Question 13 Marks
Calculate the pH of the following solutions:
0.3g of NaOH dissolved in water to give 200mL of solution.
AnswerFor 0.3 g of NaOH dissolved in water to give 200 mL of solution:
$\text{NaOH}\rightarrow\text{Na}^+_\text{(aq)}+\text{OH}^-_\text{(aq)}$
$[\text{NaOH]}=0.3\times\frac{1000}{200}=0.5\text{M}$
$[\text{OH}^-_\text{aq}]=1.5\text{M}$
$\text{Then, [H}^+]=\frac{10^{-14}}{1.5}$
$=6.66\times10^{-13}$
$\text{pH}=-\log(6.66\times10^{-13})$
$=12.18$
View full question & answer→Question 23 Marks
The concentration of hydrogen ion in a sample of soft drink is $3.8 \times 10^{-3} \mathrm{M}$. what is its pH ?
AnswerGiven,
$[\text{H}^+]=3.8\times10^{-3}\text{M}$
$=-\log[\text{H}^+]$
$=-\log(3.8\times10^{-3})$
$=-\log3.8-\log10^{-3}$
$=-\log3.8+3$
$=-0.58+3$
$=2.42$
View full question & answer→Question 33 Marks
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
AnswerGiven,
pH = 3.76
It is known that,
$\text{pH}=-\log[\text{H}^+]$
$\Rightarrow\log[\text{H}^+]=-\text{pH}$
$\Rightarrow[\text{H}^+]=\text{antilog}(-\text{pH})$
$=\text{antilog}(-3.76)$
$=1.74\times10^{-4}\text{M}$
Hence, the concentration of hydrogen ion in the given sample of vinegar is $1.74\times10^{-4}\text{M}$.
View full question & answer→Question 43 Marks
Calculate the pH of the following solutions:
1mL of 13.6M HCl is diluted with water to give 1litre of solution.
AnswerFor 1 mL of 13.6 M HCl diluted with water to give 1 L of solution:
$13.6 \times 1 \mathrm{~mL}=\mathrm{M}_2 \times 1000 \mathrm{~mL}$
(Before dilution) (After dilution)
$13.6 \times 10^{-3}=M_2 \times 1 \mathrm{~L}$
$\mathrm{M}_2=1.36 \times 10^{-2}$
${\left[\mathrm{H}^{+}\right]=1.36 \times 10^{-2}}$
$\mathrm{pH}=-\log \left(1.36 \times 10^{-2}\right)$
$=(-0.1335+2)$
$=1.866 \sim 1.87$
View full question & answer→Question 53 Marks
Equilibrium constant, $K_c$ for the reaction
$\text{N}_2\text{ (g) + 3H}_2\text{ (g)}\rightleftharpoons2\text{NH}_3\text{ (g) at 500k is 0.061}$
At a particular time, the analysis shows that composition of the reaction mixture is $3.0 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~N}_2, 2.0 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{H}_2$ and $0.5 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{NH}_3$. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
AnswerThe given reaction is:$\text{N}_2\text{ (g) + 3H}_2\text{ (g)}\rightleftharpoons2\text{NH}_3\text{ (g)}$
According to available data.
$\text{N}_2=[3.0];\text{H}_2=[2.0];\text{NH}_3=[0.50]$
$\text{Q}_{\text{c}}=\frac{[\text{NH}_3\text{(g)]}^2}{[\text{N}_2\text{(g)][H}_2\text{(g)}]^3}=\frac{[0.50]^2}{[3.0][2.0]^3}=\frac{0.25}{24}=0.0104.$
Since the value of $Q_c$ is less than that of $k_c(0.061)$, the reaction is not is a state of equilibrium. It will proceed in the forward direction till $Q_c$ becomes the same as $k_c$.
View full question & answer→Question 63 Marks
The solubility of $\mathrm{Sr}(\mathrm{OH})_2$ at 298 K is $19.23 \mathrm{~g} / \mathrm{L}$ of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Answer$\text{Molar mass of Sr(OH})_2=87.6+34$
$=121.6\text{g mol}^{-1}$
$\text{Solubility of Sr(OH)}_2\text{ in moles L}^{-1}$
$=\frac{19.23\text{g L}^{-1}}{121.6\text{g mol}^{-1}}=0.1581\text{M}$
Assuming complete dissociation,
$\text{Sr(OH)}_2\rightarrow\text{Sr}^{2+}+2\text{OH}^-$
$\therefore\ [\text{Sr}^{2+}]=0.1581\text{M},$
$[\text{OH}^-]=2\times0.1581=0.3162\text{M}$
$\text{pOH}=-\log(0.3162)=0.5,$
$\therefore\ \text{pH}=14-0.5=13.5$
View full question & answer→Question 73 Marks
A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the
partial pressure of HI(g) is 0.04 atm.
What is $\text{K}_\text{p}$ for the given equilibrium?
$\text{2HI (g)}\rightleftharpoons\text{H}_2\text{ (g) + I}_2\text{ (g)}$
Answer$\begin{matrix}&2\text{HI}_{\text{(g)}}&\rightleftharpoons&\text{H}_{2\text{(g)}}&+&\text{I}_{2\text{(g)}}\\\text{Initial pressure}&0.2\text{ atm}&&0&&0\\\text{Pressure at eqm.}&0.04\text{ atm}&&\frac{0.16}{2}&&\frac{0.16}{2}\\&&&=0.08\text{ atm}&&=0.08\text{ atm}\end{matrix}$
According to law of chemical equilibrium,
$\text{K}_{\text{p}}=\frac{\text{p}_{\text{H}_2}}{\text{p}^2_\text{HI}}=\frac{0.08\times0.08}{(0.04)^2}=\frac{0.0064}{0.0016}=4$
$\therefore\ \text{K}_{\text{p}}=4$
View full question & answer→Question 83 Marks
Calculate the pH of the resultant mixtures:
10 mL of $0.2 \mathrm{M} \mathrm{~Ca}(\mathrm{OH})_2+25 \mathrm{~mL}$ of 0.1 M HCl
Answer$\text{Moles of H}_3\text{O}^+=\frac{25\times0.1}{1000}=.0025\text{mol}$
$\text{Moles of OH}^-=\frac{10\times0.2\times2}{1000}=.0040\text{mol}$
Thus, excess of $\text{OH}^-=.0015\text{mol}$
$[\text{OH}^-]=\frac{.0015}{35\times10^{-3}}\text{mol/L}=.0428$
$\text{pOH}=-\log[\text{OH}]$
$=1.36$
$\text{pH}=14-1.36$
$=12.63$ (not matched)
View full question & answer→Question 93 Marks
A sample of pure $PCl _5$ was introduced into an evacuated vessel at 473 K . After equilibrium was attained, concentration of $PCl _5$ was found to be $0.5 \times 10^{-1} mol L ^{-1}$. If value of $K _{ c }$ is $8.3 \times 10^{-3}$, what are the concentrations of $PCl _3$ and $Cl _2$ at equilibrium?
