Acids and bases

Chemistry A2 · Equilibria. 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

An acid donates a proton and a base accepts one. When an acid loses its proton, what is left is its conjugate base. Strong acids ionise fully, so for a monoprotic acid [H⁺] equals the acid concentration. Weak acids ionise only slightly, and K_a tells us how far.

A buffer solution resists changes in pH when a small amount of acid or alkali is added. It contains a weak acid and its salt, so it has a large supply of both HA and A⁻. Added H⁺ is removed by A⁻, and added OH⁻ is removed by HA.

For a sparingly soluble salt, the solubility product, K_sp, sets the limit on the ion concentrations in a saturated solution.

Rules to remember

pH = −log[H⁺]; [H⁺] = 10^−pH; pK_a = −log K_a
K_w = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K; for a strong alkali, [H⁺] = K_w ÷ [OH⁻]
K_a = [H⁺][A⁻] ÷ [HA]; for a weak acid alone, [H⁺] = √(K_a × [HA])
Buffer: [H⁺] = K_a × [HA] ÷ [A⁻], or pH = pK_a + log([A⁻] ÷ [HA])
Buffer action: A⁻ + H⁺ → HA and HA + OH⁻ → A⁻ + H₂O. In blood: HCO₃⁻ + H⁺ ⇌ H₂CO₃.
K_sp is the product of the ion concentrations in a saturated solution, each raised to the power of its number in the formula, e.g. K_sp(PbCl₂) = [Pb²⁺][Cl⁻]²
Common ion effect: adding an ion already in the equilibrium makes the salt less soluble.

Common mistake

For a buffer, students use √(K_a × [HA]), which only works for a weak acid on its own. In a buffer [H⁺] and [A⁻] are not equal, so use [H⁺] = K_a × [HA] ÷ [A⁻].

Worked example

A buffer contains 0.15 mol dm⁻³ methanoic acid and 0.30 mol dm⁻³ sodium methanoate. K_a for methanoic acid is 1.78 × 10⁻⁴ mol dm⁻³. What is the pH?

[H⁺] = K_a × [HA] ÷ [A⁻] = 1.78 × 10⁻⁴ × (0.15 ÷ 0.30).
[H⁺] = 1.78 × 10⁻⁴ × 0.50 = 8.9 × 10⁻⁵ mol dm⁻³.
pH = −log(8.9 × 10⁻⁵) = 4.05.

Answer: pH = 4.05

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Watch: 13 | Comparison of Acids | Comparison of Bases

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