Is ammonium chloride acidic or alkaline in water, and why is sodium ethanoate alkaline even though neither salt contains an obvious OH⁻ group? This is a classic H2 Chemistry salt-hydrolysis question. The correct starting point is not the word ‘salt’; it is the acid–base behaviour of the ions after the salt dissolves. NH₄⁺ can donate a proton to water, while CH₃COO⁻ can accept one. Those opposite reactions explain their different pH values.
For families searching for an H2 Chemistry tutor in Hougang, A-Level Chemistry 9476 acid–base equilibrium lessons, JC2 salt hydrolysis pH calculations or Ka and Kb worked examples, this is a useful diagnostic: ask the student to predict the pH direction before reaching for the calculator. A student who cannot identify the reacting ion will usually struggle with buffers, titration equivalence points and mixed aqueous-equilibrium problems.
I prefer to teach this as a story about where each ion came from. A salt preserves traces of the acid and base that formed it. Once we recognise the conjugate pairs, the equations, approximations and pH calculations begin to fit together. This guide uses idealised dilute aqueous solutions at 25 °C where indicated, with illustrative equilibrium constants; always follow the values given in the question.
Browse the Hougang Chemistry learning library · Check your SEAB H2 Chemistry examination-year syllabus
A salt is not automatically neutral
Many students first encounter neutralisation as ‘acid + alkali → salt + water’. That description identifies a reaction pattern; it does not guarantee that a solution of the resulting salt has pH 7. The product ions can interact with water. The important question is whether each ion is the conjugate acid or conjugate base of a weak partner.
| Salt dissolved in water | Key dissolved ion | Hydrolysis tendency | Predicted solution |
|---|---|---|---|
| NaCl | Na⁺ and Cl⁻ are effectively spectator ions in this model | Neither undergoes important acid–base hydrolysis | Approximately neutral at 25 °C |
| NH₄Cl | NH₄⁺ is the conjugate acid of weak base NH₃ | Can transfer H⁺ to H₂O | Acidic |
| CH₃COONa | CH₃COO⁻ is the conjugate base of weak acid CH₃COOH | Can accept H⁺ from H₂O | Alkaline |
| NH₄CH₃COO | Both NH₄⁺ and CH₃COO⁻ | Both acid and base hydrolysis matter | Compare the acid and base equilibria |
This classification works as a first diagnosis. For metal ions with significant hydrolysis or unusual coordination behaviour, more specific chemistry may be needed. Here we focus on the simple conjugate-acid/conjugate-base examples that make the calculation method transparent.
Why NH₄Cl solution is acidic
Ammonium chloride dissolves to give NH₄⁺ and Cl⁻. The chloride ion is the conjugate base of the strong acid HCl, so its tendency to take a proton from water is negligible under this familiar treatment. The ammonium ion is different: it is the conjugate acid of ammonia, NH₃, a weak base.
NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)
That equilibrium produces hydronium ions, so an aqueous NH₄Cl solution is acidic. The misconception is to claim that the chloride ion makes the solution acidic ‘because the salt contains hydrochloric acid’. HCl is not floating around as intact strong-acid molecules in an ordinary dilute solution of the salt. Track the dissolved ions and their actual reaction with water.
Why sodium ethanoate solution is alkaline
Sodium ethanoate, CH₃COONa, dissolves into Na⁺ and CH₃COO⁻. Sodium is a spectator in this simplified acid–base setting. Ethanoate is the conjugate base of the weak acid ethanoic acid, CH₃COOH, so it has a measurable tendency to take a proton from water.
CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq)
The generated hydroxide ions make the solution alkaline. Students sometimes argue that ethanoate cannot be a base because its chemical formula does not end in OH. Brønsted–Lowry basicity is about accepting a proton, not wearing an OH label.
The equation that joins Ka, Kb and Kw
For a conjugate acid–base pair at a fixed temperature, the relevant constants satisfy Ka × Kb = Kw. At 25 °C in standard school-level calculations, Kw ≈ 1.0 × 10⁻¹⁴. The constant belongs to the temperature: pKw is not automatically 14 at every temperature.
For NH₄⁺ and NH₃, the acid constant of NH₄⁺ is Kw divided by the base constant of NH₃. For CH₃COOH and CH₃COO⁻, the base constant of ethanoate is Kw divided by the acid constant of ethanoic acid. This is an algebraic bridge between a weak acid and its conjugate base, not a new unrelated formula to memorise.
Worked example 1: pH of 0.100 mol dm⁻³ NH₄Cl
Suppose Kb(NH₃) = 1.8 × 10⁻⁵ and Kw = 1.0 × 10⁻¹⁴ at 25 °C. Assume a 0.100 mol dm⁻³ NH₄Cl solution and the usual dilute-solution equilibrium approximation. Start with the chemistry, then the numbers.