$PCl_5(g) \rightleftharpoons PCl_3(g)+Cl_2(g)$
AnswerLet the concentrations of both $PCl_3$ and $Cl_2$ at equilibrium $be x molL^{–1}.$ The given reaction is:
| |
$\text{PCl}_{5\text{(g)}}$ |
$\rightleftharpoons$ |
$\text{PCl}_{3\text{(g)}}$ |
$+$ |
$\text{Cl}_{2\text{(g)}}$ |
| At equilibrium |
$0.5\times10^{-1}\text{mol L}^{-1}$ |
|
$\text{xmol L}^{-1}$ |
|
$\text{xmol L}^{-1}$ |
| It is given that the value of equilibrium constant, $\text{K}_{\text{c}}\text{ is }8.3\times10^{-3}.$ |
Now we can write the expression for equilibrium as:
$\frac{[\text{PCl}_2][\text{Cl}_2]}{[\text{PCl}_2]}=\text{ K}_{\text{c}}$
$\Rightarrow\frac{\text{x}\times\text{x}}{0.5\times10^{-1}}=8.3\times10^{-3}$ $\Rightarrow\text{x}^2=4.15\times10^{-4}$$\Rightarrow\text{x}=2.04\times10^{-2}$
$=0.0204$$=0.02$ (approximately)
Therefore, at equilibrium, $[\text{PCl}_3]=[\text{Cl}_2]=0.02\text{mol L}^{-1}.$ View full question & answer→Question 103 Marks
A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
AnswerpH = 3.44
We know that,
$\text{pH}=-\log\text{[H}^+]$
$\therefore\ [\text{H}^+]=3.63\times10^{-6}$
Then,
$\text{K}_\text{h}=\frac{(3.63\times10^{-4})^2}{0.02}$ $(\because\ \text{concentration = 0.02M})$
$\Rightarrow\text{K}_\text{h}=6.6\times10^{-6}$
Now,
$\text{K}_\text{h}=\frac{\text{K}_\text{w}}{\text{K}_\text{a}}$
$\Rightarrow\text{K}_\text{a}=\frac{\text{K}_\text{w}}{\text{K}_\text{h}}=\frac{10^{-14}}{6.6\times10^{-6}}$
$=1.51\times10^{-9}$
View full question & answer→Question 113 Marks
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132 . Calculate the pH of the solution and the $\mathrm{pK}_{\mathrm{a}}$ of bromoacetic acid.
AnswerDegree of ionization, $\alpha=0.132$
Concentration, $\mathrm{c}=0.1 \mathrm{M}$
Thus, the concentration of $\mathrm{H}_3 \mathrm{O}^{+}=\mathrm{c} . \alpha$
$=0.1 \times 0.132$
$=0.0132$
$\text{pH}=-\log[\text{H}^+]$
$=-\log(0.0132)$
$=1.879:1.88$
Now,
$\text{K}_\text{a}=\text{C}\alpha^2$
$=0.1\times(0.132)^2$
$\text{K}_\text{a}=.0017$
$\text{pK}_\text{a}=2.75$
View full question & answer→Question 123 Marks
The reaction, $\text{CO (g) + 3H}_2\text{ (g)}\rightleftharpoons\text{CH}_4\text{ (g) + }\text{H}_2\text{O (g)}$ is at equilibrium at 1300K in a 1L flask. It also contain 0.30 mol of $\mathrm{CO}, 0.10 \mathrm{~molof} \mathrm{~H}_2$ and 0.02 mol of $\mathrm{H}_2 \mathrm{O}$ and an unknown amount of $\mathrm{CH}_4$ in the flask. Determine the concentration of $\mathrm{CH}_4$ in the mixture. The equilibrium constant, $\mathrm{K}_{\mathrm{c}}$ for the reaction at the given temperature is 3.90 .
AnswerLet the concentration of methane at equilibrium be x.
| |
$\text{CO}_\text{(g)}$ |
$+$ |
$\text{3H}_{2\text{(g)}}$ |
$\leftrightarrow$ |
$\text{CH}_{4\text{(g)}}$ |
$+$ |
$\text{H}_2\text{O}_\text{(g)}$ |
| At equilibrium |
$\frac{0.3}{1}=0.3\text{M}$ |
|
$\frac{0.1}{1}=0.1\text{M}$ |
|
$\text{x}$ |
|
$\frac{0.02}{1}=0.02\text{M}$ |
It is given that $\text{K}_\text{c}=3.90.$
Therefore,
$\frac{[\text{CH}_{4\text{(g)}}][\text{H}_2\text{O}_\text{(g)}]}{[\text{CO}_\text{(g)}][\text{H}_{2\text{(g)}}]}=\text{K}_\text{c}$
$\Rightarrow\frac{\text{x}\times0.02}{0.3\times(0.1)^3}=3.90$
$\Rightarrow\text{x}=\frac{3.90\times0.3\times(0.1)^3}{0.02}$
$=\frac{0.00117}{0.02}$
$=0.0585\text{M}$
$=5.85\times10^{-2}\text{M}$
Hence, the concentration of $CH_4$ at equilibrium is $5.85\times10^{-2}\text{M}$ View full question & answer→Question 133 Marks
Find out the value of $\text{K}_{\text{c}}$ for each of the following equilibria from the value of $\text{K}_{\text{p}}$:
$2\text{CaCO}_3\text{ (S)}\rightleftharpoons\text{CaO(S) + CO}_2\text{ (g)};\text{ k}_{\text{p}}=167\text{ at }1073\text{K}$
AnswerThe relation between $\text{K}_{\text{p}}$ and $\text{K}_{\text{c}}$ is given as:
$\text{K}_\text{p}=\text{K}_{\text{c}}\text{(RT})^{\Delta\text{n}}$
Here,
$\Delta\text{n}=2-1=1$
$\text{R}=0.0831\text{ bar L mol}^{-1}\text{K}^{-1}$
$\text{T}=1073\text{K}$
$\text{K}_{\text{p}}=167$
Now,
$\text{K}_\text{p}=\text{K}_{\text{c}}\text{(RT})^{\Delta\text{n}}$
$\Rightarrow167=\text{K}_{\text{c}}(0.0831\times1073)^{\Delta\text{n}}$
$\Rightarrow\text{K}_{\text{c}}=\frac{167}{0.0831\times1073}$
$=1.87(\text{approximately})$
View full question & answer→Question 143 Marks
The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction $3 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_3(\mathrm{~g})$ is $2.0 \times 10^{-50}$ at $25^{\circ} \mathrm{C}$. If the equilibrium concentration of $\mathrm{O}_2$ in air at $25^{\circ} \mathrm{C}$ is $1.6 \times 10^{-2}$, what is the concentration of $\mathrm{O}_3$ ?
