Question

$\left\lbrack H^{+} \right\rbrack = \sqrt{K_{a}.C} = \sqrt{2 \times 10^{- 12} \times 0.02} = 2 \times 10^{- 7}\text{ M}$

As [H⁺] is very small, contribution of H⁺ from water must be considered.

HA $\rightleftharpoons$ H⁺ + A^–

Equilibrium (0.02 – x) M (x + y) M x M

H₂O $\rightleftharpoons$ H⁺ + OH^–

Equilibrium (x + y) M y M

Now, $K_{a} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack A^{-} \right\rbrack}{\lbrack HA\rbrack}$

$\Rightarrow 2 \times 10^{- 12} = \frac{(x + y).x}{(0.02 - x)} \approx \frac{(x + y).x}{0.02} $$${\text{Or, }4 \times 10^{- 14} = (x + y).x(1) }{\text{and }K_{w} = \left\lbrack H^{+} \right\rbrack\left\lbrack OH^{-} \right\rbrack \Rightarrow 10^{- 14} = (x + y).y(2) }{\text{From }(1) + (2),(x + y) = \sqrt{5 \times 10^{- 14}}\text{M = }\left\lbrack H^{+} \right\rbrack }{\therefore P^{H} = - {\log\left( 5 \times 10^{- 14} \right)}^{1/2} = 6.65}$$