How to calculate molar concentration from equilibrium equations with no initial concentration given?
I will continue with the data of $[\ce{H2CO3}] = \pu{10^{-4.97} M}.$
Now, as the $K_2$ of $\ce{H2CO3}$ is very small as compared to its $K_1,$ we can assume that all the $\ce{H+}$ will come from the first dissociation of $\ce{H2CO3}.$
$$
\begin{array}{lccc}
&amp; \ce{&amp;H2CO3 &amp;&lt;=&gt; &amp;H+(aq) &amp;+ &amp;HCO3-(aq)} \\ 
&amp;\text{Initial} &amp; 10^{-4.97} &amp;&amp; 0 &amp;&amp; 0 \\
&amp;\text{Final} &amp; 10^{-4.97}(1-\alpha) &amp;&amp; 10^{-4.97}\alpha &amp;&amp; 10^{-4.97}\alpha
\end{array}
$$
$$K_1 = 10^{-6.35}$$
$$10^{-6.35} = \frac{(10^{-4.97}\alpha)^2}{10^{-4.97}(1-\alpha)}$$
As $K_1 \ll 1,$ $(1 - \alpha)\approx 1:$
$$\alpha = \sqrt{\frac{10^{-6.35}}{10^{-4.97}}}$$
$$
\begin{align}
[\ce{HCO3-}] &amp;= \ce{[H+}] = 10^{-4.97}\alpha \\
  &amp;= \sqrt{10^{-6.35}\times 10^{-4.97}} \\
  &amp;= 10^{-5.66} \approx \pu{2.19E-6 M}
\end{align} 
$$
Note, that I have not considered self ionization of water because the $\mathrm{pH}$ of this solution will be slightly less than $6$ $(\mathrm{pH} = 5.66).$ Which means even after accounting for self ionization of water, the results won't change much. This is a good approximate of what  we needed. And if you want to calculate $[\ce{CO3^2-}]$ than just calculate it in a similar way to which you calculated $[\ce{H2CO3}],$ because now you have $K_2,$ $[\ce{HCO3-}]$ and $[\ce{H+}].$