Answer:
Ka1 = 2.00x10⁻⁷, Ka2 = 5.00x10⁻¹⁰
Explanation:
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⁺] =
[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
