Hydrolysis of Salts in Advanced Problems

Introduction

A salt is often described as the "neutral" product of an acid reacting with a base, yet a solution of ammonium chloride turns universal indicator orange, and sodium ethanoate solution turns it blue-green. The reason is that ions are themselves acids and bases. When a salt dissolves, each ion may react with water, and the pH depends on which of these reactions wins. At advanced level you must predict the direction, calculate the pH, and decide which equilibria can safely be ignored.

Core explanation

Every ion has a conjugate. Chloride is the conjugate base of HCl; ammonium is the conjugate acid of NH₃; ethanoate is the conjugate base of ethanoic acid. For any conjugate pair in water at 25 °C:

Ka × Kb = Kw = 1.0 × 10⁻¹⁴

So the weaker the parent acid, the stronger its conjugate base, and vice versa.

Spectator ions. Conjugates of strong acids (Cl⁻, NO₃⁻, ClO₄⁻, HSO₄⁻ is an exception because it is itself a weak acid) and cations of strong bases (Na⁺, K⁺, Ca²⁺ in dilute solution) have negligible acid-base strength. Their Kb or Ka values are effectively zero, so they do not shift the pH.

Four classes of salt.

Salt type Example Ion that reacts Solution --- --- --- --- Strong acid + strong base NaCl neither neutral, pH 7.00 at 25 °C Strong acid + weak base NH₄Cl NH₄⁺ (acid) acidic Weak acid + strong base CH₃COONa CH₃COO⁻ (base) basic Weak acid + weak base NH₄CN, CH₃COONH₄ both compare Ka with Kb

Single reacting ion. For NH₄Cl, the relevant equilibrium is NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, with Ka = Kw ÷ Kb(NH₃) = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. The calculation is then identical to a weak-acid calculation.

Two reacting ions. In ammonium cyanide, NH₄⁺ produces H₃O⁺ while CN⁻ produces OH⁻. The two effects compete. Ka(NH₄⁺) = 5.6 × 10⁻¹⁰, but Kb(CN⁻) = Kw ÷ Ka(HCN) = 1.0 × 10⁻¹⁴ ÷ 6.2 × 10⁻¹⁰ = 1.6 × 10⁻⁵. Because Kb is much larger than Ka, the solution is basic. When the cation and anion come from partners of similar strength, as in ammonium ethanoate, the two effects nearly cancel and the pH is close to 7. For such salts the proton-transfer reaction NH₄⁺ + A⁻ ⇌ NH₃ + HA dominates, and the pH is approximately ½(pKa of the cation + pKa of the parent acid of the anion), independent of concentration over a useful range.

Hydrated metal ions. Small, highly charged cations polarise their water ligands. [Al(H₂O)₆]³⁺ has pKa near 5.0 and [Fe(H₂O)₆]³⁺ near 2.2, so their salts are distinctly acidic even though no "acid" appears in the formula.

Formulae

Ka × Kb = Kw; pKa + pKb = 14.00 at 25 °C. Single weak acid ion: [H₃O⁺] ≈ √(Ka × c). Single weak base ion: [OH⁻] ≈ √(Kb × c). Salt of weak acid HA and weak base B: pH ≈ ½(pKa(BH⁺) + pKa(HA)).

Step-by-step reasoning

1. Write the formula of each ion and identify its parent acid or base. 2. Discard spectator ions from strong acids and strong bases. 3. For each remaining ion, find Ka or Kb, converting with Ka × Kb = Kw. 4. If only one ion reacts, treat it as a simple weak acid or weak base. 5. If both react, compare Ka and Kb to predict the direction, then use the averaging expression. 6. Check that the approximation [H₃O⁺] ≪ c holds.

Visual explanation

Draw a horizontal pH axis. Place a left-pointing arrow labelled "cation acidity (Ka)" and a right-pointing arrow labelled "anion basicity (Kb)". The larger the constant, the longer the arrow, and the longer arrow wins. For NH₄Cl only the left arrow exists; for CH₃COONa only the right arrow; for NH₄CN the right arrow is far longer.

