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 H_{2}SO_{4}.

Given: K_{a2} = 1.3 x 10^{-2}

### Solution

H_{2}SO_{4} has two H^{+} (protons), so it is a diprotic acid that undergoes two sequential ionizations in water:

First ionization: H_{2}SO_{4}(aq) → H^{+}(aq) + HSO_{4}^{-}(aq)

Second ionization: HSO_{4}^{-}(aq) ⇔ H^{+}(aq) + SO_{4}^{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 HSO_{4}^{-}(aq) in the second ionization is a weak acid, so the H^{+} is in equilibrium with its conjugate base.

K_{a2} = [H^{+}][SO_{4}^{2-}]/[HSO_{4}^{-}]

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:

x^{2} + 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

**Concentration of Chemical Species**

First Ionization | H_{2}SO_{4}(aq) | H^{+}(aq) | HSO_{4}^{-}(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 | HSO_{4}^{2-}(aq) | H^{+}(aq) | SO_{4}^{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 |