Question

You were given 25.00mL of an acetic acid solution of unknown concentration. You find that it requires 37.15mL of a 0.1047M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator). Show detailed work.

Answer 1

Molarity of NaOH solution: 0.1047 M = M₁

Volume of Acid: 25 mL = V₂

Final Buret level: 37.15 mL

Initial Buret level: 0.00 mL

Volume of NaOH: 37.15-0.00 = 37.15 mL = V₁

Molarity of Acid: M₂

M₁V₁ = M₂V₂

0.1047 37.15 = M₂ 25

M₂ = 0.156 M

So, Molarity of CH₃COOH = 0.156 M

The pH at the equivalence point is determined by the titrand’s conjugate form, which in this case is a weak base. At the equivalence point the moles of acetic acid initially present and the moles of NaOH added are identical. Because their reaction effectively proceeds to completion, the predominate ion in solution is CH₃COO^–, which is a weak base. To calculate the pH we first determine the concentration of CH₃COO^–.

[CH₃COO^-] = moles of NaOH added/total volume = (0.1047 37.15)/(37.15 + 25.00) = 0.0626 M

Next, we calculate the pH of the weak base

CH₃COO⁻(aq) + H₂O(l) ⇌ OH⁻(aq) + CH₃COOH(aq)

K_b = [OH⁻][CH₃COOH] / [CH₃COO⁻] = (x)(x) / (0.0667−x)

5.71 10^{-10 ₌} x² / (0.0626−x)

x = 5.98 10^-6 M = [OH^-]

[H₃O⁺] = K_w / [OH⁻] = 1.00 10⁻¹⁴ / 5.98 10⁻⁶ = 1.67 10⁻⁹ M

pH = -log[H₃O⁺] = -log(1.67 10⁻⁹) = 8.78