Question

A 600.0 mL sample of 0.20 M HF is titrated with 0.20 M NaOH. Determine the pH of the solution after the addition of 100.0 mL of NaOH. The Ka of HF is 3.5 * 10^-4

Answer 1

no of mole of HF = M*V

                                 = 0.2*0.6    = 0.12 mol

no of mole of NaOH = 0.2*0.1 = 0.02 mol

mixer is a buffer

pH of acidic buffer = pka + log(salt/acid)

pka of HF = -log(3.5*10^-4) = 3.45

pH = 3.45 + log(0.02/(0.12-0.02))

         = 2.75