AnswerThe given reaction is:$3\text{O}_{2\text{(g)}}\leftrightarrow2\text{O}_{3\text{(g)}}$
Then, $\text{K}_\text{c}=\frac{[\text{O}_{3\text{(g)}}]^2}{[\text{O}_{2\text{(g)}}]^3}$
It is given that $\text{K}_\text{C}=2.0\times10^{-50}\text{ and }[\text{O}_{2\text{(g)}}]=1.6\times10^{-2}.$
Then, we have,
$2.0\times10^{-50}=\frac{[\text{O}_{3\text{(g)}}]^2}{[1.6\times10^{-2}]^3}$
$\Rightarrow[\text{O}_{3\text{(g)}}]^2=2.0\times10^{-50}\times(1.6\times10^{-2})^3$
$\Rightarrow[\text{O}_{3\text{(g)}}]^2=8.192\times10^{-56}$
$\Rightarrow[\text{O}_{3\text{(g)}}]=2.86\times10^{-28}\text{ M}$
Hence, the concentration of $\text{O}_3\text{ is }2.86\times10^{-28}\text{ M}.$
View full question & answer→Question 153 Marks
Find out the value of $\text{K}_{\text{c}}$ for each of the following equilibria from the value of $\text{K}_{\text{p}}$:
$2\text{NOCl (g)}\rightleftharpoons2\text{NO (g) + Cl}_2\text{ (g)};\text{ k}_{\text{p}}=1.8\times10^{-2}\text{ at }500\text{K}$
AnswerThe relation between $\text{K}_{\text{p}}$ and $\text{K}_{\text{c}}$ is given as:
$\text{K}_\text{p}=\text{K}_{\text{c}}\text{(RT})^{\Delta\text{n}}$
Here,
$\Delta\text{n}=3-2=1$
$\text{R}=0.0831\text{ bar L mol}^{-1}\text{K}^{-1}$
$\text{T}=500\text{K}$
$\text{K}_{\text{p}}=1.8\times10^{-2}$
Now,
$\text{K}_\text{p}=\text{K}_{\text{c}}\text{(RT})^{\Delta\text{n}}$
$\Rightarrow1.8\times10^{-2}=\text{K}_{\text{c}}(0.0831\times500)^1$
$\Rightarrow\text{K}_{\text{c}}=\frac{1.8\times10^{-2}}{0.0831\times500}$
$=4.33\times10^{-4}(\text{approximately})$
View full question & answer→Question 163 Marks
If 0.561g of KOH is dissolved in water to give 200mL of solution at 298K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Answer$\text{Molar mass of KOH}=56.0\text{g mol}^{-1}$
$\therefore\ \text{No. of moles of KOH}$
$=\frac{0.561\text{g}}{56.0\text{g mol}^{-1}}=0.01\text{mol}$
$\text{and conc. of KOH}$
$=\frac{0.01}{0.2\text{L}}\text{mol}=0.05\text{mol/L}$
$\text{KOH}_\text{(aq)}\rightarrow\text{K}^+_\text{(aq)}+\text{OH}^-_\text{(aq)}$
$[\text{K}^+]=[\text{OH}^-]=0.05\text{M}$
$\text{and }[\text{H}^+]=\frac{\text{K}_\text{w}}{[\text{OH}^-]}=\frac{1.0\times10^{-14}}{0.05}=2.0\times10^{-13}$
$\therefore\ \text{pH}=-\log(2.0\times10^{-13})=12.70$
View full question & answer→Question 173 Marks
What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species:
$\text{HNO}_2,\text{CN}^-,\text{HClO}_4,\text{F}^-,\text{OH}^-,\text{CO}_3^{2-}\text{and S}^{2-}$
AnswerA conjugate acid-base pair is a pair that differs only by one proton. The conjugate acid-base for the given species is mentioned in the table below.
| $\text{Species Conjugate acid-base}$ |
| $\text{HNO}_2$ |
$\text{NO}_2^-\text{(base)}$ |
| $\text{CN}-$ |
$\text{HCN}\text{(acid)}$ |
| $\text{HClO}_4$ |
$\text{ClO}_4^-\text{(base)}$ |
| $\text{F}-$ |
$\text{HF}\text{(acid)}$ |
| $\text{OH}-$ |
$\text{H}_2\text{O}\text{(acid)}/\text{O}^{2-}\text{(base)}$ |
| $\text{CO}_3^{2-}$ |
$\text{HCO}^-_3\text{(acid)}$ |
| $\text{S}^{2-}$ |
$\text{HS}^-\text{(acid)}$ |
View full question & answer→Question 183 Marks
- Give the relationship between $\text{K}_\text{a}$, c and $('\alpha')$ where '$\text{K}_\text{a}$' is acid dissociation constant, 'c’ is molar concentration, $'\alpha'$ is degree of dissociation.
- If the solubility of $\mathrm{Ca}\left(\mathrm{lO}_3\right)_2$ in water at $18^{\circ} \mathrm{C}$ is $2.1 \mathrm{~g} /$ litre. Calculate the value of solubility product.
[Molecular mass of $\left.\mathrm{Ca}\left(\mathrm{lO}_3\right)_2=390\right]$Answer
- $\text{K}_{\text{a}}=\frac{\text{c}\alpha^2}{1-\alpha}\text{ if }\alpha<<<<1\text{ then }\text{K}_{\text{a}}=\text{c}\alpha^2$
$\because1-\alpha=1$
- Solubility in $\text{mol L}^{-1}$$=\frac{2.1\text{g L}^{-1}}{390\text{g mol}^{-1}}$
$=5.4\times10^{-3}\text{mol L}^{-1}$
$\text{Ca}(\text{IO}_3)_2\rightleftharpoons\text{Ca}^{2+}+2\text{IO}^-_3$
$\text{K}_{\text{sp}}=[\text{Ca}^{2+}][\text{IO}^-_3]^2$
$\text{K}_{\text{sp}}=(5.4\times10^{-3})(2\times5.4\times10^{-3})^2$
$\text{K}_{\text{sp}}=-629.856\times10^{-9}\text{mol}^3\text{L}^{-3}$
$=6.6\times10^{-7}\text{mol}^3\text{L}^{-3}$ View full question & answer→Question 193 Marks
pH of $0.08 \mathrm{mol} \mathrm{~dm}^{-3} \mathrm{HOCl}$ solution is 2.85 . Calculate its ionisation constant.
AnswerpH of HOCl = 2.85
But, $-\text{pH}=\log\text{[H}^+]$
$\therefore-2.85=\log\text{[H}^+]$
$\Rightarrow\overline{3}.15=\log\text{[H}^+]$
$\Rightarrow\log\text{[H}^+]=1.413\times10^{-3}$
For weak mono basic acid $[\text{H}^+]=\sqrt{\text{K}_\text{a}\times\text{C}}$
$\Rightarrow\text{K}_\text{}a=\frac{[\text{H}^+]^2}{\text{C}}=\frac{(1.413\times10^{-3})^2}{0.08}$
$=24.957\times10^{-6}=2.4957\times10^{-5}$
View full question & answer→Question 203 Marks
How can you predict the following stages of a reaction by comparing the value of $K_C$ and $Q_C$?
- Net reaction proceeds in the forward direction.
- Net reaction proceeds in the backward direction.
- No net reaction occurs.
AnswerThe values of $K_C$ and $Q_C$ are self explanatory and less than or greater than one another decides the direction in which reaction will proceed as follows:
- As $\text{Q}_\text{c}<\text{K}_\text{c},$ the reaction proceeds in the forward direction.
- If $\text{Q}_\text{c}<\text{K}_\text{c},$ the reaction will proceed in the direction of reactants (reverse reaction).
