Question

11.) Glycolic acid which is a monoprotic acid and a constituent in sugar cane has a pKa of 3.90. A 25.0 mL solution of glycolic acid is titrated to the equivalence point with 35.8 mL of 0.020 M NaOH (aq). What is the pH of the resulting solution at the equivalence point?

Answer 1

Lets write Glycolic acid as HA
It will react with NaOH to form A-

mol of A- formed
= mol of NaOH reacted
= M*V
= 0.020 M * 35.8 mL
= 0.716 mmol

total volume = 25 mL + 35.8 mL = 60.8 mL

[A-] = number of mol / volume
= 0.716 mmol / 60.8 mL
= 0.118 M

use:
pKa = -log Ka
3.9 = -log Ka
Ka = 1.26E-4

use:
Kb of A- = 10^-14/Ka
Kb = 10^-14/1.26E-4
Kb = 7.94E-11

for simplicity lets write weak base as A-

A-        + H2O   ----->     AH   +   OH-
0.1180                        0         0
0.1180-x                      x         x

Kb = [AH][OH-]/[B-]
Kb = x*x/(c-x)
since kb is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((7.94E-11)*0.1180) = 3.06E-6

pOH = -log [OH-] = -log (3.06E-6) = 5.5
PH = 14 - pOH = 14 - 5.5 = 8.5