Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 29 mL of 0.17 M C₆H₅COOH(aq) with 0.16 M NaOH(aq). The K_a of benzoic acid is 6.5 x 10^-5.

Answer 1

calculate volume of NaOH

To reach equivalence point,
mol of acid = mol of base
0.17 M * 29 mL = 0.16 M * V
V = 30.8 mL

total volume = 29 + 30.8 = 59.8 mL

number of mol of C6H5COO- formed = 0.17 M * 29 mL = 4.93 mmol

[C5H5COO-] = 4.93 mmol / 59.8 mL = 0.082 M

Kb of C6H5COO- = 10^-14 / Ka
= 10^-14 / (6.5*10^-5)
= 1.54*10^-10

for simplicity lets write C6H5COO- as A-

A-        + H2O   ----->     AH   +   OH-
0.082                        0         0
0.082-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 ((0.000)*0.082) = 3.55E-6

so,
[OH-] = 3.55E-6 M;

pOH = -log [OH-] = -log (3.55E-6) = 5.45
PH = 14 - pOH = 14 - 5.45 = 8.55

Answer: 8.55