Question

A sample of carbonic acid (0.125 L, 0.154 M, pK_a1 = 6.35, pK_a2 = 10.33) was titrated with 1.53 M NaOH. Calculate the pH at the first and at the second midpoints.

Answer 1

I believe when you say the first and second mid-points, you mean the 1^st and 2^nd half equivalence points. Kindly clarify.

I shall work out the problem assuming you are referring to the 1^st and 2^nd half equivalence points.

Write out the dissociation reactions:

H₂CO₃ (aq) <====> H⁺ (aq) + HCO₃^- (aq); pK_a1 = 6.35

At the half equivalence point (1^st. mid-point), we must have

[H₂CO₃] = [HCO₃^-]

This can be easily realized because at the mid-point to the 1^st equivalence point, you must exactly half of the initial concentration of H₂CO₃ left. The remaining half concentration of H₂CO₃ is converted to HCO₃^- as per the equation:

H₂CO₃ (aq) + NaOH (aq) -----> Na⁺HCO₃^- (aq) + H₂O (l)

Therefore, [H₂CO₃]₁ = [HCO₃^-]₁ where the 1 (in subscripts) denote the 1^st. half equivalence point. Use Henderson-Hasslebach equation to write

pH = pK_a1 + log [HCO₃^-]/[H₂CO₃] = 6.35 + log (1) (since [HCO₃^-] = [H₂CO₃]).

Therefore, pH = 6.35 (ans).

After the 1^st equivalence point, the system contains only HCO₃^- which undergoes dissociation as

HCO₃^- (aq) <====> H⁺ (aq) + CO₃^2- (aq)

The reaction with NaOH is

HCO₃^- (aq) + OH^- (aq) -----> CO₃^2- (aq) + H₂O (l)

As before at half way to the 2^nd equivalence point, [HCO₃^-] = [CO₃^2-]. This is because exactly half of HCO₃^- is neutralized to CO₃^2- and the remaining half stays. Use the Henderson-Hasslebach equation to find the pH as

pH = pK_a2 + log [CO₃^2-]/[HCO₃^-] = 10.33 + log (1) = 10.33 (ans).

The answer reflects a general trend. At half way to the equivalence point, the pH of the buffer solution is equal to the pK_a.