• # question_answer 26) When $C{{O}_{2}}$ dissolves in water, the following equilibrium is established $C{{O}_{2}}+2{{H}_{2}}O{{H}_{3}}{{O}^{+}}+HC{{O}_{3}}^{-};$for which the equilibrium constant is $3.8\times {{10}^{-6}}$ and $pH=6.0$ What would be the ratio of concentration of bicarbonate ion to carbon  dioxide? A)  $3.8\times {{10}^{-12}}~$B)  $3.8$                    C)   6                         D)   $13.4$

$K=\frac{[{{H}_{3}}{{O}^{+}}]\,[HC{{O}_{3}}^{-}]}{[C{{O}_{2}}]\,{{[{{H}_{2}}O]}^{2}}}$As $pH=6.0{{[{{H}_{3}}O]}^{+}}={{10}^{-6}}$ $K=\frac{[{{H}_{3}}{{O}^{+}}]\,[HC{{O}_{3}}^{-}]}{[C{{O}_{2}}]\,{{[{{H}_{2}}O]}^{2}}}$ (${{H}_{2}}O$ is in excess, therefore its cone. remains constant) $\frac{\left[ HC{{O}_{3}}^{-} \right]}{\left[ C{{O}_{2}} \right]}=\frac{K}{\left[ {{H}_{3}}{{O}^{+}} \right]}=\frac{3.8\times {{10}^{-6}}}{{{10}^{-6}}}=3.8$