Chemistry Problem of the Week: Calculating the pH of a Solution of a Weak Acid

One of the more intimidating problems you may encounter in your chemistry course or AP test is one where you need to calculate a pH for a weak acid. Typically, you are given a weak acid, which partially dissociates. An example follows:

Given that acetic acid has an acid dissociation constant (Ka) of 1.8 · 10^-5, calculate the pH of a 0.10 M solution.

To solve this, first start with the equilibrium one would expect for the weak acid, in this case, acetic acid.

HA(aq) ↔ H+(aq) + A-(aq)

Note that you can use the ICE method (Initial, Change, Equilibrium) to calculate the equilibrium concentration of H+.

   HA(aq)      ↔      H+(aq)      +      A-(aq)

I                                                0.10 M                    0                        0

C                                                  - x                       +x                     +x

E                                              (0.10 - x)                  x                       x

Therefore, since Ka = ([H+][A-]) / [HA], then we set up the calculation as

1.8 · 10^-5 = (x^2) / (0.10 - x)

This can be solved as a very time-consuming quadratic. An easier method is to approximate by noting x will be very small compared to 0.10 in the denominator (in red). Thus, the following approximation is valid:

1.8 · 10^-5 = (x^2) / (0.10)   (the -x term in the denominator was removed)

Solving for x, we get x = [H+] = 0.0013 M. Recalling pH = -log[H+], we get

  pH = 2.9