Question

A buffer with a pH of 4.24 contains 0.39 M of sodium benzoate and 0.36 M of benzoic acid. What is the concentration of [H ] in the solution after the addition of 0.050 mol of HCl to a final volume of 1.4 L? Assume that any contribution of HCl to the volume is negligible.

Answer 1

initially

mol of acid = MV = 0.36 * 1.4 = 0.504

mol of conjugate= MV = 0.39*1.4 = 0.546

now..

after addition of 0.05 mol of H+

conjugate reacts to form acid

benzoate + H+ = benzoic acid

mol of benzoic acid formed = 0.504 + 0.05 = 0.554

mol of benzoate left after reaction = 0.546 - 0.05 = 0.496

now..

pKa for benzoic acid is 4.20 so

pH = pKa +log(benzoate/benzoic acid)

pH = 4.20 + log(0.496/0.554 )

pH = 4.15197

[H+] = 10^-pH = 10^-4.15197

[H+] = 0.00007047 M approx or 7.05*10^-5 M