Chapter 6: Equilibrium

Ans: Given the, ${\text{pH}} = 3.76$ 

It is known as:

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

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

$[{{\text{H}}^ + }] = {\text{antilog}}( - {\text{pH}})$

$[{{\text{H}}^ + }] = {\text{antilog}}( - 3.76)$

$[{{\text{H}}^ + }] = 1.74 \times {10^{ - 4}}{\text{M}}$

Hence, the concentration of hydrogen ion in the given sample of vinegar is $1.74 \times {10^{ - 4}}{\text{ M}}$. 

Ans: It is known as:

${K_b} = \dfrac{{{K_w}}}{{{K_a}}}$ 

Given, ${{\text{K}}_{\text{a}}}$ of ${\text{HF}}$ is $6.8 \times {10^{ - 4}}$

Hence, ${{\text{K}}_{\text{b}}}$ of its conjugate base ${{\text{F}}^ - }$:

${{\text{K}}_{\text{b}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}}}}{\text{}}$ 

${{\text{K}}_{\text{b}}} = \dfrac{{{{10}^{ - 14}}}}{{6.8 \times {{10}^{ - 4}}}}$

${{\text{K}}_{\text{b}}} = 1.5 \times {10^{ - 11}}$

Given ${{\text{K}}_{\text{a}}}$ of ${\text{HCOOH}}$ is $1.8$$\times$$10^{-4}$ 

Hence, ${{\text{K}}_{\text{b}}}$ of its

conjugate base ${\text{HCO}}{{\text{O}}^ - }$:

${{\text{K}}_{\text{b}}}$

=$\dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}}}}{{\text{K}}_{\text{b}}}{\text{}}$

$ = \dfrac{{{{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 4}}}}$

${{\text{K}}_{\text{b}}} = 5.6 \times {10^{ - 11}}$

Given ${{\text{K}}_{\text{a}}}$ of ${\text{HCN}}$ is $4.8$$\times $$10^{-9}$ 

Hence, ${{\text{K}}_{\text{b}}}$ of its

conjugate base ${\text{C}}{{\text{N}}^ - }$:

${{\text{K}}_{\text{b}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}}}}{\text{}}$

${{\text{K}}_{\text{b}}} = \dfrac{{{{10}^{ - 14}}}}{{4.8 \times {{10}^{ - 9}}}}$

${{\text{K}}_{\text{b}}} = 2.08 \times {10^{ - 6}}$

Answer: of phenol? What will be its degree of ionization if the Ans:is also 0.01M in sodium phenolate?

Ans: According to the reaction of ionization of phenol:

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}^ - } + {{\text{H}}_{\text{3}}}{{\text{O}}^ + }$ 

The initial concentration of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}$ is 0.05 M. 

At equilibrium, the concentration of $C_{6}H_{5}O^{ - }$ and $H_{3}O^{+ }$ will be x, then the concentration of $C_{6}H_{5}OH$ will be $(0.05 - x)$.

$K_{a} = \dfrac{{[C_{6}H_5O^{- }][H_{3}O^{+ }]}}{{[C_{6}H_{5}OH]}} $ 

${{K_a} = \dfrac{{x \times x}}{{0.05 - 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 x = \sqrt {1 \times {{10}^{ - 10}} \times 0.05} } $ 

${\quad  = \sqrt {5 \times {{10}^{ - 12}}} } $ 

${\quad  = 2.2 \times {{10}^{ - 6}}{\text{M}} = \left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right]} $ 

${{\text{Since}}\left[ {{{\text{H}}_3}{{\text{O}}^ + }} \right] = \left[ {{{\text{C}}_6}{{\text{H}}_5}{{\text{O}}^ - }} \right]} $ 

${\left[ {{{\text{C}}_6}{{\text{H}}_5}{{\text{O}}^ - }} \right] = 2.2 \times {{10}^{ - 6}}{\text{M}}}$ 

Now, let α be the degree of ionization of phenol in the presence of 0.01 M ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{ONa}}$

