Chapter 6: Equilibrium

Ans: ${\text{NaN}}{{\text{O}}_2}$ is the salt of a strong base $({\text{NaOH}})$ and a weak acid $\left( {{\text{HN}}{{\text{O}}_2}} \right)$.

${{\text{NO}}_2^ -  + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{HN}}{{\text{O}}_2} + {\text{O}}{{\text{H}}^ - }} $ 

${{K_b} = \dfrac{{\left[ {{\text{HN}}{{\text{O}}_2}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{NO}}_2^ - } \right]}}} $ 

${ \Rightarrow \dfrac{{{K_w}}}{{{K_a}}} = \dfrac{{{{10}^{ - 14}}}}{{4.5 \times {{10}^{ - 14}}}} = .22 \times {{10}^{ - 10}}} $

Now, If $x$ moles of the salt undergo hydrolysis, then the concentration of various species present in the solution, it will be:

${\left[ {NO_2^ - } \right] = .04 - x;0.04} $ 

${\left[ {HN{O_2}} \right] = x} $ 

${\left[ {O{H^ - }} \right] = x} $ 

${{K_b} = \dfrac{{{x^2}}}{{0.04}} = 0.22 \times {{10}^{ - 10}}} $ 

${{x^2} = .0088 \times {{10}^{ - 10}}} $ 

${x = .093 \times {{10}^{ - 5}}} $

${\therefore \left[ {O{H^ - }} \right] = 0.093 \times {{10}^{ - 5}}M} $ 

${\left[ {{H_3}{O^ + }} \right] = \dfrac{{{{10}^{ - 14}}}}{{.093 \times {{10}^{ - 5}}}} = 10.75 \times {{10}^{ - 9}}M} $ 

${ \Rightarrow pH =  - \log \left( {10.75 \times {{10}^{ - 9}}} \right)} $ 

 = 7.96 

Therefore, degree of hydrolysis

${ = \dfrac{x}{{0.04}} = \dfrac{{.093 \times {{10}^{ - 5}}}}{{.04}}} $ 

${ = 2.325 \times {{10}^{ - 5}}} $

Ans: ${\text{pH}} = 3.44$

We know that,

${\text{pH}} =  - \log \left[ {{{\text{H}}^ + }} \right]$ $\therefore \left[ {{H^ + }} \right] = 3.63 \times {10^{ - 4}}$

Then, ${K_b} = \dfrac{{{{\left( {3.63 \times {{10}^{ - 4}}} \right)}^2}}}{{0.02}}(\because $ concentration $ = 0.02{\text{M}})$

$ \Rightarrow {K_b} = 6.6 \times {10^{ - 6}}$

Now, ${K_b} = \dfrac{{{K_w}}}{{{K_a}}}$

$ \Rightarrow {K_a} = \dfrac{{{K_w}}}{{{K_a}}} = \dfrac{{{{10}^{ - 14}}}}{{6.6 \times {{10}^{ - 6}}}} = 1.51 \times {10^{ - 9}}$

Ans:

(i) ${\text{NaCl}}$ :

${\text{NaCl}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{NaOH}} + {\text{HCl}}$

Strong base Strong base

Therefore, it is a neutral solution.

(ii) KBr:

${\text{KBr}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{KOH}} + {\text{HBr}}$

Strong base    Strong base

Therefore, it is a neutral solution.

iii) ${\text{NaCN}}:$

${\text{NaCN}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{HCN}} + {\text{NaOH}}$

Weak acid        Strong base

Therefore, it is a basic solution.

(iv) ${\text{N}}{{\text{H}}_{\text{4}}}{\text{N}}{{\text{O}}_{\text{3}}}$

${\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_3} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{N}}{{\text{H}}_4}{\text{OH}} + {\text{HN}}{{\text{O}}_3}$

Weak acid               Strong base

Therefore, it is an acidic solution.

(v) ${\text{NaN}}{{\text{O}}_{\text{2}}}$

${\text{NaN}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{NaOH}} + {\text{HN}}{{\text{O}}_2}$

Strong base         Weak acid

Therefore, it is a basic solution.

(vi) KF

${\text{KF}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{KOH}} + {\text{HF}}$

Strong base     Weak acid

Therefore, it is a basic solution.

