A chemist titrates 60.0 mL of a 0.1935 M benzoic acid (HC (H5CO2) solution with 0.2088 M KOH solution at 25 °C. Calculate the pH at equivalence. The pKg of benzoic acid is 4.20.

pH at the equivalence point is 8.6

Explanation:

A titulation between a weak acid and a strong base, gives a basic pH at the equivalence point. In the equivalence point, we need to know the volume of base we added, so:

mmoles acid = mmoles of base

60 mL . 0.1935M = 0.2088 M . volume

(60 mL . 0.1935M) /0.2088 M = 55.6 mL of KOH

The neutralization is:

HBz + KOH  ⇄  KBz  +  H₂O

In the equilibrum:

HBz + OH⁻   ⇄  Bz⁻  +  H₂O

mmoles of acid are: 11.61 and mmoles of base are: 11.61

So in the equilibrium we have, 11.61 mmoles of benzoate.

[Bz⁻] = 11.61 mmoles / (volume acid + volume base)

[Bz⁻] = 11.61 mmoles / 60 mL + 55.6 mL = 0.100 M

The conjugate strong base reacts:

  Bz⁻  +  H₂O  ⇄  HBz + OH⁻    Kb

0.1 - x                       x        x

(We don't have pKb, but we can calculate it from pKa)

14 - 4.2 = 9.80 → pKb  → 10⁻⁹'⁸ = 1.58×10⁻¹⁰ → Kb

Kb = [HBz] . [OH⁻] / [Bz⁻]

Kb = x² / (0.1 - x)

As Kb is so small, we can avoid the quadratic equation

Kb =  x² / 0.1 → Kb . 0.1 = x²

√ 1.58×10⁻¹¹ = [OH⁻] = 3.98 ×10⁻⁶ M

From this value, we calculate pOH and afterwards, pH (14 - pOH)

- log [OH⁻] =  pOH → - log 3.98 ×10⁻⁶  = 5.4

pH = 8.6

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