Question

A student performs a titration of 50.0 mL of 0.100 M hydrofluoric acid (HF), using0.050 M sodium hydroxide (NaOH). The Ka for hydrofluoric acid is 8.8 x 10-5

What is the pH of the solution after the addition of 20.0 mL of sodium hydroxide solution?

Answer 1

Concentration of HF, 50.0 mL x 0.100 M (HF] = := 0.071M (50.0+20.0) mL 20.0 mL x 0.050 M-0014 M Concentration of OH = (NaOH) = (50.0+ 20.0) mL + - F + H2O ICE table: HF I(M): 0.071 C(M): -0.014 E(M): 0.057 OH 0.014 -0.014 0 +0.014 0.014 PK, of HF = -log K, = -log(7.2x10*) = 3.14 According to Henderson equation PR + loc (F7 (HF) 0.014 = 3.14+10g 0.057 = 2.53