The pH of a 0.002 50 mol/L solution of benzoic acid is 3.65. Calculate the Ka for benzoic acid.

Answer 1

pH = 3.65 = -log(H^+).
(H^+) = 2.24E-4

Call benzoic acid (C6H5COOH) HBz.
...........HBz ==> H^+ + Bz^-
initial..2.24E-4...0.....0
change.....-x.......x.....x
equil....2.24E-4-x...x....x

Ka = (H^+)(Bz^-)/(HBz)
Substitute from the ICE chart above into the Ka expression and solve for Ka.