When ${\mathrm{CO}}_2$ is bubbled in excess of water, the following equilibrium is established.

$$\begin{gathered} {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{HCO}}_3^{-} \\ {\mathrm{K}}_{\mathrm{c}}=3.8\times {10}^{-7},\mathrm{pH}=6. \end{gathered}$$

What would be the $\left[{\mathrm{HCO}}_3^{-}\right]/\left[{\mathrm{CO}}_2\right]$ ?

Given data: 

$$\begin{gathered} {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{HCO}}_3^{-} \\ {\mathrm{K}}_{\mathrm{c}}=3.8\times {10}^{-7},\mathrm{pH}=6. \end{gathered}$$

 Formula/Concept used: Chemical equilibrium

Detailed solution: Given,

$$\begin{gathered} {\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}\leftrightharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{HCO}}_3^{-} \\ {\mathrm{K}}_{\mathrm{C}}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{CO}}_2\right]} \\ \text{ As concentration of }{\mathrm{H}}^{+}={10}^{-\mathrm{pH}}={10}^{-6} \\ \Rightarrow 3.8\times {10}^{-7}=\left({10}^{-6}\right)\frac{{\left({\mathrm{HCO}}_3\right)}^{-}}{\left({\mathrm{CO}}_2\right)} \\ ⟹\frac{{\mathrm{HCO}}_3^{-}}{{\mathrm{CO}}_2}=0.38 \end{gathered}$$

Therefore, the correct option is (D).