Question

The pH at the equivalence point when a 0.20 M weak base (Ka = 9.1 x 10-7) is titrated with a 0.20 M strong acid is:

a) pH = 3.5

b) pH = 1.7

c) pH = 2.9

Answer 1

V L of both weak base and strong acid will react

So final volume will be 2V L

Hence concentration of salt formed will be 0.20/2 = 0.10 M
This salt in fact is weak acid ion

Ka= 9.1*10^-7

for simplicity lets write acid ion as BH+

BH+      + H2O ----->     BOH   +   H+
0.1000                    0         0
0.1000-x                  x         x

Ka = [H+][BOH]/[BH+]
Ka = x*x/(c-x)
since ka is small, x will be small and it can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((9.1E-7)*0.1000) = 3.02E-4

pH = -log [H+] = -log (3.02E-4) = 3.52

Answer: a