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}^ - } + {{\text{H}}_{\text{3}}}{{\text{O}}^ + }$ 

The concentrations are as follows:

$[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}] = 0.05 - 0.05\alpha {\text{}}$

$0.05{\text{ M}}[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}^ - }] = 0.01 + 0.05\alpha {\text{}}$

$0.01{\text{ M}}[{{\text{H}}_3}{{\text{O}}^ + }] = 0.05\alpha$

$K_{a} = \dfrac{{[C_{6}H_5O^{- }][H_{3}O^{+ }]}}{{[C_{6}H_{5}OH]}} $ 

${{\text{K}}_{\text{a}}} = \dfrac{{(0.01)(0.05\alpha )}}{{0.05}}$

$1.0 \times {10^{ - 10}} = 0.01\alpha $

$\alpha {\text{}} = 1 \times {10^{ - 8}}$

Answer: is 0.1 M in ${\text{HCl}}$  also? If the second dissociation constant of ${{\text{H}}_{\text{2}}}{\text{S}}$  is $1.2 \times {10^{ - 13}}$ , calculate the concentration of ${{\text{S}}^{2 - }}$ under both conditions.

Ans: To calculate the concentration of ${\text{H}}{{\text{S}}^ - }$ ion:

Case I (in the absence of ${\text{HCl}}$ ):

Let the concentration of ${\text{H}}{{\text{S}}^ - }$be x M:

${{\text{H}}_{\text{2}}}{\text{S}} \rightleftharpoons {{\text{H}}^{\text{ + }}} + {\text{H}}{{\text{S}}^ - }$ 

The final concentration of ${{\text{H}}_{\text{2}}}{\text{S}}$ will be $(0.1 - x)$ 

Then,

${{K_{{a_i}}} = \dfrac{{\left[ {{H^ + }} \right]\left[ {H{S^ - }} \right]}}{{\left[ {{H_2}S} \right]}}} $ 

${9.1 \times {{10}^{ - 8}} = \dfrac{{(x)(x)}}{{0.1 - x}}} $ 

${\left( {9.1 \times {{10}^{ - 8}}} \right)(0.1 - x) = {x^2}} $

Taking,  $0.1 - xM;0.1{\text{M}}$ we have $\left( {9.1 \times {{10}^{ - 8}}} \right)(0.1) = {x^2}$ 

${9.1 \times {{10}^{ - 9}} = {x^2}} $ 

$x{ = \sqrt {9.1 \times {{10}^{ - 9}}} } $ 

${ = 9.54 \times {{10}^{ - 5}}{\text{M}}} $ 

$\Rightarrow {\left[ {H{S^ - }} \right] = 9.54 \times {{10}^{ - 5}}{\text{M}}} $

Case II (in the presence of ${\text{HCl}}$):

In the presence of $0.1{\text{ M}}$ of ${\text{HCl}}$, let ${\text{[H}}{{\text{S}}^ - }]$ be y M.

 ${{\text{H}}_{\text{2}}}{\text{S}} \rightleftharpoons {{\text{H}}^{\text{ + }}} + {\text{H}}{{\text{S}}^ - }$

The final concentration of ${{\text{H}}_{\text{2}}}{\text{S}}$ will be $(0.1 - y)$ 

Now, 

${{\text{K}}_{{\text{a1}}}} =$

$\dfrac{{[HS^{-}][H^{+}]}}{{[H_2S]}}$

${K_{a1}} = \dfrac{{y(0.1 +y)}}{{(0.1 - y)}} $

$9.1 \times {10^{ - 8}} =$

$\dfrac{{y \times 0.1}}{{0.1}}  $

$y = 9.1 \times {10^{- 8}}  $

The concentration of $[{\text{H}}{{\text{S}}^ - }] = 9.1 \times {10^{ - 8}}$ 

To calculate the concentration of $[{{\text{S}}^{2 - }}]$ 

Case I (in the absence of 0.1 M of ${\text{HCl}}$):

 ${\text{H}}{{\text{S}}^ - } \rightleftharpoons {{\text{H}}^ + } + {{\text{S}}^{2 - }}$ 

$[{\text{H}}{{\text{S}}^ - }] = 9.54 \times {10^{ - 5}}{\text{ M}}$ 

Let $[{{\text{S}}^{2 - }}]$ be ${\text{X}}$ .

