First slide

Salt hydrolysis :

Question

In the equilibrium $A^{-} + H_{2} O ⇌ HA + {OH}^{-} (K_{a} = 1.0 \times 10^{- 5}) .$ The degree of hydrolysis of 0.001 M solution of the salt is

Moderate

A

$10^{- 3}$
B

$10^{- 4}$
C

$10^{- 5}$
D

$10^{- 6}$

Solution

$K_{a} = 1.0 \times 10^{- 5}$
$K_{h} = hydrolysis constant$
$K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{10^{- 5}} = 10^{- 9}$
Degree of hydrolysis (h) = $\sqrt{\frac{K_{h}}{C}}$
$= \sqrt{\frac{10^{- 9}}{0.001}} = \sqrt{10^{- 6}} = 10^{- 3}; H = 10^{- 3}$

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