You will now learn how to calculate the pH of an aqueous solution of a polyprotic acid or base.

**Synopsis**

Polyprotic acids and bases present a bit more of a complicated situation because we will have to deal with a multi-step dissociation and a different equilibrium constant for each step. The equilibrium constants become smaller at each step. Sulfuric
acid is a diprotic acid so it will have two dissociation steps and we will have two equilibrium constants, one for each step. For sulfuric acid we will have a Ka_{1} and a Ka_{2} with Ka_{1} > Ka_{2}. Phosphoric acid is
a triprotic acid and so will have three dissociation steps. The procedure is the same as before except that we now have more than one step involved.

So let's work a sulfuric acid example. Suppose that we start with 0.100 M H_{2}SO_{4} which is a very strong acid and even the Ka_{2} is 1.2 x 10^{-2}. So we have

H_{2}SO_{4}(aq) + H_{2}O(l) H_{3}O^{+}(aq) + HSO_{4}^{-}(aq).

Ka1 is very large so we will assume that the 0.100 M H_{2}SO_{4} solution gives us 0.100 M H_{3}O^{+} and 0.100 M HSO_{4}^{-} so that we will have for the second step

HSO_{4}^{-}(aq) + H_{2}O(l) H_{3}O^{+}(aq) SO_{4}^{-2}(aq)

So let's make up another table:

HSO_{4}^{-} | H_{3}O^{+} | SO_{4}^{-2} | |

Initial,M | 0.100 | 0.100 | 0 |

Change,M | -x | +x | +x |

Equilibrium,M | 0.100 -x | 0.100 + x | x |

Then for the equilibrium constant we have

K_{2} = (0.100 + x) (x)/(0.100 -x) = 1.2 x 10^{-2}

For this problem we cannot ignore x compared to 0.100 because K is so large. So we set up the quadratic equation

x^{2} + 0.112 x - 0.0012 = 0

and solve for x which gives x = 0.0379. So, we now have the equilibrium value of the hydronium ion concentration (0.1379 M) and the pH is then 0.86 rather than the 1.00 value that it would be if only the first step were considered. The sulfate ion
concentration is 0.0379 M, and the HSO_{4}^{-} concentration is 0.621 M.

This problem is unusual because we started with a strong acid. The second step is then like a weak acid problem except that the equilibrium constant is still pretty large. For a weak polyprotic acid the situation would be different for now we would have to treat each step like we treated the second step above.

For most weak acids (H_{2}A), the second equilibrium constant is usually much smaller than the first constant so that we can assume that the hydronium ion concentration overall comes solely from the first step. The [H_{3}O+] is then
equal to {(Ka1)([H_{2}A]_{o}}1/2 where [H_{2}A]_{o} is the initial concentration of the weak acid. The concentration of the anion, A_{2}^{-}, is then just equal to the Ka_{2}..

**Review Questions**

- What is the pH of a 0.100 M solution of carbonic acid (H
_{2}CO_{3})? What is the equilibrium concentration of the carbonate ion? - What is the pH of a 0.100 M solution of phosphoric acid?

*Web Author: Dr. Leon L. Combs*

*Copyright ©2000 by Dr. Leon L. Combs - ALL RIGHTS RESERVED*