Also, $[{{\text{H}}^ + }] = 9.54 \times {10^{ - 5}}{\text{ M}}$ 

${{K_{{a_2}}} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{S^{2 - }}} \right]}}{{\left[ {H{S^ - }} \right]}}} $ 

${{K_{{a_2}}} = \dfrac{{\left( {9.54 \times {{10}^{ - 5}}} \right)(X)}}{{9.54 \times {{10}^{ - 5}}}}} $ 

${1.2 \times {{10}^{ - 13}} = X = \left[ {{S^{2 - }}} \right]} $

Case II (in the presence of 0.1 M of ${\text{HCl}}$):

The concentration of ${{\text{S}}^{2 - }}$ be ${{\text{X}}^{\text{'}}}{\text{ M}}$ 

$[{\text{H}}{{\text{S}}^ - }] = 9.1 \times {10^{ - 8}}{\text{ M}}$ 

$[{{\text{H}}^ + }] = 0.1{\text{ M}}$ 

Then, 

${K_{{a_2}}} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{S^2}} \right]}}{{\left[ {H{S^ - }} \right]}}$

${1.2 \times {{10}^{ - 13}} = \dfrac{{(0.1)\left( {X'} \right)}}{{9.1 \times {{10}^{ - 8}}}}} $ 

${10.92 \times {{10}^{ - 21}} = 0.1X'} $

${\dfrac{{10.92 \times {{10}^{ - 21}}}}{{0.1}} = X'} $ 

${X' = \dfrac{{1.092 \times {{10}^{ - 20}}}}{{0.1}}} $ 

${ = 1.092 \times {{10}^{ - 19}}{\text{M}}} $ 

${ \Rightarrow {K_{{a_1}}} = 1.74 \times {{10}^{ - 5}}} $

Answer: and its pH.

Ans: For the reaction:

${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} + {{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}_{\text{3}}}{{\text{O}}^ + }$ 

The final concentration of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }$ and ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}$ is $0.05\alpha $ . Then the final concentration of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ will be $(0.05 - 0.05\alpha )$ .

${{K_a} = \dfrac{{(.05\alpha )(.05\alpha )}}{{(.05 - 0.05\alpha )}}} $ 

${ = \dfrac{{(.05\alpha )(0.05\alpha )}}{{.05(1 - \alpha )}}} $ 

${ = \dfrac{{.05{\alpha ^2}}}{{1 - \alpha }}} $

${1.74 \times {{10}^{ - 5}} = \dfrac{{.05{\alpha ^2}}}{{1 - \alpha }}} $ 

${1.74 \times {{10}^{ - 5}} - 1.74 \times {{10}^{ - 5}}\alpha  = 0.05{\alpha ^2}} $ 

${0.05{\alpha ^2} + 1.74 \times {{10}^{ - 5}}\alpha  - 1.74 \times {{10}^{ - 5}}} $ 

${D = {b^2} - 4ac} $

${ = {{\left( {1.74 \times {{10}^{ - 5}}} \right)}^2} - 4(.05)\left( {1.74 \times {{10}^{ - 5}}} \right)} $ 

${ = 3.02 \times {{10}^{ - 25}} + .348 \times {{10}^{ - 5}}} $

${\alpha  = \sqrt {\dfrac{{{K_a}}}{c}} } $ 

${\alpha  = \sqrt {\dfrac{{1.74 \times {{10}^{ - 5}}}}{{.05}}} } $ 

${ = \sqrt {\dfrac{{34.8 \times {{10}^{ - 5}} \times 10}}{{10}}} } $ 

${ = \sqrt {3.48 \times {{10}^{ - 6}}} } $

${\left[ {{\text{C}}{{\text{H}}_3}{\text{CO}}{{\text{O}}^ - }} \right]}{ = 0.05 \times 1.86 \times {{10}^{ - 3}}} $ 

${ = \dfrac{{0.93 \times {{10}^{ - 3}}}}{{1000}}} $ 

${ = .000093} $

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

${\quad  =  - \log \left( {.093 \times {{10}^{ - 2}}} \right)} $ 

${\therefore pH = 3.03} $

 Hence, the concentration of acetate ion in the Ans:is 0.00093 M and its pH is 3.03.