Ans: It is given that ${{\text{K}}_{\text{a}}}$ for ${\text{ClC}}{{\text{H}}_2}{\text{COOH}}$ is $1.35\quad {10^{ - 3}}$

$ \Rightarrow {K_b} = c{\alpha ^2}$ $\therefore \alpha  = \sqrt {\dfrac{{{K_a}}}{c}} $ $ = \sqrt {\dfrac{{1.35 \times {{10}^{ - 3}}}}{{0.1}}} \quad (\therefore $ concentration of acid $ = 0.1{\text{M}})$

$\Rightarrow K_{b}-c \alpha^{2} $

$\therefore \alpha=\sqrt{\frac{K_{a}}{c}} $

$=\sqrt{\frac{1.35 \times 10^{-3}}{0.1}} \quad(\therefore \text { concentration of acid }=0.1 \mathbf{M}) $

$\alpha=\sqrt{1.35 \times 10^{-3}} $

$=0.116 $

$\therefore\left[H^{+}\right]=c \alpha=0.1 \times 0.116 $

$\Rightarrow p H=-\log \left[H^{+}\right]=1.94$

${\text{ClC}}{{\text{H}}_2}{\text{COONa}}$ is the salt of a weak acid i.e., ${\text{ClC}}{{\text{H}}_2}{\text{COOH}}$ and a strong base i.e., ${\text{NaOH}}$.

${{\text{ClC}}{{\text{H}}_2}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}_2}{\text{O}} \leftrightarrow {\text{ClC}}{{\text{H}}_2}{\text{COOH}} + {\text{O}}{{\text{H}}^ - }} $ 

${{K_b} = \dfrac{{\left[ {{\text{ClC}}{{\text{H}}_2}{\text{COOH}}} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{ClC}}{{\text{H}}_2}{\text{CO}}{{\text{O}}^ - }} \right]}}} $ 

${{K_b} = \dfrac{{{K_w}}}{{{K_a}}}} $ 

${{K_b} = \dfrac{{{{10}^{ - 14}}}}{{1.35 \times {{10}^{ - 3}}}}} $ 

${\quad  = 0.740 \times {{10}^{ - 11}}} $

Also, ${K_b} = \dfrac{{{x^2}}}{{0.1}}$ (where ${\text{x}}$ is the concentration of ${\text{O}}{{\text{H}}^ - }$and $\left. {{\text{ClC}}{{\text{H}}_2}{\text{COOH}}} \right)$

${0.740 \times {{10}^{ - 11}} = \dfrac{{{x^2}}}{{0.1}}} $ 

${0.074 \times {{10}^{ - 11}} = {x^2}} $ 

${ \Rightarrow {x^2} = 0.74 \times {{10}^{ - 12}}} $ 

${x = 0.86 \times {{10}^{ - 6}}} $ 

${\left[ {O{H^ - }} \right] = 0.86 \times {{10}^{ - 6}}} $ 

${\therefore \left[ {{H^ + }} \right] = \dfrac{{{K_w}}}{{0.86 \times {{10}^{ - 6}}}}} $ 

${ = \dfrac{{{{10}^{ - 14}}}}{{0.86 \times {{10}^{ - 6}}}}} $ 

${\left[ {{H^ + }} \right] = 1.162 \times {{10}^{ - 8}}} $ 

${pH =  - \log \left[ {{H^ + }} \right]} $

$ = 7.94$

Ans: Ionic product,

${{K_w} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]} $ 

${\operatorname{Let} \left[ {{H^ + }} \right] = x} $

${{\text{ Since }}\left[ {{H^ + }} \right] = \left[ {O{H^ - }} \right],{K_w} = {x^2}.} $ 

${ \Rightarrow {K_w}{\text{ at }}310\;{\text{K is }}2.7 \times {{10}^{ - 14}}} $ 

${\therefore 2.7 \times {{10}^{ - 14}} = {x^2}} $ 

${ \Rightarrow x = 1.64 \times {{10}^{ - 7}}} $ 

${ \Rightarrow \left[ {{H^ + }} \right] = 1.64 \times {{10}^{ - 7}}} $ 

${ \Rightarrow pH =  - \log \left[ {{H^ + }} \right]} $ 

${ =  - \log \left[ {1.64 \times {{10}^{ - 7}}} \right]} $ 

${ = 6.78} $

Hence, the ${\text{pH}}$ of neutral water is 6.78.

