Question

When 25.0 mL of 0.100 M NaOH was added to 50.0 mL of a 0.100 M solution of a weak acid, HX, the pH of the mixture reached a value of 3.56. What is the value of Ka for the weak acid?

Answer 1

Total volume after addition, Vt = 50.0 mL + 25.0 mL = 75.0 mL = 0.075 L

Moles of NaOH before reaction = MxV = 0.100 M x 0.025 L = 0.0025 mol

Moles of weak acid(HA) before reaction = MxV = 0.100 M x 0.050 L = 0.005 mol

Given the pH of the solution after reaction, pH = - log[H⁺(aq)] = 3.56

=> [H⁺(aq)] = 10^-3.56 = 2.754x10^-4 M

The acid base neutralization reaction is

-------------- HA + OH^-(aq) --------> A^-(aq) + H2O

Init.mol: 0.005, 0.0025, ----------- 0

eqm.mol:(0.005-0.0025), (0.0025-0.0025), 0.0025

----------- = 0.0025, = 0.00, ------------ 0.0025

Hence [HA] = [A^-(aq)] =  0.0025 mol / 0.075 L = 0.0333 M

Nw this solution will act as a buffer solution. Hence applying Hendersen equation

pH = pKa + log[A^-(aq)] / [HA] = pKa + log1 = pKa

=> pH = pKa = 3.56

=> Ka = 10^-3.56 = 2.754x10^-4 M (answer)