Question

A 0.20 M sodium chlorobenzoate (NaC₇H₄ClO₂) solution has a pH of 8.65. Calculate the pH of a 0.20 M chlorobenzoic acid (HC₇H₄ClO₂) solution.

Short answer

pH of a 0.20 M chlorobenzoic acid (HC₇H₄ClO₂) solution is 2.35.

Step-by-step solution

Sodium Chlorobenzoate Dissociation

Species present are Naand C₇H₄ClO₂^-. Na⁺ does not undergo hydrolysis.

H⁺ ion can be calculated as:

$\begin{array}{l}{\text{[H}}^{+}\text{]}={10}^{\text{-pH}}\\ {\text{[H}}^{+}\text{]}={10}^{-8.65}\\ {\text{H}}^{+}\text{]}=2.24\times {10}^{-9}\\ {\text{[OH}}^{-}\text{]}=\frac{{\text{K}}_{\mathrm{w}}}{{\text{[H}}^{+}\text{]}}=\frac{1\times {10}^{-14}}{2.24\times {10}^{-9}}\\ {\text{[OH}}^{-}\text{]}=4.46\times {10}^{-6}\end{array}$

Reaction involved

$\begin{array}{l}{\text{C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}{{}^{\text{-}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{\hspace{0.17em}HC}}_{\text{7}}{\text{H}}_{\text{4}}{{\text{ClO}}_{\text{2}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ 0.200\\ 0.2-\text{x\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}}4.46\times {10}^{-6}4.46\times {10}^{-6}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{b}}=\frac{\text{[x][x]}}{\text{[0.2-x]}}\\ {\text{K}}_{\text{b}}=\frac{{(4.46\times {10}^{-6})}^2}{\text{[0.2]}}\\ {\text{K}}_{\text{b}}=1\times {10}^{-\text{'}10}\end{array}$

Calculation of pH

First of all lets calculate Ka

${\text{K}}_{\text{a}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}=\frac{1\times {10}^{-14}}{1\times {10}^{-10}}=1\times {10}^{-4}$

$\begin{array}{l}{\text{HC}}_{\text{7}}{\text{H}}_{\text{4}}{{\text{ClO}}_{\text{2}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{\hspace{0.17em}C}}_{\text{7}}{\text{H}}_{\text{4}}{\text{ClO}}_{\text{2}}{{}^{-}}_{\text{(aq)}}{\text{+\hspace{0.17em}H}}_{\text{3}}{{\text{O}}^{\text{+}}}_{\text{(aq)}}\\ 0.200\\ 0.2-\text{x\hspace{1em}\hspace{0.17em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}x\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{a}}=\frac{\text{[x][x]}}{\text{[0.2-x]}}=\frac{{\text{x}}^2}{\text{[0.2]}}=1\times {10}^{-\text{'}4}(0.2-\text{x}\approx 0.2)\\ \text{x}=4.48\times {10}^{-\text{'}3}\end{array}$

$\begin{array}{l}{\text{pH=-log[H}}^{\text{+}}\text{]}\\ {\text{pH=-log(4.48×10}}^{\text{-'3}}\text{)}\\ \text{pH=2.35}\end{array}$