Question

The degree of hydrolysis in hydraulic equilibrium, $A^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons HA\left(aq\right)+O{H}^{-}\left(aq\right)$ at salt concentration $0.0001\text{M}$ is: ($K_a=1.0\times {10}^{-6})$ (Multiply your answer by $100$)

Answer 1

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