When $CO_2$ dissolves in water,the following equilibrium is established:
$CO_2 + 2H_2O \rightleftharpoons H_3O^{+} + HCO_3^-$
For which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^-]$ to $[CO_2]$ would be:
$CO_2 + 2H_2O \rightleftharpoons H_3O^{+} + HCO_3^-$
For which the equilibrium constant is $3.8 \times 10^{-7}$ and $pH = 6.0$. The ratio of $[HCO_3^-]$ to $[CO_2]$ would be:
- A$3.8 \times 10^{-13}$
- B$3.8 \times 10^{-1}$
- C$6.0$
- D$13.4$
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