Calculate the pH of a 0.080 M carbonic acid solution,H₂CO₃₍_aq₎,that has the stepwise dissociation constants K_a1 = 4.3 × 10^-7 and K_a2 = 5.6 × 10^-11. A) 1.10 B) 3.73 C) 6.37 D) 10.25

Question

Quizplus · Accepted Answer

Stepwise dissociation reactions of carbonic acid:
H2CO3 ⇌ H+ + HCO3-
Ka1 = 4.3 × 10-7
HCO3- ⇌ H+ + CO32-
Ka2 = 5.6 × 10-11
To find the pH of a 0.080 M carbonic acid solution, we need to calculate the concentration of H+ ions in the solution at equilibrium.
Let's assume that x moles of H+ ions are generated after the first dissociation. Then the concentration of H+ ions is x M, the concentration of HCO3- is (0.080 - x) M, and the concentration of H2CO3 is (0.080 - x) M (since it's a weak acid, we can assume that the amount of H+ ions from autoionization of water is negligible compared to the amount generated from the dissociation of carbonic acid).
Using the ka1 expression: 
ka1 = [H+][HCO3-]/[H2CO3]
4.3 × 10-7 = x(0.080 - x) / (0.080 - x)
x2 = 4.3 × 10-7 × 0.080
x = 1.23 × 10-4 M
This is the concentration of H+ ions generated from the first dissociation.
Using the ka2 expression:
ka2 = [H+][CO32-]/[HCO3-]
5.6 × 10-11 = x(0.080 - x - 1.23 × 10-4) / (0.080 - 1.23 × 10-4)
x2 = 5.6 × 10-11 × (0.080 - 1.23 × 10-4)
x = 6.06 × 10-7 M
This is the concentration of H+ ions generated from the second dissociation.
The total concentration of H+ ions in solution is the sum of the concentrations from both dissociations: 
[H+] = 1.23 × 10-4 + 6.06 × 10-7 
pH = -log[H+] = 3.73
Therefore, the answer is B.

Calculate the PH of a 0