Polyprotic Acid Example Chemistry Problem

How To Work a Polyprotic Acid Problem

Sulfuric acid is an example of a common polyprotic acid.
Sulfuric acid is an example of a common polyprotic acid. LAGUNA DESIGN / Getty Images
Updated July 06, 2016

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: Ka2 = 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) + SO42-(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.

Ka2 = [H+][SO42-]/[HSO4-]

Ka2 = 1.3 x 10-2

Ka2 = (0.10 + x)(x)/(0.10 - x)

Since Ka2 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 IonizationH2SO4(aq)H+(aq)HSO4-(aq)
Initial0.10 M0.00 M0.00 M
Change-0.10 M+0.10 M+0.10 M
Final0.00 M0.10 M0.10 M
    
    
Second IonizationHSO42-(aq)H+(aq)SO42-(aq)
Initial0.10 M0.10 M0.00 M
Change-x M+x M+x M
At Equilibrium(0.10 - x) M(0.10 + x) Mx M