The dissociation constant of a substituted benzoic acid at $25^{\circ} \mathrm{C}$ is $1.0 \times 10^{-4}$.…

The dissociation constant of a substituted benzoic acid at $25^{\circ} \mathrm{C}$ is $1.0 \times 10^{-4}$. The $\mathrm{pH}$ of $0.01 \mathrm{M}$ solution of its sodium salt is

Solution

$ \text { The hydrolysis reaction of conjugate base of acid is } $ $ \begin{gathered} A^{-}(a q)+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HO}^{-}+\mathrm{HA} \\ K_h=\frac{K_w}{K_a}=\frac{10^{-14}}{10^{-4}}=10^{-10} \end{gathered} $ Since, degree of hydrolysis is negligible; $ \left[\mathrm{OH}^{-}\right]=\sqrt{K_h C}=10^{-6} \cdot p[\mathrm{OH}]=6 \text { and } \mathrm{pH}=14-6=8 $

Asked in: JEE Advanced 2009 (Paper 2)

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