The degree of hydrolysis in hydraulic equilibrium, $A^{-} (a q) + H_{2} O (l) ⇌ H A (a q) + O H^{-} (a q)$ at salt concentration $0.0001 M$ is: ( $K_{a} = 1.0 \times 10^{- 6})$ (Multiply your answer by $100$ )

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Solution

By using the formula,
$K_{h} = \frac{K_{w}}{K_{a}}$
Substituting the values, we get
$K_{h} = \frac{10^{- 14}}{1 \times 10^{- 6}} = 10^{- 8}$

Again, $K_{h} = C α^{2}$
$\Rightarrow α = \sqrt{\frac{K_{h}}{C}} = \sqrt{\frac{1 \times 10^{- 8}}{0.0001}}$
$\Rightarrow α = 1 \times 10^{- 2}$
So multiplying with $100$ we will get the solution to be $1$ .

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