Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\begin{array}{cccccccc}{\text{H}}_2{\text{CO}}_3& \rightleftharpoons & {\text{H}}^{+}& +& {\text{HCO}}_3^{-}\text{;}& {\text{K}}_1& =& \text{4.2}\times 10^{-7}\\ {\text{HCO}}_3^{-}& \rightleftharpoons & {\text{H}}^{+}& +& {\text{CO}}_3^{2-}\text{;}& {\text{K}}_2& =& \text{4.8}\times 10^{-11}\end{array}$

$\because {\text{K}}_1\gg {\text{K}}_2$, so${\text{H}}_2{\text{CO}}_3$ionises more than${\text{HCO}}_3^{-}$and hence, contribution of${\text{H}}^{+}$is mostly due to ionization of carbonic acid. Thus, the concentrations of${\text{H}}^{+}$and${\text{HCO}}_3^{-}$are approximately equal.