$H_{2} C O_{3}(a q)+H_{2} O(l) \rightleftharpoons H C O_{3}(a q)+H_{3} O_{x}^{+}(a q)$

$\quad 0.034-x\quad \quad \quad \quad \quad \quad \quad x\quad \quad \quad  \quad \quad x$

$K_{1}=\frac{\left.\left[H C O_{3}^{-}\right] | H_{3} O^{+}\right]}{\left|H_{2} C O_{3}\right|}$

$=\frac{x \times x}{0.034-x}$

$\Rightarrow 4.2 \times 10^{-7}=\frac{x^{2}}{0.034}$

$\Rightarrow x=1.195 \times 10^{-4}$

As $H_{2} C O_{3}$ is a weak acid so the concentration of

$H_{2} C O_{3}$ will remain 0.034 as $0.034>>x$

$x=\left[H^{+}\right]=\left[H C O_{3}\right]$

$=1.195 \times 10^{-4}$

Now, $H C O_{3}(a q)+H_{2} O(l) \rightleftharpoons C O_{3_{y}}^{2-}(a q)+H_{3} O_{y}^{+}(a q)$

As $H C O_{3}$ is again a weak acid (weaker than $H_{2} C O_{3}$ )

with $x>>y$

$K_{2}=\frac{\left[c o_{3}^{2}\right]\left|H_{3} O^{+}\right|}{\left[H C o_{3}\right]}$

$=\frac{y \times(x+y)}{(x-y)}$

Note : $\left[H_{3} O^{+}\right]=H^{+}$ from first step $(x)$ and from second step $(y)=(x+y)$

$[\text { As } x>>y \text { so } x+y \simeq x \text { and } x-y \simeq x]$

So, $K_{2} \simeq \frac{y \times x}{x}=y$

$\Rightarrow K_{2}=4.8 \times 10^{-11}$

$y=y=\left[C O_{3}^{2}\right]$

So the concentration of $\left[H^{+}\right]=\left[H C O_{3}^{-}\right]=$ concentrations obtained from the first step. As the dissociation will be very low in second step so there will be no change in these concentrations. Thus the final concentrations are $\left[H^{+}\right]=\left[H C O_{3}^{-}\right]=1.195 \times 10^{-4}$

$\left[\mathrm{CO}_{3}^{2-}\right]=4.8 \times 10^{-11}$