- If $\text{Q}_\text{c}<\text{K}_\text{c},$ no net reaction occurs.
View full question & answer→Question 213 Marks
For the reaction, $\text{CH}_4(\text{g})+2\text{H}_2\text{S(g)}\rightleftharpoons\text{CS}_2\text{(g)}+4\text{H}_2(\text{g})$ at 1173K, the magnitude of the equilibrium constant, $K_c$ is 3.6. For each of the following compositions, decide whether reaction mixture is at equilibrium. If it is not, decide in which direction the reaction should go?
- $[\text{CH}_4]=1.07\text{M},[\text{H}_2\text{S}]=1.20\text{M},[\text{CS}_2]$
$=0.90\text{M},[\text{H}_2]=1.78\text{M}$
- $[\text{CH}_4]=1.45\text{M},[\text{H}_2\text{S}]=1.29\text{M},[\text{CS}_2]$
$=1.25\text{M},[\text{H}_2]=1.75\text{M}$ Answer
- $\text{CH}_4(\text{g})+2\text{H}_2\text{S(g)}\rightleftharpoons\text{CS}_2\text{(g)}+4\text{H}_2(\text{g})$
$\text{Q}_{\text{c}}=\frac{[\text{CS}_4][\text{H}_2]^4}{[\text{CH}_4][\text{H}_2\text{S}]^2}=\frac{[0.90][1.78]^4}{[1.07][1.20]^2}$
$=\frac{10.03\times0.9}{1.44\times1.07}=\frac{9.027}{1.54}$
$=5.86\text{ K}_{\text{c}}=3.6$
Since $Q_c > K_c$, the equilibrium will shift in backward direction.
- $\text{Q}_{\text{c}}=\frac{[1.25][1.75]^4}{[1.45][1.29]^2}$
$=\frac{9.38\times1.25}{1.45\times1.66}=\frac{11.725}{2.407}=4.87$
Since $Q_c > K_c$, the equilibrium will shift in backward direction. View full question & answer→Question 223 Marks
Calculate the pH of a solution containing 0.1 M acetic acid and 0.1 M benzoic acid, $\mathrm{K}_{\mathrm{a}}$ for $\mathrm{CH}_3 \mathrm{COOH}$ and $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}$ are $1.8 \times 10^{-5}$ and $6.5 \times 10^{-5}$ respectively.
Answer$[\text{H}_3\text{O}^+]=\sqrt{\text{K}_{\text{a}_1}\text{C}_1+\text{K}_{\text{a}_2}\text{C}_2}$]
$=\sqrt{(1.8\times10^{-5})(0.1)+(6.5\times10^{-5})(0.1)}$
$=\sqrt{1.8\times10^{-6}+6.5\times10^{-6}}$
$=\sqrt{8.3\times10^{-6}}$
$=2.88\times10^{-3}\text{M}$
$\text{pH}=-\log[\text{H}_3\text{O}^+]$
$=-\log(2.88\times10^{-3})$
$=3-0.4594=2.5406$
View full question & answer→Question 233 Marks
Neutral solutions have pH = 7 at 298K. A sample of pure water is found to have pH < 7. Does it mean that it is acidic? Explain.
Answer$\mathrm{pH}<7$ for pure $\mathrm{H}_2 \mathrm{O}$ shows that water is at a temperature higher than 298 K . It is neutral at all temperatures. At higher temperature. $\mathrm{H}_2 \mathrm{O}$ dissociates more to give large concentrations of $\mathrm{H}^{+}$ions and $\mathrm{OH}^{-}$ions. Hence, $\mathrm{pH}<7$. However, $\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]$at all temperatures.
View full question & answer→Question 243 Marks
Why common salt is added to precipitate out soap from the solution during its manufacture?
AnswerSoaps are sodium salts of higher fatty acids having general formula RCOONa. On adding common salt, $\text{Na}^+$ ion concentration increases and due to common ion effect the equilibrium (I) as shown below shifts in the backward direction leading to the precipitation of soap.
$\text{NaCl}(\text{s})\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{Na}^+(\text{aq})+\text{Cl}^-\text{(aq)}$
$\text{RCOONa(s)}\rightleftharpoons\text{RCOO}^-(\text{aq})+\text{Na}^+(\text{aq}).$
View full question & answer→Question 253 Marks
Following data is given for the reaction: $\text{CaCO}_3\text{(s)}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ }\text{CaO(s)}+\text{CO}_2\text{(g)}$
$\Delta_\text{f}\text{H}^\ominus[\text{CaO}\text{(s)}]=-635.1\text{KJ mol}^{-1}$
$\Delta_\text{f}\text{H}^\ominus[\text{CO}_2\text{(g)}]=-393.5\text{KJ mol}^{-1}$
$\Delta_\text{f}\text{H}^\ominus[\text{CaCO}_3\text{(s)}]=-1206.9\text{KJ mol}^{-1}$
Predict the effect of temperature on the equilibrium constant of the above reaction.
Answer$\Delta_\text{f}\text{H}^\circ=\Delta_\text{r}\text{H}^\circ[\text{CaO}]+\Delta_\text{f}\text{H}^\circ[\text{CO}_2]-\Delta_\text{f}\text{H}^\circ[\text{CaCO}_3]$
$=[-635.1]+[-393.5]-[-1206.9]=178.3\text{KJ mol}^{-1}$
Thus, the reaction is endothermic.
Hence, according to Le Chatelier’s principle, on increasing the temperature, the equilibrium will proceed in the forward direction.
View full question & answer→Question 263 Marks
What is the effect of temperature on the reactions? Give reason.
- $\text{N}_2\text{(g)+3}\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})+\text{Heat}$
- $\text{N}_2\text{(g)+}\text{O}_2(\text{g})\rightleftharpoons2\text{NO}(\text{g})+\text{Heat}$
Answer
- The increase in temperature will favour backward reaction because the reaction is exothermic. On increasing temperature, equilibrium will shift in opposite direction to counter balance the effect of increasing temperature, i.e., backward reaction.
- The increase in temperature favours forward reaction because the reaction is endothermic.
View full question & answer→Question 273 Marks
MY and $\mathrm{NY}_3$ two nearly insoluble salts, have same $\mathrm{K}_{\text {sp }}$ values $6.2 \times 10^{-3}$ of non temperature. Calculate solubility of each salt. Which has more solubility.
