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Question: Calculate the pH of \[1.0 \times {10^{ - 3}}M\] sodium phenolate \[{\text{NaO}}{{\text{C}}_{\text{6}...

Calculate the pH of 1.0×10−3M1.0 \times {10^{ - 3}}M sodium phenolate NaOC6H5{\text{NaO}}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}} . Ka{K_a} for C6H5OH{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} is 1.0×10−101.0 \times {10^{ - 10}} .

Explanation

Solution

The hydrolysis of sodium phenolate in the aqueous medium gives phenol and hydroxide ions. The expression for the hydrolysis constant is Kh=[C6H5OH] × [OH−][C6H5O−]{{\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 Kh=KwKa{{\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.

Initial concentration C  Equilibrium concentration C(1−h)                      Ch                          Ch   {\text{Initial concentration C }} \\\ {\text{Equilibrium concentration C}}\left( {1 - {\text{h}}} \right){\text{ $\;\;\;\;\;\;\;\;\;\;$ Ch $\;\;\;\;\;\;\;\;\;\;\;\;$ Ch }} \\\

Write the expression for the hydrolysis constant.
Kh=Ch × ChC(1−h){{\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.

Kh=Ch × ChC (∵(1−h)≈1) Kh= Ch2  {{\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 Kh=KwKa=10−1410−10=10−4{{\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.

Kh=Ch2 10−4=(1.0×10−3)h2 h2=10−4(1.0×10−3)=10−1  {{\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.
h = 0.316{\text{h = 0}}{\text{.316}}
Calculate hydroxide ion concentration

[OH−]=Ch2 [OH−]=(1.0×10−3)×0.316 [OH−]=3.16×10−4 M  \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:

[H+]=Kw[OH−] [H+]=1×10−143.16×10−4 [H+]=3.16×10−11 M  \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:

pH=−log⁡10[H+] pH=−log⁡10(3.16×10−11) M pH=10.43  {\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 Kw=[H+][OH−]=1×10−14{K_w} = \left[ {{{\text{H}}^ + }} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right] = 1 \times {10^{ - 14}} .
pH and pOH are also related as pH + pOH = 14{\text{pH + pOH = 14}} .
Once hydroxide ion is calculated, one can also calculate pOH and then calculate pH.
pOH=−log⁡10[OH−]{\text{pOH}} = - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]
pH = 14−pOH{\text{pH = 14}} - {\text{pOH}}