Question

The interaction between carbon dioxide and water results in formation of carbonic acid, H2CO3, which lowers the pH of water through the reactions

H2CO3(aq)+H2O(l) <---> HCO3-(aq) + H3O+(aq) Ka1= 4.3*10^-7

HCO3-(aq) +H2O(l) <---> CO3 2- (aq) + H3O+ (aq) Ka2=5.6*10^-11

This dissocociation causes "normal" rainwater to have a pH of approximately 5.7. However, H3O+(aq) can also come from other sources, particularly from acids resulting from interactions of water with industrial pollutants.

Scientists studying a particular lake found that its pH was 4.8, and the total carbonate concentration (that of H2CO3, HCO-, and CO3 2-) was 4.5*10^-3M. Determine the concentration of each individual carbonate species.

Answer 1

The pH of the solution is 4.8; we define pH as pH = -log [H⁺]; therefore, [H⁺] = antilog(-pH) = 10^-4.8 = 1.5849*10^-5 M.

Find out the fraction of species as

f(H₂CO₃) = [H⁺]²/D where D = [H⁺]² + K_a1*[H⁺] + K_a1*K_a2 = (1.5849*10^-5)² + (4.3*10^-7)*(1.5849*10^-5) + (4.3*10^-7)*(5.6*10^-11) = 2.5800*10^-10.

Therefore, f(H₂CO₃) = (1.5849*10^-5)²/(2.5800*10^-10) = 0.9736.

f(HCO₃^-) = K_a1*[H⁺]/D = (4.3*10^-7)*(1.5849*10^-5)/(2.5800*10^-10) = 0.0264.

f(CO₃^2-) = K_a1*K_a2/D = (4.3*10^-7)*(5.6*10^-11)/(2.5800*10^-10) = 9.3333*10^-7.

[H₂CO₃] = f(H₂CO₃)*(total carbonate concentration) = 0.9736*(4.5*10^-3 M) = 4.3812*10^-3 M (ans).

[HCO₃^-] = f(HCO₃^-)*(total carbonate concentration) = 0.0264*(4.5*10^-3 M) = 1.188*10^-4 M (ans).

[CO₃^2-] = f(CO₃^2-)*(total carbonate concentration) = 9.3333*10^-7*(4.5*10^-3 M) = 4.199985*10^-9 M ≈ 4.20*10^-9 M (ans).