Ans: Moles of ${{\text{H}}_3}{{\text{O}}^ + } = \dfrac{{25 \times 0.1}}{{1000}} = .0025\;{\text{mol}}$

Moles of $O{H^ - } = \dfrac{{10 \times 0.2 \times 2}}{{1000}} = .0040\;{\text{mol}}$

Thus, excess of $O{H^ - } = .0015\;{\text{mol}}$

1. $10\;{\text{mL}}$ of $0.01{\text{M}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 10\;{\text{mL}}$ of $0.01{\text{MCa}}{({\text{OH}})_2}$

Ans: Moles of ${{\text{H}}_3}{{\text{O}}^ + } = \dfrac{{2 \times 10 \times 0.1}}{{1000}} = .0002\;{\text{mol}}$

Moles of $O{H^ - } = \dfrac{{2 \times 10 \times 0.1}}{{1000}} = .0002\;{\text{mol}}$

Since there is neither an excess of ${{\text{H}}_3}{{\text{O}}^ + }$or ${\text{O}}{{\text{H}}^ - }$

The solution is neutral. Hence, ${\text{pH}} = 7$.

1. $10\;{\text{mL}}$ of $0.1{\text{M}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 10\;{\text{mL}}$ of $0.1{\text{MKOH}}$

Ans: Moles of ${{\text{H}}_3}{O^ + } = \dfrac{{2 \times 10 \times 0.1}}{{1000}} = .002\;{\text{mol}}$

Moles of $O{H^ - } = \dfrac{{10 \times 0.1}}{{1000}} = 0.001\;{\text{mol}}$

Excess of ${{\text{H}}_3}{{\text{O}}^ + } = .001\;{\text{mol}}$

Thus, $\left[ {{H_3}{O^ + }} \right] = \dfrac{{.001}}{{20 \times {{10}^{ - 3}}}} = \dfrac{{{{10}^{ - 3}}}}{{20 \times {{10}^{ - 3}}}} = .05$ $\therefore pH =  - \log (0.05)$ $ = 1.30$

Ans:

1. Silver chromate:${\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4} \to 2{\text{A}}{{\text{g}}^ + } + {\text{CrO}}_4^{2 - }$

Then,

${K_{sp}} = {\left[ {A{g^ + }} \right]^2}\left[ {CrO_4^{2 - }} \right]$

Let the solubility of ${\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4}$ be ${\text{s}}$.

$ \Rightarrow \left[ {{\text{A}}{{\text{g}}^ + }} \right] = 2\;{\text{s and }}\left[ {{\text{CrO}}_4^{2 - }} \right] = s$

Then,

${{K_{sp}} = {{(2s)}^2} \cdot s = 4{s^3}} $ 

${ \Rightarrow 1.1 \times {{10}^{ - 12}} = 4{s^3}} $ 

${.275 \times {{10}^{ - 12}} = {s^3}} $ 

${s = 0.65 \times {{10}^{ - 4}}M} $

$ Molarity of Ag^{+} = 2s = 2x0.65x10^{ - 4} = 1.30x10^{ - 4}M $

Molarity of $CrO_4^{2 - } = {\text{s}} = 0.65 \times {10^{ - 4}}{\text{M}}$

1. Barium Chromate:

${\text{BaCr}}{{\text{O}}_4} \to {\text{B}}{{\text{a}}^{2 + }} + {\text{CrO}}_4^{2 - }$

Then, ${K_{sp}} = \left[ {B{a^{2 + }}} \right]\left[ {CrO_4^{2 - }} \right]$

Let the solubility of ${\text{BaCr}}{{\text{O}}_4}$ be ${\text{s}}$.

So, $\left[ {B{a^{2 + }}} \right] = {\text{s}}$ and $\left[ {{\text{CrO}}_4^{2 - }} \right] = {\text{s}} \Rightarrow {K_{sp}} = {s^2}$

$ \Rightarrow 1.2 \times {10^{ - 10}} = {s^2}$

$ \Rightarrow s = 1.09 \times {10^{ - 5}}M$

Molarity of $B{a^{2 + }} = $ Molarity of $CrO_4^{2 - } = s = 1.09 \times {10^{ - 5}}M$

1. Ferric Hydroxide:

${{\text{Fe}}{{({\text{OH}})}_3} \to {\text{F}}{{\text{e}}^2} + 3{\text{O}}{{\text{H}}^ - }} $ 

${{K_{sp}} = \left[ {{\text{F}}{{\text{e}}^{2 + }}} \right]{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}^3}} $

