Question

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

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

Answer 1

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

The expression of $K_c$ j for the given reaction will be:

$K_c=\frac{[H{]}^{+}{[HCO}_3{]}^{-}}{\left[C{O}_2\right]}$

$\therefore 3.8\times {10}^{-7}=\frac{\left({10}^{-6}\right)\left[HC{O}_3^{-}\right]}{\left[C{O}_2\right]}$
$\frac{\left[HC{O}_3^{-}\right]}{\left[C{O}_2\right]}=0.38$

Hence, option A is correct.