Question

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibrium equation CO2(g) CO2(aq) K=0.032 The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO2. For a CO2 partial pressure of 1.9x10-4 bar in the atmosphere, what is the pH of water in equilibrium with the atmosphere? (For carbonic acid Ka1 = 4.46x10-7 and Ka2 = 4.69x 10-11).

Answer 1

Explanation:

The reaction equation will be as follows.

           $CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)$

Calculate the amount of $CO_{2}$ dissolved as follows.

             $CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}$

It is given that $K_{CO_{2}}$ = 0.032 M/atm and $P_{CO_{2}}$ = $1.9 \times 10^{-4}$ atm.

Hence, $[CO_{2}]$ will be calculated as follows.

           $[CO_{2}]$ = $K_{CO_{2}} \times P_{CO_{2}}$          

                           = $0.032 M/atm \times 1.9 \times 10^{-4}atm$

                           = $0.0608 \times 10^{-4}$

or,                        = $0.608 \times 10^{-5}$

It is given that $K_{a} = 4.46 \times 10^{-7}$

As,      $K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}$

          $4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}$  

               $[H^{+}]^{2}$ = $2.71 \times 10^{-12}$

                      $[H^{+}]$ = $1.64 \times 10^{-6}$

Since, we know that pH = $-log [H^{+}]$

So,                      pH = $-log (1.64 \times 10^{-6})$

                                 = 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.