pH of benzoic acid from a weighed sample

C3L6 · UKChO Round 1 · IChO · unreviewed

A weighed mass of a weak acid with known M and Kₐ is made up to a known volume; find the pH without the small-x approximation.

Practise this with fresh numbers

Worked example

A student weighs 0.387 g of benzoic acid (Kₐ = 6.3 × 10⁻⁵, M = 122.12 g mol⁻¹) into a 818 cm³ volumetric flask and makes up to the mark. Find the exact pH (ignore water autoionisation).

1. Moles from mass: solve for n
$n = \dfrac{m}{M}$
= 0.003169 mol
2. Concentration from moles and volume: solve for c
$c = \dfrac{n}{V}$
= 0.003874 mol dm⁻³
3. Weak acid (without the approximation): solve for [H⁺]
$[\mathrm{H^+}] = \dfrac{-K_\mathrm{a} + \sqrt{K_\mathrm{a}^2 + 4K_\mathrm{a}c}}{2}$
= 4.635 × 10⁻⁴ mol dm⁻³
4. pH: solve for pH
$\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$
= 3.334

Answer: pH = 3.33

Method

n = m/M, c = n/V, then solve Kₐ = x²/(c − x) for x = [H⁺] and take pH = −log₁₀ x.

Common mistakes

That uses √(Kₐc). At this dilution the acid is appreciably dissociated, so solve the quadratic.

Formulas used

Moles from mass
$n = \dfrac{m}{M}$
Concentration from moles and volume
$c = \dfrac{n}{V}$
Weak acid (without the approximation)
$K_\mathrm{a} = \dfrac{x^2}{c - x},\quad x = [\mathrm{H^+}]$
pH
$\mathrm{pH} = -\log_{10}\dfrac{[\mathrm{H^+}]}{c^\circ}$