Question

calculate the ph of a 0.040 benzoic acid

Answer 1

Have you got the Ka of benzoic acid?

Answer 2

Ka for benzoic acid C6H5CO2H is 6.5E-5

Let A- represent benzoate ion and HA = benzoic acid
A- + H2O <--> HA + OH-
Kb = [HA][OH-]/[A-] = Kw/Ka = 1.54x10^-10
let X = mol/L of OH- formed by hydrolysis of A-
1.54x10^-10 = X^2 / (0.04-X)
X = [OH-] = 2.48x10^-6
pOH = 5.6
pH = 14.0-5.6 = 8.4 (2 s.f.)