Question

6. A student titrated 50.0 mL of the 0.10 M unknown diprotic H2A with 0. 10 M NAOH. After 25.0 mL of NaOH was added, the pH of the resulting solution was 6.70. After 50.0 mL of NaOH was added, the pH of the solution was 8.00. What are the values of Ka1 and Ka2?

Answer 1

Write down the step-wise and complete acid-base reactions between H₂A and NaOH as below.

H₂A (aq) + NaOH (aq) -------> NaHA (aq) + H₂O (l)

NaHA (aq) + NaOH (aq) ---------> Na₂A (aq) + H₂O (l)

The overall acid-base equation is

H₂A (aq) + 2 NaOH (aq) -------> Na₂A (aq) + 2 H₂O (l)

Write down the ionic equations for the ionization reactions as

H₂A (aq) + OH^- (aq) --------> HA^- (aq) + H₂O (l)     pK_a1

HA^- (aq) + OH^- (aq) --------> A^2- (aq) + H₂O (l)       pK_a2

Millimols H₂A = (50.0 mL)*(0.10 M) = 5.0 mmol.

Millimols NaOH corresponding to 25.0 mL = (25.0 mL)*(0.10 M) = 2.5 mmol.

Millimols NaOH corresponding to 50.0 mL = (50.0 mL)*(0.10 M) = 5.0 mmol.

When 25.0 mL of NaOH is added to 50.0 mL of H₂A, half of H₂A is neutralized to HA^- and half of the original H₂A remains in the solution.

The pH of the solution is given by the Henderson-Hasslebach equation as

pH = pK_a1 + log [HA^-]/[H₂A]

= pK_a1 ([HA^-] = [H₂A] since the original H₂A is exactly half neutralized).

Since pH = 6.70, hence,

pK_a1 = 6.70

=====> K_a1 = antilog (-6.70)

=====> K_a1 = 1.995*10^-7 ≈ 2.00*10^-7 (ans).

When 50.0 mL of 0.10 M NaOH is added to 50.0 mL of 0.10 M H₂A, H₂A is neutralized to HA^-. HA^- is amphoteric in nature and thus, the pH of the solution is given as

pH = ½*(pK_a1 + pK_a2)

======> 8.00 = ½*(6.70 + pK_a2)

======> 16.00 = 6.70 + pK_a2

======> pK_a2 = 9.30

======> K_a2 = antilog (-9.30)

======> K_a2 = 5.012*10^-10 ≈ 5.01*10^-10 (ans).