Question

A 0.351 L solution of 1.31 M ascorbic acid (Ka1 = 7.9e-5 and Ka2 = 1.6e-12) is titrated with 1.65 M NaOH. What will the pH of the solution be when 0.4933 L of the NaOH has been added?

Answer 1

mmol of acid = MV = 0.351*1.31 = 0.45981

mmol of base = M V= 1.65*0.4933 = 0.813945

therefore;

mmol of H+ = 2*mmol of acid = 2*0.45981 = 0.91962 mmol of H+

therefore,

H+ + OH- =H2O

mmol of H+ left = 0.91962 - 0.813945 = 0.105675

mmol of conjugate formed = 0.813945

therefore...

this is

HA- + OH- <--> H2O + A-2

which is a buffer, since HA- and A-2 are both present

weak acid + conjugate base

pH = pKa2 + log(A-2/HA)

pKa2 = -log(Ka2) = -log(1.6*10^-12) = 11.7958

now..

pH = 11.795+ log(0.813945 /0.105675)

pH = 12.681