A 0.100 M solution of weak acid HA ionizes 2.00%. What will be its percent ionization if 10.0 mL of this solution is diluted with 30.0 mL of water?

Answer 1

This is a problem designed to illustrate how the % ionization of a weak acid changes with its concentration. We will determine the % ionization of the diluted sample and compare it to the % ionization (2.00%) of the concentrated sample. We should see that the diluted solution has a greater % ionization than the more concentrated acid. Before doing this, we will have to use the data provided to find the Ka of the weak acid.

HA ==> H⁺ + A^- if it ionizes 2% then from a 0.1 M solution you will have 0.02 M H⁺ and 0.02 M A^- and 0.08 M HA

Ka = [H⁺][A^-]/[HA] = (0.02)(0.02) / 0.08

Ka = 0.005

Diluted sample: (10.0 ml)(0.100 M) = (40.0 ml)(x M) and x = 0.025 M

Ka = 0.005 = (x)(x)/0.025-x

x² = 0.000125 - 0.005x

x² + 0.005x - 0.000125 = 0

x = 0.00896 M = [H⁺] = [A^-]

% ionization = 0.00896 / 0.025 (x100%) = 35.8% ionization (compare this to the 2% of the original sample).