Question

A student was given a 0.15 M solution of an unknown diprotic acid H2A and asked to determine the Ka1 and Ka2 values for the diprotic acid. The student titrated 46 mL of the 0.15 M H2A with 0.15 M KOH. After 23.0 mL of KOH was added, the pH of the resulting solution was 6.37. After 46 mL of KOH was added, the pH of the resulting solution was 8.34. What are the values of Ka1 and Ka2 for the diprotic acid?

Answer 1

Answer:

Ka1 = 4.27x10⁻⁷, Ka2 = 2.78x10⁻¹⁶

Explanation:

The number of moles that had reacted when the pH was 6.37 was:

nH₂A = 0.046 L * 0.15 mol/L = 6.9x10⁻³ mol

nKOH = 0.023 L * 0.15 mol/L = 3.45x10⁻³ mol

Thus, H₂A is in excess, and the base will not react with HA⁻ to form A⁻², so the first reaction is absolute, and after the reaction:

nH₂A = 6.9x10⁻³ - 3.45x10⁻³ = 3.45x10⁻³

nHA⁻ = 3.45x10⁻³

Because the volume will be the same, we can use the number of moles instead of the concentrations. So:

6.37 = pKa1 + log(3.45x10⁻³/3.45x10⁻³)

pKa1 = 6.37

Ka1 = $10^{-6.37}$

Ka1 = 4.27x10⁻⁷

After 46 mL of KOH was added:

nKOH = 0.046 * 0.15 = 6.9x10⁻³

So, all the H₂A reacts to form HA⁻, and the absolute equilibrium is:

HA⁻ ⇄ H⁺ + A⁻²

6.9x10⁻³ 0 0 Initial

-x +x x Reacts

6.9x10⁻³-x x x Equilibrium

pH = -log[H⁺]

8.34 = -log[H⁺]

[H⁺] = $10^{-8.34}$

[H⁺] = 4.57x10⁻⁹ M

The total volume is 46 mL + 46 mL = 92 mL = 0.092 L

x = 0.092 L * 4.57x10⁻⁹ = 4.20x10⁻¹⁰ mol

Thus, nHA⁻ = 6.9x10⁻³ - 4.20x10⁻¹⁰ ≅ 6.9x10⁻³

Then,

8.34 = pKa2 + log(4.20x10⁻¹⁰/6.9x10⁻³)

8.34 = pKa2 - 7.216

pKa2 = 15.556

Ka2 = $10^{-15.556}$

Ka2 = 2.78x10⁻¹⁶