A 0.60 M solution of a weak acid, HA, has a pH of 3.72. What is the percentage ionization of the acid? What is the value of Ka for the acid?

Answer 1

pH = -log(H^+)
pH = 3.72; (H^+) = approx 2E-4 but that's an approximation. You should do this more accurately as well as all of the calculations that follow.

%ion = [(H^+)/Msample]*100

...........HA ==> H^+ + A^-
I.........0.6.....0......0
C...........-x....x......x
E........0.6-x....x......x

You know x is approx 2E-4, Substitute that for x into the Ka expression for HA and solve for Ka.