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
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S
|
S
|
|
At equilibrium in sulphuric acid
|
1 -S
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S
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(S + 0.01)
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$K_{sp}$ 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) |
$\text{K}_{\text{c}}^{\frac{1}{2}}$ |
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