Question

The titration curve for a 25.0 ml sample of benzoic acid shows that 10.0 ml of 0.100M NaOH is required to reach the equivalence point.

1. What is the concentration of the original benzoic acid sample?

2. What is the pH at the equivalence point?

3. Calculate the pH after addition of 5.00 ml of NaOH and the pOH after addition of 20.0 ml NaOH.

Answer 1

In equivalence point:

HB + NaOH ---> H2O + Na+ and B-

B- and H2O <----< HB+ and OH-

there will be more OH- ions, expect basic solution

Kb = [HB+][OH-] / [B-]

NOTE, volume has changed...

VT = 25 ml + V2... Since no notes on the NaOH solution is made, assume the same volume

Vt = 25 ml

Kb = [HB+][OH-] / [B-]

[HB+] = [OH-] = x

[B-] = 0.15

Substitute and solve for x

Kb = [HB+][OH-] / [B-]

Kb = Kw/Ka =(10^-14)/(4.5*10^-4) = 2.22*10^-11

2.22*10^-11 = x*x / (0.15)

3.32*10^-12 = x^2

X = 1.825*10^-6

[OH-] = x = 1.825*10^-6

pOH = -log(1.825*10^-6) = 5.74

pH = 14- pOH = 14-5.74 = 8.26

which is basic as expected