The dissociation of carbonic acid in blood is given by: $$ \mathrm{H}_{2} \mathrm{CO}_{3} \rightleftharpoons \mathrm{H}^{+} + \mathrm{HCO}_{3}^{-} $$ Using the Henderson-Hasselbalch equation for a buffer system: $$ \mathrm{pH} = \mathrm{pK}_{\mathrm{a}} + \log \frac{[\mathrm{HCO}_{3}^{-}]}{[\mathrm{H}_{2} \mathrm{CO}_{3}]} $$ Given, $\mathrm{pH} = 7.4$ and $\mathrm{K}_{\mathrm{a}} = 4.3 \times 10^{-7}$. First, calculate $\mathrm{pK}_{\mathrm{a}}$: $$ \begin{aligned} \mathrm{pK}_{\mathrm{a}} &= -\log(\mathrm{K}_{\mathrm{a}}) \\ &= -\log(4.3 \times 10^{-7}) \\ &\approx 6.367 \end{aligned} $$ Now, substitute the values into the Henderson-Hasselbalch equation: $$ \begin{aligned} 7.4 &= 6.367 + \log \frac{[\mathrm{HCO}_{3}^{-}]}{[\mathrm{H}_{2} \mathrm{CO}_{3}]} \\ \log \frac{[\mathrm{HCO}_{3}^{-}]}{[\mathrm{H}_{2} \mathrm{CO}_{3}]} &= 7.4 - 6.367 \\ \log \frac{[\mathrm{HCO}_{3}^{-}]}{[\mathrm{H}_{2} \mathrm{CO}_{3}]} &= 1.033 \\ \frac{[\mathrm{HCO}_{3}^{-}]}{[\mathrm{H}_{2} \mathrm{CO}_{3}]} &= 10^{1.033} \\ \frac{[\mathrm{HCO}_{3}^{-}]}{[\mathrm{H}_{2} \mathrm{CO}_{3}]} &\approx 10.79 \end{aligned} $$ Thus, the ratio of $[\mathrm{HCO}_{3}^{-}]$ to $[\mathrm{H}_{2} \mathrm{CO}_{3}]$ is $10.79:1$.