Answer$\text{MY}\rightleftharpoons\text{M}^++\text{Y}^-\\'\text{s}'\ \ \ \ \ \ \ \ '\text{s}'\ \ \ \ \ \ \ \ '\text{s}'$
$\text{K}_{\text{sp}}=[\text{M}^+][\text{Y}^-]=\text{s}^2$
$\Rightarrow\text{s}=\sqrt{\text{K}_{\text{sp}}}$
$=\sqrt{6.2\times10^{-13}}$
$=\sqrt{62\times10^{-14}}$
$=7.9\times10^{-7}\text{mol L}^{-1}$
$[\because\sqrt{62}=7.9\text{ and }\sqrt{10^{-14}}=10^{-7}]$
$\text{NY}_3\rightleftharpoons\text{M}^{+3}+\text{3Y}^-\\'\text{s}'\ \ \ \ \ \ \ \ \ \ '\text{s}'\ \ \ \ \ \ \ \ \ '\text{3s}'$
$\text{K}_{\text{sp}}=[\text{M}^+][\text{Y}^-]$
$=\text{s}\times(3\text{s})^3=27\text{s}^4$
$\text{s}^4=4\sqrt{\frac{\text{K}_{\text{sp}}}{27}}$
$=4\sqrt{\frac{62\times10^{-14}}{24}}$
$=4\sqrt{2.296\times10^{-14}}$
$4\sqrt{229.6\times10^{-16}}$
$3.85\times10^{-4}\text{mol L}^{-1}$
View full question & answer→Question 283 Marks
A certain buffer is made by mixing sodium formate and formic acid in water. With the help of equations explain how this buffer neutralizes addition of a small amount of an acid or a base?
Answer$\text{HCOONa}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{HCOO}^-+\text{Na}^+$
$\text{HCOOH}\rightleftharpoons\text{HCOO}^-+\text{H}^+$
$\mathrm{HCOO}^{-}$is common ion in the above acidic buffer. When small amount of $\mathrm{H}^{+}$ions is added, these $\mathrm{H}^{+}$ions combine with $\mathrm{HCOO}^{-}$which are in excess to form HCOOH back and $\left[\mathrm{H}^{+}\right]$remains practically same, so pH remains constant. When small amount of $\mathrm{OH}^{-}$ions are added, $\mathrm{OH}^{-}$ions will take up ${\mathrm{H}^{+}}$and association of HCOOH will increase so as to maintain concentration of $\mathrm{H}^{+}$ions. So, pH is not affected.
View full question & answer→Question 293 Marks
Explain the following:
- Common ion effect.
- Bronsted Lowry concepts of acids and bases.
Answer
- Common Ion Effect: The decrease in concentration of one of the ions by adding other ion as common ion is called common ion effect, e.g.
$\text{H}_2\text{S}\rightleftharpoons\text{2H}^++\text{S}^{2-}$
$\text{HCl}\xrightarrow{\ \ \ \ \ \ \ \ }\text{H}^++\text{Cl}^-$
In presence of HCl conc. of $\mathrm{S}^{2-}$ will decrease due to common ion effect.
ii. According to Bronsted theory: Acids are proton donors whereas bases are proton acceptors, e.g., HCl is an acid whereas $\mathrm{Cl}^{-}$is a base. View full question & answer→Question 303 Marks
Match Column (I) with Column (II).
| Column I |
Column II |
| i. |
Equilibrium. |
a. |
$\Delta\text{G}>0, \text{ K}<1$ |
| ii. |
Spontaneous reaction. |
b. |
$\Delta\text{G}=0$ |
| iii. |
Non spontaneous reaction. |
c. |
$\Delta\text{G}^\ominus=0$ |
| |
|
d. |
$\Delta\text{G}<0, \text{ K}>1$ |
AnswerMatch Column (I) with Column (II).
| Column I |
Column II |
| i. |
Equilibrium. |
c. |
$\Delta\text{G}^\ominus=0$ |
| ii. |
Spontaneous reaction. |
d. |
$\Delta\text{G}<0, \text{ K}>1$ |
| iii. |
Non spontaneous reaction. |
a. |
$\Delta\text{G}>0, \text{ K}<1$ |
View full question & answer→Question 313 Marks
pH of a solution of a strong acid is 5.0. What will be the pH of the solution obtained after diluting the given solution a 100 times?
Answer$\mathrm{pH}=5$
${\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}}$
On 100 times dilution.
$\left[\mathrm{H}^{+}\right]=10^{-7} \mathrm{~mol} \mathrm{~L}^{-1}$
On calculating the pH using the equation $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]$, value of pH comes out to be 7 . It is not possible. This indicates that solution is very dilute. Hence,
Total hydrogen ion concentration $=\left[\mathrm{H}^{+}\right]$
$=\left[\text { Contibution of } \mathrm{H}_3 \mathrm{O}^{+} \text {ion concentration of Acid }\right]+\left[\text { Contibution of } \mathrm{H}_3 \mathrm{O}^{+}\right. \text {ion concentration of Water] }$
$=10^{-7}+10^{-7} \text {. }$
$\mathrm{pH}=2 \times 10^{-7}=7-\log 2=7-0.3010=6.6990$
View full question & answer→Question 323 Marks
Calculate
- Dissociation constant of conjugate base of HF, $K_a = 6.8 \times 10^{-4}$.
- Dissociation constant of conjugate acid of $\text{CO}^{2-}_3,\text{K}_{\text{b}}=2.1\times10^{-4}$
Answer
- $\text{HF}\rightleftharpoons\text{H}^++\text{F}^-$
$K_a (acid) \times k_b (conjugate ~base)$ = 1.0 × 10-14
$\text{K}_{\text{b}}(\text{F}^-)=\frac{1.0\times10^{-14}}{6.8\times10^{-14}}$
$=1.47\times10^{-11}$
- $\text{CO}^{2-}_3+\text{H}^+\rightleftharpoons\text{HCO}_3^-$
$\text{K}_{\text{b}}(\text{CO}^{2-}_3)\times\text{K}_{\text{a}}(\text{HCO}^-_3)=1\times10^{-14}$
$\text{K}_{\text{a}}(\text{HCO}^-_3)=\frac{1\times10^{-14}}{2.1\times10^{-4}}=4.76\times10^-11{}$ View full question & answer→Question 333 Marks
The molar solubility of lead iodate, $Pb(I0_3)_2$ is $4.0 \times 10^{-5} mol/ L at 25^\circ C$. Calculate its $K_{sp}$.
$\text{Pb}(\text{IO}_3)_2\rightleftharpoons\text{Pb}^{2+}+\text{2IO}_3$
Answer$\therefore \mathrm{K}_{\text {sp }}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{IO}_3^{-}\right]^2$
$\because \text { The molar solubility of } \mathrm{Pb}\left(\mathrm{IO}_3\right)_2 \text { is } 4.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$
$\therefore\left[\mathrm{~Pb}^{2+}\right]=4.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$
${\left[\mathrm{IO}_3^{-}\right]=2 \times 4.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}}$
$\therefore \mathrm{~K}_{\text {sp }}=\left(4.0 \times 10^{-5}\right)\left(8.0 \times 10^{-5}\right)^2$
$=2.56 \times 10^{-13}$
View full question & answer→Question 343 Marks
Some processes are given below. What happens to the process if it is subjected to a change given in the brackets?
-
(Pressure is increased)
- Dissolution of NaOH in water (Temperature is increased.)
- $\text{N}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO}(\text{g})-180.7\text{KJ}$ (Pressure is increased and temperature is decreased.)
Answer
- Equilibrium will shift in the forward direction, i.e. more of ice will melt.
- Solubility will decrease because it is an exothermic process.
- Pressure has no effect. Decrease of temperature will shift the equilibrium in the backward direction.
View full question & answer→Question 353 Marks
What is pOH? What is its value for neutral water at 25°C.