Let ${\text{s}}$ be the solubility of ${\text{Fe}}{({\text{OH}})_3}$

Molarity of $O{H^ - } = 3s = 4.17 \times {10^{10}}M$

1. Lead Chloride:

${{\text{PbC}}{{\text{l}}_2} \to {\text{P}}{{\text{b}}^{2 + }} + 2{\text{C}}{{\text{l}}^ - }} $ 

${{K_{SP}} = \left[ {{\text{P}}{{\text{b}}^{2 + }}} \right]{{\left[ {{\text{C}}{{\text{l}}^ - }} \right]}^2}} $

Let ${{\text{K}}_{{\text{SP}}}}$ be the solubility of ${\text{PbC}}{{\text{l}}_2}$.

${\left[ {P{B^{2 + }}} \right] = s{\text{ and }}\left[ {C{l^ - }} \right] = 2s} $ 

${{\text{ Thus, }}{K_{sp}} = s.{{(2s)}^2}} $ 

${\quad  = 4{s^3}} $ 

${ \Rightarrow 1.6 \times {{10}^{ - 5}} = 4{s^3}} $ 

${ \Rightarrow 0.4 \times {{10}^{ - 5}} = {s^3}} $ 

${4 \times {{10}^{ - 6}} = {s^3} \Rightarrow 1.58 \times {{10}^{ - 2}}M = S.1}$

Molarity of $P{B^{2 + }} = s = 1.58 \times {10^{ - 2}}M$

Molarity of chloride $ = 2s = 3.16 \times {10^{ - 2}}{\text{M}}$

1. Mercurous Iodide:

${{\text{H}}{{\text{g}}_2}{{\text{I}}_2} \to {\text{H}}{{\text{g}}^{2 + }} + 2{I^ - }} $ 

${{K_{sp}} = \left[ {{\text{Hg}}_2^{2 + }} \right]{{\left[ {{I^ - }} \right]}^2}} $

Let ${\text{s}}$ be the solubility of ${\text{H}}{{\text{g}}_2}{{\text{I}}_2}$.

$ \Rightarrow \left[ {{\text{Hg}}_2^{2 + }} \right] = s{\text{ and }}\left[ {{I^ - }} \right] = 2s$

Thus, ${K_{sp}} = s{(2s)^2} \Rightarrow {K_{sp}} = 4{s^3}$

$4.5 \times {10^{ - 29}} = 4{s^3}$

${1.125 \times {{10}^{ - 29}} = {s^3}} $ 

${ \Rightarrow s = 2.24 \times {{10}^{ - 10}}{\text{M}}} $

Molarity of ${\text{Hg}}_2^{2 + } = s = 2.24 \times {10^{ - 10}}{\text{M}}$

Molarity of ${I^ - } = 2s = 4.48 \times {10^{ - 10}}M$

Ans: Let ${\text{s}}$ be the solubility of ${\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4}$

Then, ${\text{A}}{{\text{g}}_2}{\text{Cr}}{{\text{O}}_4} \to {\text{A}}{{\text{g}}^{2 + }} + 2{\text{CrO}}_4^{2 - }$

${{K_{sp}} = {{(2s)}^2} \cdot s = 4{s^3}} $ 

${1.1 \times {{10}^{ - 12}} = 4{s^3}} $ 

${s = 6.5 \times {{10}^{ - 5}}M} $

Let ${{\text{s}}^\prime }$ be the solubility of ${\text{AgBr}}$.

${AgB{r_{(s)}} \leftrightarrow A{g^ + } + B{r^ - }} $ 

${{K_{sp}} = {s^{\prime 2}} = 5.0 \times {{10}^{ - 13}}} $ 

${\therefore {s^\prime } = 7.07 \times {{10}^{ - 7}}{\text{M}}} $

Therefore, the ratio of the molarities of their saturated solution is

$\dfrac{s}{{{s^\prime }}} = \dfrac{{6.5 \times {{10}^{ - 5}}M}}{{7.07 \times {{10}^{ - 7}}M}} = 91.9$

Ans: When 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., $0.001{\text{M}}$. Then,

${{\text{Nal}}{{\text{O}}_3} \to {\text{N}}{{\text{a}}^ + } + {\text{lO}}_3^ - } $ 

