Question

Using the definition of Ka and the Debeye-Huckel expression for the mean ionic activity
coefficient, calculate the pH of a 0.10 M acetic acid solution in (a) pure water (b) 0.1 M
NaCl and (c) in 0.2 M MgSO4. Assume consistent results after 3 iterations in each case

Answer 1

pH of acetic acid

(a) pure water

CH3COOH <==> CH3COO- + H+

Ka = 1.8 x 10^-5 = x^2/0.1

x = [H+] = 1.34 x 10^-3 M

pH = -log[H+] = 2.87

(b) In 0.1 M NaCl

ionic strength (u) = 0.1

activity of [H+] = inv.log[(-0.51.Z^2.sq.rt.(u))/(1+3.3,r.sq.rt.(0.1)] = inv/log(-0.51.sq.rt.(0.1)/(1+3.3 x 0.9 x sq.rt.(0.1))]

                        = 0.826

activity of [CH3COO-] = inv/log(-0.51.sq.rt.(0.1)/(1+3.3 x 0.45 x sq.rt.(0.1))] = 0.770

1.8 x 10^-8 = x^2(0.826 x 0.77)/0.1

x = [H+] = 1.68 x 10^-3 M

pH = -log(0.826 x 1.68 x 10^-3) = 2.86

(b) In 0.2 M MgSO4

ionic strength (u) = 0.8

activity of [H+] = inv/log(-0.51.sq.rt.(0.8)/(1+3.3 x 0.9 x sq.rt.(0.8))]

                        = 0.750

activity of [CH3COO-] = inv/log(-0.51.sq.rt.(0.8)/(1+3.3 x 0.45 x sq.rt.(0.8))] = 0.636

1.8 x 10^-8 = x^2(0.750 x 0.636)/0.1

x = [H+] = 1.94 x 10^-3 M

pH = -log(0.750 x 1.94 x 10^-3) = 2.84