Ans: Let the organic compound be ${\text{HA}}$.

${\text{HA}} \rightleftharpoons {{\text{H}}^ + } + {{\text{A}}^ - }$ 

Concentration of  ${\text{HA}}$ is 0.01 M and its pH is 4.15.

${ - \log \left[ {{H^ + }} \right] = 4.15} $ 

${\left[ {{H^ + }} \right] = 7.08 \times {{10}^{ - 5}}} $ 

${{\text{Now,}}{K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{[HA]}}} $ 

${\left[ {{H^ + }} \right] = \left[ {{A^ - }} \right] = 7.08 \times {{10}^{ - 5}}} $ 

${[HA] = 0.01} $

Now, 

${{\text{K}}_{\text{a}}} = \dfrac{{[{{\text{H}}^ + }][{{\text{A}}^ - }]}}{{[{\text{HA}}]}}[{{\text{H}}^ + }] = [{{\text{A}}^ - }] = 7.08 \times {10^{ - 5}}[{\text{HA}}] = 0.01{\text{}}$

Then,

${{K_a} = \dfrac{{\left( {7.08 \times {{10}^{ - 5}}} \right)\left( {7.08 \times {{10}^{ - 5}}} \right)}}{{0.01}}} $ 

${{K_a} = 5.01 \times {{10}^{ - 7}}} $ 

${p{K_a} =  - \log {K_a}} $ 

${\quad  =  - \log {K_a}} $ 

${\quad  = \left( {5.01 \times {{10}^{ - 7}}} \right)} $ 

${p{K_a} = 6.3001}$ 

Ans: ${{\text{H}}_{\text{2}}}{\text{O}} + {\text{HCl}} \rightleftharpoons {{\text{H}}_{\text{3}}}{{\text{O}}^ + } + {\text{C}}{{\text{l}}^ - }$ 

Since ${\text{HCl}}$ is completely ionized.

$[{{\text{H}}_{\text{3}}}{{\text{O}}^ + }] = [{\text{HCl}}] \Rightarrow [{{\text{H}}_{\text{3}}}{{\text{O}}^ + }] = 0.003{\text{}}$ 

${\text{pH}} = {\text{}} - \log [{{\text{H}}_{\text{3}}}{{\text{O}}^ + }]$

${\text{pH}} = {\text{}} - \log (0.003)$

${\text{pH}} = 2.52$

Hence, the pH of the solution is

2.52.

1. 0.005 M ${\text{NaOH}}$  

Ans: ${\text{NaOH(aq)}} \rightleftharpoons Na^{+} (aq) + OH^{-}(aq)$ 

$[OH^{-}] = [NaOH] \Rightarrow [OH^{-}] =0.005$ 

${\text{pOH}} = {\text{}} - \log [{\text{O}}{{\text{H}}^ - }]$

${\text{pOH}} = {\text{}} - \log (0.005)$

${\text{pOH}} = 2.30$

${\text{pH}} = 14 - {\text{pOH}}$

${\text{pH}} = 14 - 2.30$

${\text{pH}} = 11.70$

Hence, the pH of the solution is

11.70.

1.  0.002 M ${\text{HBr}}$ 

Ans: ${\text{HBr}} +H_{2}O \rightleftharpoons H_{3}O^{+} + Br^{-}$ 

$[H_{3}O^{+}] = [HBr] \Rightarrow [H_{3}O^{+}] =0.002$ 

${\text{pH}} = {\text{}} - \log [{{\text{H}}_{\text{3}}}{{\text{O}}^ + }]$

${\text{pH}} = {\text{}} - \log (0.002)$

${\text{pH}} = 2.69$

Hence, the pH of the solution is

2.69.

