Astudent 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.

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

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being