Question

The dissociation constant of a substituted benzoic acid at ${25}^{o}C$ is $1\times {10}^{-4}$. The $pH$ of a $0.01M$ solution of its sodium salt is:

Answer 1

$BH\to B^{-}+H^{+}$

Let $x$ be the dissociated $BH.Ka=1\ast {10}^{-}4$

$p^H$ of $0.01M{C}_6{H}_{5}COONa$

$C-6H_{5}CO{O}^{-}+H_{2}O\to C_6{H}_{5}COOH+O{H}^{-}$

$K_h=\frac{Kw}{Ka}=\frac{0.01h^2}{1-h}$

$\frac{{10}^{-}14}{{10}^{-}4}={10}^{-}2h^2$

$\left[OH\right]=0.01h$

$=0.01\ast {10}^{-}4={10}^{-}6$

$\to \left[H^{+}\right]={10}^{-}8$

$p^H=8$