A student was given a 50.0 mL of 0.10 M solution of an unknown diprotic acid, H2A, which was titrated with 0.10 M NaOH. After a total of 25.0 mL of NaOH was added, the resulting solution has a pH of 6.70.
After a total of 50.0 mL of NaOH was added, the pH of the solution was 8.00.

Determine the values of Ka1 and Ka2 for the diprotic acid.

A diprotic acid is a substance that can release 2 H⁺ when in aqueous solution. Because it is a weak acid, the ionization will be reversible. So, the acid has two equilibrium reactions, each one with its equilibrium constant:

H₂A ⇄ H⁺ + HA⁻ Ka1 = ([HA⁻]*[H⁺]/[H₂A])

HA⁻ ⇄ H⁺ + A⁻ Ka2 = ([A⁻]*[H⁺]/[HA⁻])

First, the number of moles of H₂A was:

n = 0.10 mol/L *0.05L = 0.005 mol

And then was added NaOH:

n = 0.1 mol/L * 0.025 L = 0.0025 mol

So, all the NaOH will reacts, then, the number of moles will be:

H₂A = 0.005 - 0.0025 = 0.0025 mol

HA⁻ = 0.0025 mol (the stoichiometry is 1:1:1)

The concentrations of H₂A and HA⁻ will be the same, so Ka1 = [H⁺], and

pH = -log[H⁺]

6.70 = -log[H⁺]

[H⁺] = [tex]10^{-6.70}[/tex]

[H⁺] = 2.00x10⁻⁷ M

Ka1 = 2.00x10⁻⁷

After the addition of 50.0 mL of NaOH the second equilibrium must dominate the reaction. For the expression of Ka1:

Ka1 = ([HA⁻]*[H⁺]/[H₂A])

[HA⁻]*[H⁺] = Ka1*[H₂A]

[HA⁻] = (Ka1*[H₂A])/[H⁺]

Ka2 = ([A⁻]*[H⁺]/[HA⁻])

Ka2 = ([A⁻]*[H⁺]²)/(Ka1*[H₂A])

By the stoichiometry, [H₂A] = [A⁻], so:

Ka2 = [H⁺]²/Ka1

pH = -log[H⁺]

8.00 = -log[H⁺]

[H+] = 10⁻⁸

Ka2 = (10⁻⁸)²/(2.00x10⁻⁷)

Ka2 = 5.00x10⁻¹⁰