Question

Calculate the pH and concentrations of the following species in a 0.025 M carbonic acid solution. (Given Ka1 = 4.3x10-7 and Ka2 = 5.6x10-11)

a) H2CO3

b) HCO3-

c) CO32-

d) H+

e) OH

Answer 1

There are two equilibrium expressions that need to be considered. Since we are dealing with a diprotic acid, we have two acid dissociation reactions requiring our attention. First, one should notice that in general, K_a2 is smaller than K_a1. This is usually the case. In fact, K_a2 is usually 1000 times smaller than K_a1. Thus, it is possible to obtain a satisfactory estimate of the pH of a polyprotic acid solution by considering only K_a1

For H₂CO₃, we can take the following approach:

H₂CO₃(aq) + H₂O(l) <====> H₃O⁺(aq) + HCO₃^-(aq)

K_a1 = 4.3 x 10^-7 = [H₃O⁺][HCO₃^-]/[H₂CO₃]

From the weak acid handout, we know that

[H⁺]² = K_ac

so, [H⁺]² = (4.3 x 10^-7)(0.025)

[H⁺] = 1.04 x 10^{-4 .}

pH = - log[H+] = - log(1.04 x 10^-4) = 3.98.

For the second deprotonation:

HCO₃^-(aq) + H₂O(l) <====> H₃O⁺(aq) + CO₃^2-(aq)

K_a2 = 5.6 x 10^-11 = [H₃O⁺][CO₃^2-]/[HCO₃^-]

From the first deprotonation we know that [H⁺]=[HCO₃^-]=1.04 x 10^-4 (We can assume that these will also be the equilibrium values after the second deprotonation since K_a1 is at least 1000 times larger than K_a2)

so,

[CO₃^2-] = 5.6 x 10^-11

Final answer:

[H₃O⁺] = [H⁺]= 1.04 x 10^-4 M

[HCO₃^-] = 1.04 x 10^-4 M

[CO₃^2-] = 5.6 x 10^-11 M

[H₂CO₃] = 0.025 M - 1.04 x 10^-4 M = 0.025 M

pH = 3.98