AnswerpOH is defined as negative logarithm of $\mathrm{OH}^{-}$concentration. $\mathrm{pOH}=7 \log \left[\mathrm{OH}^{-}\right] \mathrm{pOH}=7$ for neutral water at $25^{\circ} \mathrm{C}$.
$\left[\because \mathrm{pH}=\mathrm{pOH}=7\right.$ at $25^{\circ} \mathrm{C} \mathrm{pH}+\mathrm{pOH}=14$ at $25^{\circ} \mathrm{C}$.
View full question & answer→Question 363 Marks
Solid $\mathrm{Ba}\left(\mathrm{NO}_3\right)$, is gradually dissolved in a $1.0 \times 10^{-4} \mathrm{M} \mathrm{Na}_2 \mathrm{CO}_3$ solution. At what concentration of $\mathrm{Ba}^{2+}$ will a precipitate begin to form. $\left(\mathrm{K}_{\mathrm{sp}}\right.$ for $\left.\mathrm{BaCO}_3=5.1 \times 10^{-9}\right)$
Answer$\text{BaCO}_3(\text{s})\rightleftharpoons\text{Ba}^{2+}+\text{CO}^{2-}_{3}$
$\text{Na}_2\text{CO}_3\xrightarrow{\ \ \ \ \ \ \ \ }2\text{Na}^++\text{CO}^{2-}_3$
$[\text{Na}_2\text{CO}_3]=1.0\times10^{-4}\text{M}$
$\text{K}_{\text{sp}}=[\text{Ba}^{2+}][\text{CO}_3^{2-}]$
$5.1\times10^{-9}=[\text{Ba}^{2+}][1\times10^{-4}\text{M}]$
$\text{[Ba}^{2+}]=\frac{5.1\times10^{-9}}{1\times10^{-4}}$
$=5.1\times10^{-5}\text{M}$
At $5.1 \times 10^{-5} \mathrm{M}$ concentration of $\left[\mathrm{Ba}^{2+}\right], \mathrm{I} . \mathrm{P}\left(\right.$ ionic product) will become equal to $\mathrm{K}_{\mathrm{sp}}$ and precipitate will begin to from.
View full question & answer→Question 373 Marks
i. Why is $NH _4 Cl$ added before addition of $NH _4 OH$ in qualitative analysis of $3^{\text {rd }}$ group?
ii. Which will be added to precipitate soap
(RCOONa)? NaCl or KCl and why?'
AnswerIt is done so as to decrease conc. of $OH^-$ due to common ion effect so that only group 3 radicals get precipitated and higher group radicals do not.
NaCl will be added to precipitate soap.
$\text{RCOO}^-+\text{Na}^+\rightleftharpoons\text{RCOONa}$
$\text{Na}^++\text{Cl}^-\rightleftharpoons\text{NaCl}$
Due to common ion effect of $Na^+$, soap (RCOONa) will get precipitated completely.
View full question & answer→Question 383 Marks
For the reaction, $\mathrm{N}_2(\mathrm{g})+3 \mathrm{H}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{g})$, the partial pressures of $\mathrm{N}_2$ and $\mathrm{H}_2$ are 0.80 and 0.40 atmosphere respectively at equilibrium. The total pressure of the system is 2.80 atmosphere. What is $\mathrm{K}_{\mathrm{p}}$ for the above reaction?
Answer$\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g}),$ Given, at equilibrium, $\text{p}_{\text{N}_2}=0.80$ atmosphere, $\text{p}_{\text{H}_2}=0.40$ atmosphere $\text{p}_{\text{N}_2}+\text{p}_{\text{H}_2}+\text{p}_{\text{NH}_3}=2.80$ atmosphere$\therefore\text{p}_{\text{NH}_3}=2.80-(0.80+0.40)=1.60$ atmosphere
From, $\text{K}_{\text{p}}=\frac{\text{p}_{\text{NH}_3}^2}{\text{p}_{\text{N}_3}\times\text{p}^3_{\text{H}_2}}$ $=\frac{(1.60)^2}{0.80\times(0.40)^3}=50.0$
View full question & answer→Question 393 Marks
Calculate the molar solubility of $Ni ( OH )_2$ in 0.10 M NaOH . The ionic product of $Ni ( OH )_2$ is $2.0 \times 10^{-15}$.
Answer$\text{Ni(OH)_2}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{Ni}^{2+}+2\text{OH}^-$
Ionic product $= [Ni^{2+}] [OH^-]^2$
$[OH^-] = 0.1M, [Ni^{2+}] = ?$
$2.0 \times 10^{-15} = [Ni^{2+}] (0.1)^2$
$\Rightarrow[\text{Ni}^{2+}]=\frac{2\times10^{-15}}{10^{-2}}$
$=2\times10^{-13}\text{M}$
Solubility of $Ni(OH)_2$ will be equal to $[Ni^{2+}] = 2 \times 10^{-13}M.$
View full question & answer→Question 403 Marks
Urine has a pH of 6.0. If a patient eliminates 1300mL of urine per day, how many gram equivalents of the acid he eliminates per day?
Answer$\because \text { pH }=6.0 . \therefore\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=10^{-6} \mathrm{M}$
$\text { i.e. }[\text { Acid }]=10^{-6} \mathrm{M}=10^{-6} \mathrm{~N}$
Thus, 1000 mL of the urine contain acid $=10^{-6} \mathrm{~g}$ e.q
$\therefore 1300 \mathrm{~mL}$ pf the urine will contain acid $=1.3 \times 10^{-6} \mathrm{~g}$ eq.
View full question & answer→Question 413 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: $\text{CO(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{CH}_4\text{(g)}+\text{H}_2\text{O(g)}$
Answer$\Delta\text{G}=\Delta\text{G}^\ominus+\text{RT InQ}$
$\Delta\text{G}^\ominus$ = Change in free energy as the reaction proceeds.
$\Delta\text{G}$ = Standard free energy change.
Q = Reaction quotient.
R = Gas constant.
T = Absolute temperature.
Since $\Delta\text{G}^\ominus=-\text{RT InK}$
$\therefore\Delta\text{G}=-\text{RT InK}+\text{RT InQ}=\text{RT In}\frac{\text{Q}}{\text{K}}$
If $\text{Q}<\text{K},\Delta\text{G}$ will be negative. Reaction proceeds in the forward direction.
If $\text{Q}=\text{K},\Delta\text{G}=0$, no net reaction.
[Hint: Next relate Q with concentration of $CO, H_2, CH_4$ and $H_2O$ in view of reduced volume (increased pressure). Show that $\text{Q}<\text{K}$ and hence the reaction proceeds in forward direction.]
View full question & answer→Question 423 Marks
What is meant by reaction quotient? Give its expression.
AnswerReaction quotient: It is defined as product of molar concentration of products each raised to the power equal to their respective stoichiometric coefficients in balanced chemical equation divided by the product of molar concentration of the reactants raised to the power equal to their individual stoichiometric coffecients at any stage of reaction,
$\text{e.g.}\text{ aA}+\text{bB}\rightleftharpoons\text{cC}+\text{dD}$
$\text{Q}_{\text{c}}=\frac{[\text{C}^{\text{c}}][\text{D}]^{\text{d}}}{[\text{A}]^{\text{a}}[\text{B}]^{\text{b}}}$
View full question & answer→Question 433 Marks
- Sulphuric acid is a very strong acid, yet it can also act as base in some reactions. Explain how?
