Calculating pH - syllabus problem 6D.9

[Eleanor_C_1I]

6D.9 The percentage deprotonation of benzoic acid in a 0.110 M solution is 2.4%. What is the pH of the solution and the Ka of benzoic acid?

Would anyone be able to explain why we would take the -log(2.6x10^-3) in this problem? I understand how to find Ka but am not sure where the solutions manual gets the 2.6x10^-3 M from when finding the pH of benzoic acid.

[Irene Oh 2A]

You're given the percentage deprotonation of the acid and for this problem, the expression for that would be the concentration of H3O+ over the initial concentration of benzoic acid, and then multiply that by 100. We put "X" for the [H3O+], so it becomes 2.4 = (X/.110) x 100. If you solve for x, you get 2.6x 10^-3, and since that is the [H3O], you plug that into the -log( ) equation to find the pH