Question

If 20.0 ml of glacial acetic acid (pure hc2h3o2) is diluted to 1.40 l with water, what is the ph of the resulting solution? the density of glacial acetic acid is 1.05 g/ml.

Answer 1

By using the formula, mass = density x volume, we calculate mass in grams

20.0 mL CH₃COOH x (1.05 g / mL) = 21.0 g CH₃COOH

To find the moles, molar mass of CH₃COOH = 60.05g/mol

21.0 g CH₃COOH x (1 mole CH₃COOH / 60.05 g CH₃COOH) = 0.350 moles CH₃COOH

To find molarity,

[CH₃COOH] = moles CH₃COOH / L of solution = 0.350 / 1.40 = 0.250 M CH₃COOH

When CH₃COOH is dissolved in water, it produces small and equal amounts of H₃O⁺+ and C₂H₃O₂⁻.

Molarity , CH₃COOH + H₂O <==> H₃O⁺ + C₂H₃O₂⁻

Initial 0.250 0 0
Change -x x x
Equilibrium 0.250-x x x

Kₐ = [H₃O⁺][C₂H₃O₂⁻] / [HC₂H₃O₂] = (x)(x) / (0.250-x) = 1.8 x 10⁻⁵

Since Kₐ is relatively small, we can neglect the -x term after 0.250 to simplify

x² / 0.250 = 1.8 x 10⁻⁵

x² = 4.5 x 10⁻⁶

x = 2.1 x 10⁻³ = [H₃O⁺]

pH = -log [H₃O⁺] = -log (2.1 x 10⁻³) = 2.68