Solution:
Benzoic acid is C6H5COOH
In finding pH
C6H5COOH(aq) <=> C6H5COO^- + H^+ pKa = 4.19, pKa = -logKa so Ka = 10^(-4.19)
Ka = 6.45 x 10^-6
[C6H5COO^-] = x = [H^+]; [C6H5COOH] = 0.5 - x (we are able to make an estimate of [C6H5COOH] = 0.5.
Ka = [H^+][C6H5COO^-]/[C6H5COOH] = x^2/(0.5 - x) = 6.45 x 10^-6
Now,
According to the quadratic equation. x^2 = 3.23 x 10^-5 - 6.45 x 10^-6x
x^2 + (6.45 x 10^-6)x - 3.23 x 10^-5 = 0
enter a = 1, b = 0.00000645, c = 0.0000323
x = 5.68 x 10^-3 = 0.00568 M expression is [C6H5COOH] = 0.5 M is the correct answer.
[H^+] = 0.00568 M, so pH = -log(0.00568 M ) = 2.25
This is the required solution.