Ionic equilibria

MDCAT Chemistry · Chemistry · Chemical equilibrium. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Weak acids and bases ionise only partly in water, and sparingly soluble salts dissolve only slightly. Each of these is an equilibrium with its own constant: K_a, K_b or the solubility product K_sp.

Adding an ion that is already part of the equilibrium pushes it backwards. This common ion effect makes a salt less soluble and a weak acid less ionised.

A buffer keeps its pH almost constant when a little acid or alkali is added. It contains a weak acid with its salt, or a weak base with its salt. One part removes added H⁺ and the other removes added OH⁻.

Rules to remember

Arrhenius: an acid gives H⁺ in water, a base gives OH⁻. Brønsted–Lowry: an acid donates a proton, a base accepts one. Lewis: an acid accepts an electron pair, a base donates one
pH = −log[H⁺]; pOH = −log[OH⁻]; pH + pOH = pK_w = 14 at 25 °C (K_w = 1.0 × 10⁻¹⁴ mol² dm⁻⁶)
K_a = [H⁺][A⁻] ÷ [HA]; pK_a = −log K_a; a larger K_a (smaller pK_a) means a stronger acid
Weak acid of concentration c: [H⁺] = √(K_a × c)
K_sp for A_xB_y = [A]^x[B]^y; with solubility s: AB gives s², AB₂ or A₂B gives 4s³
A precipitate forms only when the ionic product is greater than K_sp
Buffer: pH = pK_a + log([salt] ÷ [acid]); blood is kept at pH 7.35 to 7.45, mainly by H₂CO₃ / HCO₃⁻

Common mistake

Students forget that one formula unit can give two ions of the same kind. For AB₂ the ion B has concentration 2s, so K_sp = s × (2s)² = 4s³, and 0.0050 mol dm⁻³ Ba(OH)₂ has [OH⁻] = 0.010 mol dm⁻³.

Worked example

A sparingly soluble salt MX₂ has K_sp = 3.2 × 10⁻¹¹ mol³ dm⁻⁹. What is its solubility in pure water?

MX₂(s) ⇌ M²⁺(aq) + 2X⁻(aq). If the solubility is s, then [M²⁺] = s and [X⁻] = 2s.
K_sp = [M²⁺][X⁻]² = s × (2s)² = 4s³.
s³ = (3.2 × 10⁻¹¹) ÷ 4 = 8.0 × 10⁻¹².
s = cube root of 8.0 × 10⁻¹² = 2.0 × 10⁻⁴.

Answer: 2.0 × 10⁻⁴ mol dm⁻³

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