Question

What is the percent ionization of a solution prepared by dissolving 0.0183 mol of chloroacetic acid, HC2H2O2Cl) in 1.30 L of water? For chloroacetic acid, Ka = 1.4 × 10–3. I thought that you would find the molarity of chloroacetic acid, which turns out to be 0.01407. Multiply that by Ka, and then square root it. That will be the [H+] concentration, and to find the percent ionization, you would divide it by the orginal concentration and multiply it by 100%. The problem is I am not getting the right answer and I don't know what I am dong incorrectly :/