1. 0.002 M ${\text{KOH}}$ 

Ans: $KOH (aq) \rightleftharpoons K^{+}(aq) + OH^{ -}(aq)$ 

$[OH^{-}] = [KOH]$

$\Rightarrow [OH^{-}] = 0.002$

${\text{pOH}} = {\text{}} - \log [{\text{O}}{{\text{H}}^ - }]$

${\text{pOH}} = {\text{}} - \log (0.002)$

${\text{pOH}} = 2.69$

${\text{pH}} = 14 - {\text{pOH}}$

${\text{pH}} = 14 - 2.69$

${\text{pH}} = 11.31$

Hence, the pH of the solution is 11.31.

Ans: For 2g of ${\text{TlOH}}$ dissolved in water to give 2 L of solution:

${\left[ {TIO{H_{(aq)}}} \right] = \dfrac{2}{2}g/L} $ 

${ = \dfrac{2}{2} \times \dfrac{1}{{221}}M} $ 

${ = \dfrac{1}{{221}}M} $

${\text{TlOH (aq}}) \to {\text{T}}{{\text{l}}^ + }({\text{aq}}) + {\text{O}}{{\text{H}}^ - }{\text{(aq}})$ 

${TIO{H_{(aq)}} \to Tl_{(aq)}^ +  + OH_{(aq)}^ - } $ 

${\left[ {OH_{(aq)}^ - } \right] = \left[ {TIO{H_{(aq)}}} \right] = \dfrac{1}{{221}}M} $ 

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

${{{10}^{ - 14}} = \left[ {{H^ + }} \right]\left( {\dfrac{1}{{221}}} \right)} $ 

${221 \times {{10}^{ - 14}} = \left[ {{H^ + }} \right]} $

${ \Rightarrow pH =  - \log \left[ {{H^ + }} \right] =  - \log \left( {221 \times {{10}^{ - 14}}} \right)} $ 

${ =  - \log \left( {2.21 \times {{10}^{ - 12}}} \right)} $ 

${ = 11.65} $

1. 0.3 g of ${\text{Ca(OH}}{{\text{)}}_{\text{2}}}$  dissolved in water to give 500 mL of solution. 

Ans: Molecular weight of ${\text{Ca(OH}}{{\text{)}}_{\text{2}}}$ is 74g.

So, 0.3g ${\text{Ca(OH}}{{\text{)}}_{\text{2}}}$ = $\dfrac{{0.3}}{{74}}{\text{ mole}}$ 

${\text{Molar Concentration  =  }}\dfrac{{{\text{Moles in Solution}}}}{{{\text{Volume in Solution}}}}$ 

Molar concentration of ${\text{Ca(OH}}{{\text{)}}_{\text{2}}}$ = $\dfrac{{\dfrac{{0.3}}{{74}}{\text{ mole}}}}{{0.5{\text{ L}}}} = 8.11 \times {10^{ - 3}}{\text{ M}}$ 

${\text{Ca(OH}}{)_2} \to {\text{C}}{{\text{a}}^{ + 2}} + 2{\text{O}}{{\text{H}}^ - }$ 

$[{\text{O}}{{\text{H}}^ - }] = 2[{\text{Ca(OH}}{)_2}]{\text{}}$ 

$[{\text{O}}{{\text{H}}^ - }] = 2 \times (8.11 \times {10^{ - 1}}){\text{ M}}$

$[{\text{O}}{{\text{H}}^ - }] = 16.22 \times {10^{ - 1}}{\text{M}}$

${\text{pOH}} = {\text{}} - \log (16.22 \times {10^{ - 1}})$

${\text{pOH}} = 3 - 1.2101$

${\text{pOH}} = 1.79$

${\text{pH}} = 14 - 1.79$

${\text{pH}} = 12.21$

1. 0.3 g of ${\text{NaOH}}$ 

dissolved in water to give 200 mL of solution.

