Question

Calculate the pH of \[1.0 \times {10^{ - 3}}M\] sodium phenolate \[{\text{NaO}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\] . \[{K_a}\] for \[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}\] is \[1.0 \times {10^{ - 10}}\] .

Answer 1

Hint: The hydrolysis of sodium phenolate in the aqueous medium gives phenol and hydroxide ions. The expression for the hydrolysis constant is \[{{\text{K}}_{\text{h}}} = \dfrac{{\left[ {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}} \right]{\text{ $\times$ }}\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}^ - }} \right]}}\]. The hydrolysis constant is related to the acid dissociation constant and the ionic product of water through the relation \[{{\text{K}}_{\text{h}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}}}}\] .

Complete answer:
Let C be the initial concentration of sodium phenolate and h be its degree of hydrolysis. Write the equilibrium for the hydrolysis of sodium phenolate in the aqueous solution.

\[{\text{Reaction at Equilibrium: }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{{\text{O}}^ - }{\text{ + }}{{\text{H}}_2}{\text{O }} \rightleftharpoons {\text{ }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH + O}}{{\text{H}}^ - } \\
  {\text{Initial concentration C }} \\
  {\text{Equilibrium concentration C}}\left( {1 - {\text{h}}} \right){\text{ $\;\;\;\;\;\;\;\;\;\;$ Ch $\;\;\;\;\;\;\;\;\;\;\;\;$ Ch }} \\
 \]
Write the expression for the hydrolysis constant.
\[{{\text{K}}_{\text{h}}} = \dfrac{{{\text{Ch $\times$ Ch}}}}{{{\text{C}}\left( {{\text{1}} - {\text{h}}} \right)}}\]
Assume the degree of hydrolysis to be much smaller than 1.
\[
  {{\text{K}}_{\text{h}}} = \dfrac{{{\text{Ch $\times$ Ch}}}}{{\text{C}}}{\text{ }}\left( {\because \left( {{\text{1}} - {\text{h}}} \right) \approx 1} \right) \\
  {{\text{K}}_{\text{h}}} = {\text{ C}}{{\text{h}}^{\text{2}}} \\
 \]
But \[{{\text{K}}_{\text{h}}} = \dfrac{{{{\text{K}}_{\text{w}}}}}{{{{\text{K}}_{\text{a}}}}} = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 10}}}} = {10^{ - 4}}\]
Substitute values in the expression for the hydrolysis constant.
\[
  {{\text{K}}_{\text{h}}} = {\text{C}}{{\text{h}}^2} \\
  {10^{ - 4}} = \left( {1.0 \times {{10}^{ - 3}}} \right){{\text{h}}^2} \\
  {{\text{h}}^2} = \dfrac{{{{10}^{ - 4}}}}{{\left( {1.0 \times {{10}^{ - 3}}} \right)}} = {10^{ - 1}} \\
 \]
Take square roots on both sides.
\[{\text{h = 0}}{\text{.316}}\]
Calculate hydroxide ion concentration
\[
  \left[ {{\text{O}}{{\text{H}}^ - }} \right] = {\text{C}}{{\text{h}}^2} \\
  \left[ {{\text{O}}{{\text{H}}^ - }} \right] = \left( {1.0 \times {{10}^{ - 3}}} \right) \times {\text{0}}{\text{.316}} \\
  \left[ {{\text{O}}{{\text{H}}^ - }} \right] = 3.16 \times {10^{ - 4}}{\text{ M}} \\
 \]

Calculate hydrogen ion concentration:
\[
  \left[ {{{\text{H}}^ + }} \right] = \dfrac{{{K_w}}}{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}} \\
  \left[ {{{\text{H}}^ + }} \right] = \dfrac{{1 \times {{10}^{ - 14}}}}{{3.16 \times {{10}^{ - 4}}}} \\
  \left[ {{{\text{H}}^ + }} \right] = 3.16 \times {10^{ - 11}}{\text{ M}} \\
 \]
Calculate pH of the solution:
\[
  {\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}^ + }} \right] \\
  {\text{pH}} = - {\log _{10}}(3.16 \times {10^{ - 11})}{\text{ M}} \\
  {\text{pH}} = 10.43 \\
 \]

Hence, the pH of the solution is 10.43.

Note: The ionic product of water is \[{K_w} = \left[ {{{\text{H}}^ + }} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right] = 1 \times {10^{ - 14}}\] .
pH and pOH are also related as \[{\text{pH + pOH = 14}}\] .
Once hydroxide ion is calculated, one can also calculate pOH and then calculate pH.
\[{\text{pOH}} = - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]\]
\[{\text{pH = 14}} - {\text{pOH}}\]