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 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

**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 |