Question

$CO_{3}^{2 -} + H_{2}O \rightleftharpoons HCO_{3}^{-} + OH^{-}; $$$(0.5 - x)\text{ M}(x - y)\text{ M}(x + y)\text{ M}$$

$K_{h_{t}} = \frac{Kw}{K_{a_{2}}} = 2 \times 10^{- 4}$

$\underset{(x - y)\text{ M}}{HCO_{3}^{-}} + H_{2}O \rightleftharpoons \underset{y\text{ M}}{H_{2}CO_{3}} + \underset{(x + y)\text{ M}}{OH^{-}}; $$$K_{h_{2}} = \frac{Kw}{K_{a_{1}}} = 2.5 \times 10^{- 9}$$

Now, $2 \times 10^{- 4} = \frac{(x - y).(x + y)}{(0.5 - x)} \approx \frac{x.x}{0.5} \Rightarrow x = 10^{- 2}$

and $2.5 \times 10^{- 9} = \frac{y.(x + y)}{(x - y)} \approx \frac{y.x}{x} \Rightarrow y = 2.5 \times 10^{- 9}$

Now, $h = \frac{x}{0.5} = 0.02$

$P^{OH} = - \log(x + y) - \log\left( 10^{- 2} \right) = 2.0 \Rightarrow P^{H} = 12.0 $$$\text{and }\left\lbrack H_{2}CO_{3} \right\rbrack = y = 2.5 \times 10^{- 9}\text{ M}$$