Finding the pH of a solution of Benzoic acid

Question

What is the pH of a 0.050 M solution of benzoic acid (pKa = 4.2)? Please assist by showing the step-by-step computation.

Answer 1

Remember benzoic acid is a weak acid. The dissociation follows the following reaction:

C₆H₅COOH ⇔ H₃O⁺ + C₆H₅COO^-

Remember pKa = -log Ka

If the pKa is 4.2, then the Ka = 10^-4.2 = 6.31e-05

The formula for Ka is:

Ka = [C₆H₅COO^-][H₃O⁺]/[C₆H₅COOH]

Using the "ICE table", we can assume x mol/L of [C₆H₅COO^-] and x mol/L [H₃O⁺] dissociate, leaving 0.050-x M of [C₆H₅COOH]

Thus

Ka = 6.31e-05 = x²/(0.050-x)

Here we make an assumption that due to C₆H₅COOH being a weak acid, only a miniscule amt dissociates, meaning x is very small. Therefore we "approximate" that (0.050-x)≈0.050 on the denominator, simplifying into:

Ka = 6.31e-05 = x²/0.050

Leaving the rest of the solution to you...from here should be relatively straightforward. Please book a session with me to review in more depth!