The pH of a 0.010 M aqueous solution of a weak monoprotic acid, HX, is 4.5. What is the value of the acid ionization constant?

Answer 1

pH = -log(H^+) = 4.5
(H^+) = 3.16E-5

.............HX ==> H^+ + X^-
I..........0.10M...0.....0
C............-x.....x.....x
E..........0.10-x...x.....x

x = 3.16E-5. Substitute into Ka expression and solve for Ka.