$K_{a}$ for butyric acid is $2 \times 10^{- 5}$ . Calculate pH and hydroxyl ion concentration of $0.2 M$ aqueous solution of sodium butyrate.

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Solution

Butyrate ion hydrolysis in solution into butyric acid and ${O H}^{-}$ ions. Let $x$ mole of butyrate ion be hydrolysed.
$C_{3} H_{7} C O O^{-} + H_{2} O ⇌ C_{3} H_{7} C O O H + {O H}^{-}$
$(0.2 - x)$ $x$ $x$
$K_{h} = \frac{x^{2}}{(0.2 - x)} = \frac{x^{2}}{0.2}$ ( $x$ being small is neglected as compared to $0.2$ )
We know,
$K_{h} = \frac{K_{w}}{K_{a}}$
so, $\frac{K_{w}}{K_{a}} = \frac{x^{2}}{0.2}$
$\frac{10^{- 14}}{2 \times 10^{- 5}} = \frac{x^{2}}{0.2}$
or $x = 10^{- 5} m o l L^{- 1}$
$x = 10^{- 5} m o l L^{- 1}$
$[{O H}^{-}] = 10^{- 5} M$
$[H^{+}] = \frac{10^{- 14}}{10^{- 5}} = 10^{- 9} M$
$p H = - log [H^{+}] = log [10^{- 9}] = 9$

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