When $$C{O_2}$$  dissolves in water, the following equilibrium is established $$C{O_2} + 2{H_2}O \rightleftharpoons {H_3}{O^ + } + HCO_3^ - $$       for which the equilibrium constant is $$3.8 \times {10^{ - 7}}$$   and $$pH = 6.0.$$   The ratio of $$\left[ {HCO_3^ - } \right]$$  to $$\left[ {C{O_2}} \right]$$  would be

Solution :
$$\eqalign{ & C{O_2} + 2{H_2}O \rightleftharpoons {H_3}{O^ + } + HCO_3^ - ; \cr & {K_c} = \frac{{\left[ {{H_3}{O^ + }} \right]\left[ {HCO_3^ - } \right]}}{{\left[ {C{O_2}} \right]}}; \cr & \frac{{\left[ {HCO_3^ - } \right]}}{{\left[ {C{O_2}} \right]}} \cr & = \frac{{3.8 \times {{10}^{ - 7}}}}{{{{10}^{ - 6}}}} \cr & = 3.8 \times {10^{ - 1}} \cr} $$