Carbon dioxide dissolves in water to form carbonic acid, which is primarily diss

Question

Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO_2. Dissolved CO_2 satisfies the equilibrium equation CO_2(g) equivalent to CO_2(aq) K = 0.032 The acid dissociation constants listed in most standard reference texts for carbonic acid actually apply to dissolved CO_2. For a CO_2 partial pressure of 7.8 times 10^-4 bar in the atmosphere, what is the pH of water in equilibrium with the atmosphere? (For carbonic acid K _a1 = 4.46 times 10^-7 and K_a2 = 4.69 times 10^-11).

Explanation / Answer

Given the partial pressure of CO2(g) in atmosphere, P(CO2) = 7.8x10-4 bar

CO2(g) <------- > CO2(aq) or H2CO3; K = 0.032

K = 0.032 = [H2CO3] / P(CO2) =[H2CO3] / 7.8x10-4 bar

=> [H2CO3] = 0.032 x 7.8x10-4 = 2.50x10-5 M

For the first dissociation of H2CO3

------------------ H2CO3 -----> HCO3-(aq) + H+(aq) ; Ka1 = 4.46x10-7

Init.Conc(M):2.50x10-5 , 0 0

eqm.conc(M):(2.50x10-5 -y), y, y

Ka1 = 4.46x10-7 = [H+(aq)]x[HCO3-(aq)] / [H2CO3] = y2 / (2.50x10-5 -y)

=>  y2 + 4.46x10-7y - 1.115x10-11 = 0

=> y = 3.12x10-6 M

Hence [H+(aq)] = y =  3.12x10-6 M

Since the second dissociation constant of H2CO3 is very small, the amount of H+(aq) produced in second dissociation can be neglected. Hence

[H+(aq)] = y =  3.12x10-6 M

=> pH = - log [H+(aq)] = - log (3.12x10-6 M) = 5.51 (answer)

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