What is the ph of a 0.500 m solution of benzoic acid, pka = 4.19?

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.