General Chemistry 2 Class

Question

A 0.100 M solution of bromoacetic acid 1BrCH2COOH2 is 13.2% ionized. Calculate [BrCH2COO-], [H+], [BrCH2COOH], and Ka for bromoacetic acid.

Answer 1

BrCH2COOH2 should be BrCH2COOH

BrCH2COOH => BrCH2COO- + H+

If you begin with 0.1 M BrCH2COOH and it is 13.2% ionized, you will have

0.132 x 0.1 M = 0.0132 M H+ and 0.0132 M BrCH2COO- leaving 0.0868 M BrCH2COOH

Ka = [BrCH2COO-][H+] / [BrCH2COOH]

Ka = (0.0132)(0.0132)/0.0868)

Ka = 2.0x10^-3