Question

When ${CO}_2$ dissolves in water, the following equilibrium is established
${CO}_2+2H_{2}O\rightleftharpoons H_3{O}^{+}+HC{O_3}^{-}$ ;
for which the equilibrium constant is $3.8\times {10}^{-6}$ and $pH=\mathrm{6.0.}$ What would be the ratio of concentration of bicarbonate ion to carbon dioxide?

• A. $38\times {10}^{-12}$
• B. $3.8$
• C. $6$
• D. $13.4$

Answer 1

The correct option is B $3.8$

Given,

$C{O}_2+2H_{2}O\leftrightharpoons H_3{O}^{+}+HC{O}_3^{-}$

$K_C=\frac{\left[H_3{O}^{+}\right]\left[HC{O}_3^{-}\right]}{\left[C{O}_2\right]}$

As concentration of $H^{+}={10}^{-pH}={10}^{-6}$

$⟹3.8\times {10}^{-6}=\left({10}^{-6}\right)\frac{(HC{O}_3{)}^{-}}{\left(C{O}_2\right)}$

$⟹\frac{HC{O}_3^{-}}{C{O}_2}=3.8$