- All Bronsted acids are not Lewis acids. Explain.
Answeri. Sulphuric acid $\left(\mathrm{H}_2 \mathrm{SO}_4\right)$ is weaker acid with respect to perchloric acid $\left(\mathrm{HClO}_4\right) . \mathrm{H}_2 \mathrm{SO}_4$ can take up a proton from $\mathrm{HClO}_4$ to form $\mathrm{H}_2 \mathrm{SO}_4$. Hence, it acts as a base in this reaction.
ii. Bronsted acids can donate $\mathrm{H}^{+}$easily but they may not be able to donate electrons e.g. $\mathrm{HCl}, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{HNO}_3$.
Therefore all Bronsted acids are not Lewis acids.
View full question & answer→Question 443 Marks
For the reaction : $\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)}$ Equilibrium constant $\text{K}_\text{c}=\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$ Some reactions are written below in Column I and their equilibrium constants in terms of $\text{K}_\text{c}$ are written in Column II. Match the following reactions with the corresponding equilibrium constant.
| Column I (Reaction) |
Column II (Equilibrium constant) |
| i. |
$2\text{N}_2\text{(g)}+\text{6H}_2\text{(g)}\rightleftharpoons\text{4NH}_3\text{(g)}$ |
a. |
$2\text{K}_\text{c}$ |
| ii. |
$2\text{NH}_3\text{(g)}\rightleftharpoons\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}$ |
b. |
$\text{K}_\text{c}^\frac{1}{2}$ |
| iii. |
$\frac{1}{2}\text{N}_2\text{(g)}+\frac{3}{2}\text{H}_2\text{(g)}\rightleftharpoons\text{NH}_3\text{(g)}$ |
c. |
$\frac{1}{\text{K}_\text{c}}$ |
| |
|
d. |
$\text{K}_\text{c}^2$ |
Answer
| Column I (Reaction) |
Column II (Equilibrium constant) |
| i. |
$2\text{N}_2\text{(g)}+\text{6H}_2\text{(g)}\rightleftharpoons\text{4NH}_3\text{(g)}$ |
d. |
$\text{K}_\text{c}^2$ |
| ii. |
$2\text{NH}_3\text{(g)}\rightleftharpoons\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}$ |
c. |
$\frac{1}{\text{K}_\text{c}}$ |
| iii. |
$\frac{1}{2}\text{N}_2\text{(g)}+\frac{3}{2}\text{H}_2\text{(g)}\rightleftharpoons\text{NH}_3\text{(g)}$ |
b. |
$\text{K}_\text{c}^\frac{1}{2}$ |
View full question & answer→Question 453 Marks
Match standard free energy of the reaction with the corresponding equilibrium constant.
| Column I |
Column II |
| i. |
$\Delta\text{G}^\ominus>0$ |
a. |
$\text{K}>1$ |
| ii. |
$\Delta\text{G}^\ominus<0$ |
b. |
$\text{K}=1$ |
| iii. |
$\Delta\text{G}^\ominus=0$ |
c. |
$\text{K}=0$ |
| |
|
d. |
$\text{K}<1$ |
Answer
| Column I |
Column II |
| i. |
$\Delta\text{G}^\ominus>0$ |
d. |
$\text{K}<1$ |
| ii. |
$\Delta\text{G}^\ominus<0$ |
a. |
$\text{K}>1$ |
| iii. |
$\Delta\text{G}^\ominus=0$ |
b. |
$\text{K}=1$ |
View full question & answer→Question 463 Marks
Why solution of sugar in water does not conduct electricity whereas that of common salt in water does?
AnswerCommon salt $(\mathrm{NaCl})$ is an electrolyte which in the aqueous solution gives $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions. Hence, it conducts electricity. Sugar is sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ which is a non-electrolyte and does not give ions in the solution. Hence, it does not conduct electricity.
View full question & answer→Question 473 Marks
What will be the value of $\Delta\text{G}\text{ and }\Delta\text{G}^\circ$ for the reaction, $\text{A}+\text{B}\rightleftharpoons\text{C}+\text{D}\text{ at }27^\circ\text{C}$ for which $\text{K}=10^2$. Predict the extent of reaction.
Answer$\Delta\text{G}=0$ at equilibrium
$\Delta\text{G}^\circ=-2.303\text{ RT }\log\text{ K}$
$\text{R}=8.314\text{Jk}^{-1}\text{mol}^{-1};$]
$\text{T}=27^\circ\text{C}+273=300\text{K};$
$\text{K}=10^2$
$\Delta\text{G}^\circ=-2.303\times8.314\times300\log10^2$
$[\because\log10^2=2\log10=2\times1=2]$
$\Delta\text{G}^\circ=-19.147\times300\times2$
$\Delta\text{G}^\circ=-57.441\times2\times100\text{J}$
$\Delta\text{G}^\circ=114.882\times10^2\text{J}$
$=-11.49\text{kJ mol}^{-1}$
The reaction will be almost complete because concentration of products is 100 times more then reactants.
View full question & answer→Question 483 Marks
The $K_{s p}$ values of two slightly soluble salts $A B$ and $\mathrm{PQ}_2$ are each equal to $4.0 \times 10^{-18}$. Which salt is more soluble?
AnswerFor $A B$ salt, $K_{s p}=\left[A^{+}\right]\left[B^{-}\right]$If $x$ is the solubolity, then
$x^2=4.0 \times 10^{-18}$
$\text { or } x=2.0 \times 10^{-9}$
For $\mathrm{PQ}_2$ salt, $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{P}^{2+}\right]\left[\mathrm{Q}^{-}\right]^2$
If y is the solubility, then
$4 y^3=K_{\text {sp }}=4.0 \times 10^{-18}$
or $y=1.0 \times 10^{-6}$
Since $y$ is more than $x, P Q_2$ salt has more solubility.
View full question & answer→Question 493 Marks
Why do we pass $\mathrm{H}_2 \mathrm{S}$ gas in acidic medium in group 2?
AnswerIt is done so as to reduce concentration of $\left[\mathrm{SP}^{2-}\right]$ by common ion effect so that only group 2 radicals get precipitated whereas higher group radicals do not as $\mathrm{K}_{\text {sp }}$ of group II sulphides is low as compared to $\mathrm{K}_{\text {sp }}$ of sulphides of higher groups.
View full question & answer→Question 503 Marks
If $\mathrm{K}_{\mathrm{w}}=49 \times 10^{-14}$, what will be neutral pH of $\mathrm{H}_2 \mathrm{O}$ ?
