Question

When ${CO}_2$ dissolves in water, the following equilibrium is established,
${CO}_2+2H_{2}O\rightleftharpoons H_3{O}^{+}+H{CO}_3^{-}$
for which the equilibrium constant is $3.8\times {10}^{-7}$ and $pH=6.0$.
The ratio of $\left[H{CO}_3^{-}\right]$ to $\left[{CO}_2\right]$ would be:

• A. $3.8\times {10}^{-13}$
• B. $3.8\times {10}^{-1}$
• C. $6.0$
• D. $13.4$

Answer 1

The correct option is B $3.8\times {10}^{-1}$

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

$pH=6$ means $\left[H_3{O}^{+}\right]={10}^{-6}M$.

So $3.8\times {10}^{-7}=\frac{\left[{10}^{-6}\right]\left[HC{O}_3^{-}\right]}{\left[C{O}_2\right]}$

So $\frac{\left[HC{O}_3^{-}\right]}{\left[C{O}_2\right]}=3.8\times {10}^{-1}$.