Question

When $C{O}_2$ dissolves in water, the following equilibrium is establishes, $C{O}_2+2H_{2}O\rightleftharpoons H_3{O}^{+}+HC{O}_3^{-}$, for which the equilibrium constant is $3.8\times {10}^{-7}$ and $pH=6.0$. The ratio of $\left[HC{O}_3^{-}\right]/\left[C{O}_2\right]$ is

• A. $3.8\times {10}^{-18}$
• B. $3.8$
• C. $0.38$
• D. $13.8$

Answer 1

The correct option is C $0.38$

$\begin{array}{cc}K=\frac{\left[H_{3}O\right]\left[HC{O}_3^{-}\right]}{\left[C{O}_2\right]{\left[H_{2}O\right]}^2}& K=\frac{\left[H_{3}O\right]\left[HC{O}_3^{-}\right]}{\left[C{O}_2\right]}\\ P^n=6;\left[H^{+}\right]={10}^{-6}& \\ 3.8\times {10}^{-7}={10}^{-6}\frac{\left[HC{O}_3^{-}\right]}{\left[C{O}_2\right]}=1\frac{\left[HC{O}_3^{-}\right]}{\left[C{O}_2\right]}=\frac{3.8\times {10}^{-7}}{{10}^{-6}}\Rightarrow 3.8\times {10}^{-1}\Rightarrow 0.38& \end{array}$

Option C is correct.