Real-world analogy

A salt solution is like a tug-of-war on a rope marked with a pH scale. Each ion is a team. Spectator ions are teams who sit down and do not pull. When only one team pulls, the rope moves its way. When two teams of equal strength pull, the rope stays near the middle.

Real-world example

Aluminium sulfate is used in water treatment and in gardening to acidify soil for plants such as blueberries and hydrangeas. The hydrated Al³⁺ ion releases protons, lowering the pH, while sulfate is too weak a base to compensate. Sodium carbonate, by contrast, is used as washing soda because the carbonate ion is a moderately strong base and makes water alkaline.

Why?

Why does a salt of a weak acid make water basic? The anion is a base strong enough to take a proton from water, leaving OH⁻ behind. The conjugate of a strong acid cannot do this, because its acid form gives up protons completely; its tendency to accept a proton is negligible.

Common misconception

"Salts formed in neutralisation are always neutral." Only salts of a strong acid with a strong base give pH 7. The equivalence point of any titration involving a weak partner lies away from 7 precisely because the salt formed undergoes hydrolysis.

Worked example

Question: Calculate the pH of 0.10 mol dm⁻³ sodium ethanoate at 25 °C. Ka(CH₃COOH) = 1.8 × 10⁻⁵.

Reasoning: Na⁺ is a spectator. Kb(CH₃COO⁻) = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. [OH⁻] ≈ √(5.6 × 10⁻¹⁰ × 0.10) = √(5.6 × 10⁻¹¹) = 7.5 × 10⁻⁶ mol dm⁻³. pOH = 5.13, so pH = 14.00 − 5.13 = 8.87. Check: 7.5 × 10⁻⁶ is far below 0.10, so the approximation holds.

Answer: pH ≈ 8.87.

Quick check

1. Is a solution of potassium nitrate acidic, basic or neutral, and why? Answer: Neutral, because K⁺ comes from a strong base and NO₃⁻ from a strong acid, so neither ion reacts with water.

Exam focus

Show the Ka to Kb conversion explicitly; examiners award a mark for it. State which ion is a spectator and why. For salts of two weak partners, a qualitative comparison of Ka and Kb is often all that is required, but be ready to use the averaging formula.

Advanced insight

The averaging formula for salts like NH₄CN comes from combining the charge balance with both equilibria and assuming the proton-transfer reaction between the ions dominates over reactions with water. It fails in very dilute solutions, where water autoionisation becomes comparable, and when the two pKa values differ so much that one ion reacts almost completely. The same mathematics underlies the pH of amphiprotic ions such as HCO₃⁻.

Summary

Ions from weak acids are bases and ions from weak bases are acids, with Ka × Kb = Kw linking each pair. Spectator ions from strong partners do not change the pH. Salts with one reactive ion are treated as simple weak acids or bases. When both ions react, compare Ka with Kb; the pH lies near the average of the two pKa values. Hydrated highly charged metal ions are significant acids.

Practice questions

1. Calculate the pH of 0.10 mol dm⁻³ NH₄Cl. Kb(NH₃) = 1.8 × 10⁻⁵. Answer: Ka = 5.6 × 10⁻¹⁰; [H₃O⁺] = √(5.6 × 10⁻¹¹) = 7.5 × 10⁻⁶ mol dm⁻³; pH ≈ 5.13. 2. Predict whether NH₄CN solution is acidic or basic, giving numerical reasons. Answer: Basic, because Kb(CN⁻) = 1.6 × 10⁻⁵ is much larger than Ka(NH₄⁺) = 5.6 × 10⁻¹⁰. 3. Estimate the pH of ammonium ethanoate solution, given pKa(NH₄⁺) = 9.25 and pKa(CH₃COOH) = 4.74. Answer: pH ≈ ½(9.25 + 4.74) ≈ 7.0, essentially neutral. 4. Explain why iron(III) chloride solution is acidic. Answer: The small, highly charged Fe³⁺ polarises its water ligands so [Fe(H₂O)₆]³⁺ releases H⁺ (pKa about 2.2); Cl⁻ is a spectator.