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

\(\quad 0.034-x\quad \quad \quad \quad \quad \quad \quad x\quad \quad \quad  \quad \quad x\)

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

\(=\frac{x \times x}{0.034-x}\)

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

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

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 \(\left[H^{+}\right]=\left[H C O_{3}^{-}\right]=1.195 \times 10^{-4}\)

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