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

Carbonic acid (H2CO3) has two dissociation constants, Ka1 and Ka2. The first dissociation goes as follows:
H2CO3 ⇌ H+ + HCO3-
Ka1 = 4.3 × 10^-7
The second dissociation goes as follows:
HCO3- ⇌ H+ + CO32-
Ka2 = 5.6 × 10^-11
The pH of the solution can be calculated using the following formula:
pH = pKa + log ([A-]/[HA])
For the first dissociation,
pKa1 = -log (Ka1) = -log (4.3 × 10^-7) = 6.37
[HA] = [H2CO3] = 0.080 M
[A-] = [HCO3-] = x
pH1 = 6.37 + log (x/0.080)
For the second dissociation,
pKa2 = -log (Ka2) = -log (5.6 × 10^-11) = 10.25
[HA] = [HCO3-] = x
[A-] = [CO32-] = x
pH2 = 10.25 + log (x/x)
Since the second dissociation constant is much smaller than the first, we can assume that x is approximately equal to the concentration of HCO3-. Hence, pH = pH1 = 3.73. Therefore, the best choice is B) 3.73.

Calculate the PH of a 0