Question

The interaction between carbon dioxide and water results in formation of carbonic acid, H2CO3, which lowers...

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.

Homework Answers

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 + Ka1*[H+] + Ka1*Ka2 = (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-) = Ka1*[H+]/D = (4.3*10-7)*(1.5849*10-5)/(2.5800*10-10) = 0.0264.

f(CO32-) = Ka1*Ka2/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).

[CO32-] = f(CO32-)*(total carbonate concentration) = 9.3333*10-7*(4.5*10-3 M) = 4.199985*10-9 M ≈ 4.20*10-9 M (ans).

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