Calculate the pH after the addition of 55 mL of KOH (aq)?

Question

for the titration of 50 mL of 0.020 M aqueous salicylic acid with 0.020 M KOH (aq)..

For salycylic acid, pKa= 2.97

I couldnt get this answer 10.98....can u plz show me

Answer 1

Since salicylic acid is a monoprotic acid the titration reaction is:

( HA= salicylic acid)

HA + OH- -----> A- + H20

50mL of .02M acid gives you .0010mol of the acid and 55ml of .020M KOH gives you .0011mol OH-. The OH- completely deprotonates the acid, leaving .0010 mol A- in solution along with .0001 mol OH-. The OH- remaining in solution will control the ph of the solution since  the amount of A- reacting with water to produce addition OH- is negligable in this case(kb=9.33x10^-12).

.0001mol OH-/.105L = 9.52 x10^-4M OH-

Recall that 1x10-14=[OH-][H+]..... so [H+]=1.05x10^-11 making the pH 10.98