Question

A 25.0 mL sample of 0.150 M formic acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The K_a of formic acid is 1.8 ⋅ 10^-4.

Answer 1

1 mol of NaOH reacts with 1mol of formic acid

Molarity of Formic acid (M₁) = 0.150 M

Volume of formic acid (V₁) = 25.0 mL

Molarity of NaOH (M₂) = 0.150 M

Let the volume of NaOH required at equivalence point be V₂

From Molarity equation

M₁V₁ = M₂V₂

V₂ = 0.150 M * 25.0 mL / 0.150 M = 25 mL

Total volume of solution = 25 + 25 = 50.0 mL = 0.075 L

[HCOO^-] = Molarity *volume = 0.150M * 0.025 L/0.050L = 0.075 moles

K_b = K_w/K_a = (10^-14 ) /(1.8 *10^-4) = 5.56 *10^-11

HCOO^- + H₂O <--> HCOOH + OH^-

K_b = [HCOOH][OH^-] / [HCOO^-]

5.56 *10^-11 = x² / (0.075)

x = 2.04 *10^-6 M = [OH^-]

pOH = -log [OH^-] = -log (2.04 *10^-6 M) = 5.69

pH = 14 - pOH = 14 -5.69 = 8.31