${0.001{\text{M}}}{0.001{\text{M}}} $ 

${{\text{Cu}}{{\left( {{\text{Cl}}{{\text{O}}_3}} \right)}_2} \to {\text{C}}{{\text{u}}^{2 + }}}{ + 2{\text{ClO}}_3^ - } $ 

${0.001{\text{M}}} {0.001M} $

Now, the solubility equilibrium for copper iodate can be written as:

$Cu(lO_{3})_{2} \to Cu_{(aq)}^{2 + } + 2lO_{3(aq)}^{ -} $

Ionic product of copper iodate:

${ = \left[ {C{u^{2 + }}} \right] \times {{\left[ {lO_3^ - } \right]}^2}} $ 

${ = (0.001){{(0.001)}^2}} $ 

${ = 1 \times {{10}^{ - 9}}} $

Since the ionic product $(1 \times \left. {{{10}^{ - 9}}} \right)$ is less than ${{\text{K}}_{{\text{sp}}}}\left( {7.4 \times {{10}^{ - 8}}} \right)$, precipitation will not occur.

Ans: ${{\text{ Since pH}} = 3.19{\text{, }}} $ 

${\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = 6.46 \times {{10}^{ - 4}}{\text{M}}} $ 

${{{\text{C}}_6}{{\text{H}}_5}{\text{COOH}} + {{\text{H}}_2}{\text{O}} \leftrightarrow {{\text{C}}_6}{{\text{H}}_5}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}_3}{\text{O}}} $ 

${{K_a} = \dfrac{{\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{CO}}{{\text{O}}^ - }} \right]\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]}}{{\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{COOH}}} \right]}}} $ 

${\dfrac{{\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{COOH}}} \right]}}{{\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{CO}}{{\text{O}}^ - }} \right]}} = \dfrac{{\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]}}{{{K_a}}} = \dfrac{{6.46 \times {{10}^{ - 4}}}}{{6.46 \times {{10}^{ - 5}}}} = 10} $

Let the solubility of ${{\text{C}}_6}{{\text{H}}_5}{\text{COOAg}}$ be ${\text{xmol}}/{\text{L}}$.

Then,

${\left[ {A{g^ + }} \right] = x} $ 

${\left[ {{C_6}{H_5}COOH} \right] + \left[ {{C_6}{H_5}CO{O^ - }} \right] = x} $ 

${10\left[ {{C_6}{H_5}CO{O^ - }} \right] + \left[ {{C_6}{H_5}CO{O^ - }} \right] = x} $ 

${\left[ {{C_6}{H_5}CO{O^ - }} \right] = \dfrac{x}{{11}}} $ 

${{K_{sp}}\left[ {A{g^ + }} \right]\left[ {{C_6}{H_5}CO{O^ - }} \right]} $ 

${2.5 \times {{10}^{ - 13}} = x\left( {\dfrac{x}{{11}}} \right)} $ 

${x = 1.66 \times {{10}^{ - 6}}\;{\text{mol}}/{\text{L}}} $

Thus, the solubility of silver benzoate in a ${\text{pH}}\quad 3.19$ solution is $1.66 \times {10^{ - 6}}\;{\text{mol}}/{\text{L}}$. Now, let the solubility of ${{\text{C}}_6}{{\text{H}}_5}{\text{COOAg}}$ be ${x^\prime }{\text{mol}}/{\text{L}}$.

Then, $\left[ {A{g^ + }} \right] = {x^\prime }M$ and $\left[ {{C_6}{H_5}CO{O^ - }} \right] = {x^\prime }M$

${{K_{sp}} = \left[ {{\text{A}}{{\text{g}}^ + }} \right]\left[ {{{\text{C}}_6}{{\text{H}}_5}{\text{CO}}{{\text{O}}^ - }} \right]} $ 

${{K_{sp}} = {{\left( {{x^\prime }} \right)}^2}} $ 

${{x^\prime } = \sqrt {{K_{sp}}}  = \sqrt {2.5 \times {{10}^{ - 13}}}  = 5 \times {{10}^{ - 7}}\;{\text{mol}}/{\text{L}}} $ 

${\therefore \dfrac{x}{{{x^\prime }}} = \dfrac{{1.66 \times {{10}^{ - 6}}}}{{5 \times {{10}^{ - 7}}}} = 3.32} $

Hence, ${{\text{C}}_6}{{\text{H}}_5}{\text{COOAg}}$ is approximately 3.317 times more soluble in a low ${\text{pH}}$ solution.