Answer$\text{K}_{\text{w}}=49\times10^{}-14[\text{H}_3\text{O}^+][\text{OH}^-]$
$=49\times10^{-14}$
$[\text{H}_3\text{O}^+]=[\text{OH}^-]$
$\Rightarrow[\text{H}_3\text{O}^+]^2=49\times10^{-14}$
$\Rightarrow[\text{H}_3\text{O}^+]=7\times10^{-7}\text{ mol L}^{-1}$
$\text{pH}=-\log[\text{H}_3\text{O}^+]$
$=-\log7\times10^{-7}$
$=-\log7-\log10^{-7}$
$=-\log7+7\log10$
$\text{pH}=-0.8451+7.000$
$=6.1549$
View full question & answer→Question 513 Marks
A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the inoisation constant of pyridine.
Answer$\text{C}_6\text{H}_5\text{N}+\text{HCl}^-+\text{H}_2\text{O}\rightleftharpoons\text{C}_6\text{H}_5\text{N}^+\text{HOH}^-+\text{HCl}$
(Solution is acidic due to hydrolysis)
$\text{pH}=7-\frac{\text{pK}_\text{b}}{2}-\frac{\log\text{C}}{2}$
$3.44=7-\frac{\text{pK}_{\text{b}}}{2}-\frac{\log0.02}{2}$
$\text{pK}_{\text{b}}=8.82$
$\text{pK}_{\text{b}}=8.82=-\log\text{ pK}_{\text{b}}$
$\log\text{K}_{\text{b}}=-8.82=\bar{9}.18$
$\text{K}_{\text{b}}=\text{antilog }\bar{9}.18$
$=1.5\times10^{-9}$
View full question & answer→Question 523 Marks
On the basis of Le Chatelier principle explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction.
$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)} \ \ \ \ \ \ \ \Delta\text{H}= – 92.38\text{kJ mol}^{–1}$
What will be the effect of addition of argon to the above reaction mixture at constant volume?
Answer$\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)} \ \ \ \ \ \ \ \Delta\text{H}= – 92.38\text{kJ mol}^{–1}$ Effect of temperature: All chemical reactions are accompanied by the liberation or uptake if heat. If we regard heat as a reactant or product in an endothermic or exothermic reaction respectively, we can use the Le Chatelier principle to predict the direction in which an increase or decrease in temperature will shift the equilibrium state. Therefore for the give reaction we can write, $\text{N}_2\text{(g)}+\text{3H}_2\text{(g)}\rightleftharpoons\text{2NH}_3\text{(g)} \ \ \ \ \ \ \ \Delta\text{H}= – 92.38\text{kJ mol}^{–1}$ The Le Chatelier principle tells us that a net reaction will occur in the direction that will partially counteract this change. Since the reaction is exothermic, a shift of the equilibrium to the left will take place. Thus, by decreasing the temperature yield of $NH_3$ can be increased. Effect of pressure:
$\Delta\text{n}_\text{g}=\text{n}_\text{g}\text{(products)}-\text{n}_\text{g}\text{(reactants)}$ $= 2 - (1 + 3) = 2 - 4 = -2$ Since $\Delta\text{n}_\text{g}$ is negative, if the pressure is increased,The equilibrium will shift in the forward direction according to the Lr Chatelier principle, because doing this will decrease the total number of moles of gases and hence the pressure which will decrease the effect of increasing pressure. Thus, high pressure would increase the yield of ammonia. Addition of argon at constant volume The addition of argon at constant volume would not change the partial pressures of any substance, $N_2, H_2$ or $NH_3$. Hence, the equilibrium is not disturbed and there would not be any effect of addition of argon at constant Volume.
View full question & answer→Question 533 Marks
$2\text{NO}(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g}),\Delta\text{H}=-117\text{J}$
Predict the effect of an increase in concentration of NO on the equilibrium concentration of $\text{NO}_2$.
Predict the effect of pressure decrease as a result of increased volume on the equilibrium concentration of $\text{NO}_2$.
Answer$2\text{NO}(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{NO}_2(\text{g});\Delta\text{H}=-117\text{J}$
- If we increase the concentration of $\text{NO}_2$ the rate of forward reaction will increase, i.e. more $\text{NO}_2$ will be formed.
- Decrease in pressure will favour backward reaction, i.e. less $\text{NO}_2$ will be formed because number of moles of reactants are more than products.
View full question & answer→Question 543 Marks
$\begin{matrix}\text{A }\rightleftharpoons\text{ B }&\text{K}_1=1&\text{B }\rightleftharpoons\text{ C}&\text{K}_2=2\\\text{C }\rightleftharpoons\text{ D}&\text{K}_3=3&\text{D }\rightleftharpoons\text{ E}&\text{K}_4=4\end{matrix}$
What is value of K for $\text{A }\rightleftharpoons\text{ E}?$
Answer$\text{A }\rightleftharpoons\text{ E K}=\frac{[\text{E}]}{[\text{A}]}$ $\text{K}_1=\frac{[\text{B}]}{[\text{A}]},\text{K}_2=\frac{[\text{C}]}{[\text{B}]},$$\text{K}_3=\frac{[\text{D}]}{[\text{C}]}\text{ and }\text{K}_4=\frac{[\text{E}]}{[\text{D}]}$
$\because\text{K}=\text{K}_1\times\text{K}_2\times\text{K}_3\times\text{K}_4=\frac{[\text{E}]}{[\text{A}]}$ $=1\times2\times3\times4=24$
View full question & answer→Question 553 Marks
Write two equations to show the amphiprotic (acid as well as base) property of water.
Answer$\begin{matrix}\text{HCl}&+&\text{H}_2\text{O}&\rightleftharpoons&\text{H}_3\text{O}^+&+&\text{Cl}^-\\\text{Acid}&&\text{Base}_2&&\text{Acid}_2&&\text{Base}_1\\\end{matrix}$
$\begin{matrix}\text{NH}_3&+&\text{H}_2\text{O}&\rightleftharpoons&\text{NH}^+_4&+&\text{OH}^-\\\text{Base}_1&&\text{Acid}_2&&\text{Acid}_1&&\text{Base}_2\end{matrix}$
In first equation $\mathrm{H}_2 \mathrm{O}$ acts as base and in second it acts as acid, so $\mathrm{H}_2 \mathrm{O}$ is amplification, i.e., can gain $\mathrm{H}^{+}$and lose $\mathrm{H}^{+}$.
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Which of the following reactions involve homogeneous equilibrium and which involve heterogeneous equilibrium?
- $\text{Ag}_2\text{O}(\text{s})+2\text{HNO}_3(\text{aq})\rightleftharpoons2\text{AgNO}_3(\text{aq})+\text{H}_2\text{O}(\text{l})$
- $\text{C(s)}+\text{CO(}_2(\text{g})\rightleftharpoons2\text{CO(g)}$
- $\text{CH}_3\text{COOC}_2\text{H}_5(\text{aq})+\text{H}_2\text{O}(\text{l})\\\rightleftharpoons\text{CH}_3\text{COOH}(\text{aq})+\text{C}_2\text{H}_5\text{OH}(\text{aq})$
- $2\text{SO}_2(\text{g})+\text{O}_2(\text{g})\rightleftharpoons2\text{SO}_3(\text{g})$
Answer
- Heterogeneous equilibrium.
- Heterogeneous equilibrium.
- Homogeneous equilibrium.
- Homogeneous equilibrium.
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Match the following graphical variation with their description.
