Polyprotic Acid Example Chemistry Problem

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Updated February 12, 2018

A polyprotic acid is an acid that can donate more than one hydrogen atom (proton) in an aqueous solution. To find the pH of this type of acid, it's necessary to know the dissociation constants for each hydrogen atom. This is an example of how to work a polyprotic acid chemistry problem.

Polyprotic Acid Chemistry Problem

Determine the pH of a 0.10 M solution of H₂SO₄.

Given: K_a2 = 1.3 x 10^-2

Solution

H₂SO₄ has two H⁺ (protons), so it is a diprotic acid that undergoes two sequential ionizations in water:

First ionization: H₂SO₄(aq) → H⁺(aq) + HSO₄^-(aq)

Second ionization: HSO₄^-(aq) ⇔ H⁺(aq) + SO₄^2-(aq)

Note that sulfuric acid is a strong acid, so its first dissociation approaches 100%. This is why the reaction is written using → rather than ⇔. The HSO₄^-(aq) in the second ionization is a weak acid, so the H⁺ is in equilibrium with its conjugate base.

K_a2 = [H⁺][SO₄^2-]/[HSO₄^-]

K_a2 = 1.3 x 10^-2

K_a2 = (0.10 + x)(x)/(0.10 - x)

Since K_a2 is relatively large, it's necessary to use the quadratic formula to solve for x:

x² + 0.11x - 0.0013 = 0

x = 1.1 x 10^-2 M

The sum of the first and second ionizations gives the total [H⁺] at equilibrium.

0.10 + 0.011 = 0.11 M

pH = -log[H⁺] = 0.96

Learn More

Introduction to Polyprotic Acids

Strength of Acids and Bases

Concentration of Chemical Species

First Ionization	H₂SO₄(aq)	H⁺(aq)	HSO₄^-(aq)
Initial	0.10 M	0.00 M	0.00 M
Change	-0.10 M	+0.10 M	+0.10 M
Final	0.00 M	0.10 M	0.10 M


Second Ionization	HSO₄^2-(aq)	H⁺(aq)	SO₄^2-(aq)
Initial	0.10 M	0.10 M	0.00 M
Change	-x M	+x M	+x M
At Equilibrium	(0.10 - x) M	(0.10 + x) M	x M