Ans: 1 mole of  ${\text{NaOH}}$ = 40g

Since, 40 gm of ${\text{NaOH}}$ = 1 mole

of ${\text{NaOH}}$

So, 0.3 gm of $NaOH= $\frac{0.3}{40}$

Molar Concentration=$\dfrac{{{\text{Moles in Solution}}}}{{{\text{Volume in Solution}}}}$

Molar concentration = $\dfrac{{\dfrac{{0.3}}{{40}} Moles}}{{0.2L}} = 3.75 \times 10^{-1} M$ 

$[OH^{-}] = 3.75 \times {10^{ - 2}}{\text{ M}}$ 

${\text{pOH}} = {\text{}} - \log (3.75 \times {10^{ - 2}})$

${\text{pOH}} = 1.43$

${\text{pH}} = 14 - 1.43$

${\text{pH}} = 12.57$

1. 1mL of 13.6 M ${\text{HCl}}$  is diluted with water to give 1 litre of solution.

Ans: For this question:

$M_{1}V_{1}= M_{2}V_{2}$ 

$13.6 \times {\text{1 mL}} = {{\text{M}}_{\text{2}}} \times 100{\text{0 mL}}$

$13.6 \times {10^{ - 3}} = {{\text{M}}_{\text{2}}} \times {\text{1 L}}$

${{\text{M}}_2} = 1.36 \times {10^{ - 2}}$

$[{{\text{H}}^ + }] = 1.36 \times {10^{ - 2}}$

${\text{pH}} = {\text{}} - \log (1.36 \times {10^{ - 2}})$

${\text{pH}} = {\text{}} - 0.1335 + 2$

${\text{pH}} = 1.87$

Ans: Degree of ionization, $\alpha {\text{}} = 0.132$ 

Concentration, c = 0.1 M

Thus, the concentration of

${H_3}{O^+}$=$c.\alpha$ 

$[{H_3}{O^+}]$ = $0.1\times 0.132$ 

$[{{\text{H}}_{\text{3}}}{{\text{O}}^ + }] = 0.0132$

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

${\text{pH}} = {\text{}} - \log (0.0132)$

${\text{pH}} = 1.88$

Now,

${{\text{K}}_{\text{a}}} = C{\alpha ^2}{\text{}}$

${{\text{K}}_{\text{a}}} = 0.1 \times {(0.132)^2}$

${{\text{K}}_{\text{a}}} = 0.0017{\text{p}}$

${{\text{K}}_{\text{a}}} = 2.75$

Ans: ${{{\text{K}}_{\text{b}}} = 4.27\quad {{10}^{ - 10}}} $ 

${{\text{c}} = 0.001{\text{M}}} $ 

${{\text{pH}} = ?\quad } $ 

${\alpha  = ?} $ 

${{k_a} = C{\alpha ^2}} $ 

${4.27 \times {{10}^{ - 10}} = 0.001 \times {\alpha ^2}} $ 

${4270 \times {{10}^{ - 10}} = {\alpha ^2}} $ 

${65.34 \times {{10}^{ - 5}} = \alpha  = 6.53 \times {{10}^{ - 4}}} $

Then, concentration of anion:

$[{\text{anion}}] = c.\alpha {\text{}} = 0.01 \times 65.34 \times {10^{ - 5}}{\text{}}$

$[{\text{anion}}] = 0.65 \times {10^{ - 5}}$

${pOH =  - \log \left( {.065 \times {{10}^{ - 5}}} \right)} $ 

${ = 6.187} $ 

${pH = 7.813} $

${{K_a} \times {K_b} = {K_w}} $ 

${\therefore 4.27 \times {{10}^{ - 10}} \times {K_a} = {K_w}} $ 

${{K_a} = \dfrac{{{{10}^{ - 14}}}}{{4.27 \times {{10}^{ - 10}}}}} $ 

${ = 2.34 \times {{10}^{ - 5}}} $

Thus, the ionization constant of the conjugate acid of aniline is $2.34 \times {10^{ - 5}}$ .