text{M}$
$\text{Since}[\text{oAc}^-]=[\text{H}^+],$
$[\text{H}^+]=.00093=.093\times10^{-2}.$
$\text{pH}=-\log[\text{H}^+]$
$=-\log(.093\times10^{-2})$
$\therefore\ \text{pH}=3.03$
Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03. View full question & answer→Question 1075 Marks
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
AnswerThe hydrogen ion concentration in the given substances can be calculated by using the given relation:
$\text{pH}=-\log[\text{H}^+]$
- pH of milk = 6.8
Since, $\text{pH}=-\log[\text{H}^+]$
$6.8=-\log[\text{H}^+]$
$\log[\text{H}^+]=-6.8$
$[\text{H}^+]=\text{anitlog}(-6.8)$
$=1.5\times10^{-7}\text{M}$
- pH of black coffee = 5.0
Since, $\text{pH}=-\log[\text{H}^+]$
$5.0=-\log[\text{H}^+]$
$\log[\text{H}^+]=-5.0$
$[\text{H}^+]=\text{anitlog}(-5.0)$
$=10^{-5}\text{M}$
- pH of tomato juice = 4.2
Since, $\text{pH}=-\log[\text{H}^+]$
$4.2=-\log[\text{H}^+]$
$\log[\text{H}^+]=-4.2$
$[\text{H}^+]=\text{anitlog}(-4.2)$
$=6.31\times10^{-5}\text{M}$
- pH of lemon juice = 2.2
Since, $\text{pH}=-\log[\text{H}^+]$
$2.2=-\log[\text{H}^+]$
$\log[\text{H}^+]=-2.2$
$[\text{H}^+]=\text{anitlog}(-2.2)$
$=6.31\times10^{-3}\text{M}$
- pH of egg white = 7.8
Since, $\text{pH}=-\log[\text{H}^+]$
$7.8=-\log[\text{H}^+]$
$\log[\text{H}^+]=-7.8$
$[\text{H}^+]=\text{anitlog}(-7.8)$
$=1.58\times10^{-8}\text{M}$ View full question & answer→