- Identify the acid: NH₄⁺ is the species that can donate H⁺; Cl⁻ is a spectator.
- Find Ka(NH₄⁺): Kw / Kb(NH₃) = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰.
- Represent the equilibrium: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. If x is the equilibrium hydronium concentration, Ka = x²/(0.100 − x), ignoring water’s initial contribution where justified.
- Check the approximation: x is very small compared with 0.100, so Ka ≈ x²/0.100.
- Solve: [H₃O⁺] ≈ √(5.56 × 10⁻¹⁰ × 0.100) = 7.45 × 10⁻⁶ mol dm⁻³.
- Convert to pH: pH ≈ −log₁₀(7.45 × 10⁻⁶) = 5.13.
A useful reality check is that 5.13 is below 7, as predicted for ammonium chloride under these assumptions. A pH around 8.9 would signal that the wrong equilibrium or pH/pOH conversion was used. A calculator cannot rescue a chemically inverted setup.
Worked example 2: pH of 0.100 mol dm⁻³ sodium ethanoate
Suppose Ka(CH₃COOH) = 1.8 × 10⁻⁵ at 25 °C. Ethanoate is a weak base in water. Use the same concentration, 0.100 mol dm⁻³, so that we can compare like with like.
- Identify the base: CH₃COO⁻ accepts a proton from water and generates OH⁻.
- Find Kb(CH₃COO⁻): Kw / Ka(CH₃COOH) = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰.
- Set up the equilibrium: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, with Kb = x²/(0.100 − x).
- Approximate and solve: [OH⁻] ≈ √(5.56 × 10⁻¹⁰ × 0.100) = 7.45 × 10⁻⁶ mol dm⁻³.
- Find pOH: pOH = −log₁₀(7.45 × 10⁻⁶) ≈ 5.13.
- Find pH: pH + pOH = 14.00 at 25 °C, so pH ≈ 8.87.
Notice the pleasing symmetry: because we deliberately used the same 1.8 × 10⁻⁵ constant for NH₃ and ethanoic acid and the same salt concentration, we obtained the same magnitude of weak hydrolysis but on opposite sides of neutral. That numerical mirror is specific to the given values, not a general statement that every ammonium salt has the exact opposite pH of every carboxylate salt.
How to test whether the square-root approximation is justified
The familiar approximation x ≪ C should be checked, not treated as a law. In our NH₄⁺ example, x/C is about 7.45 × 10⁻⁵, far below 5%, so replacing (0.100 − x) by 0.100 is safe at the displayed precision. If the initial solution is extremely dilute, or the acid/base ionisation is relatively large, solve the quadratic or use the full relevant equilibrium conditions rather than mechanically applying √(KC).
At very low concentrations, water’s autoionisation may cease to be negligible. That is a modelling limit worth recognising, even if your school questions concentrate on cases where the standard approximation is warranted. A clear exam answer identifies assumptions explicitly when the data make them important.
When the salt contains both an acidic and a basic ion
Ammonium ethanoate illustrates a more demanding case: NH₄⁺ can donate a proton, and CH₃COO⁻ can accept one. You cannot classify it using a shortcut such as ‘all ammonium salts are acidic’ without considering the other ion. The direction of pH depends on the relative acid and base equilibria and the specific solution conditions. If Ka and Kb effects balance closely, the pH may be near neutral; do not assume this without evidence.
Salt hydrolysis is not the same thing as a buffer
A buffer typically contains appreciable quantities of a weak acid and its conjugate base, or a weak base and its conjugate acid. A solution prepared from NH₄Cl alone supplies mostly NH₄⁺ with very little NH₃ created by hydrolysis; it is not automatically an effective ammonia buffer. Sodium ethanoate alone is likewise not a standard ethanoic-acid/ethanoate buffer at a useful composition. Adding the appropriate weak conjugate partner changes the calculation and the resistance to added acid or alkali.
For the separate question of how paired species resist pH changes, read why buffer solutions resist pH changes. For why a titration equivalence point need not be pH 7, see the equivalence-point pH guide. These are related ideas, but each solves a different student mistake.
A diagnostic table for tutor feedback
| Student statement | What went wrong | Better question |
|---|---|---|
| ‘NaCl and NH₄Cl are both salts, so both are neutral.’ | Assumed neutralisation fixes product pH | Which ion can react with water? |
| ‘Cl⁻ is acidic because it comes from HCl.’ | Confused strong parent acid with the weak conjugate base | What is the conjugate acid–base strength relationship? |
| ‘Ethanoate contains no OH, so it cannot be a base.’ | Defined basicity by formula appearance | Can ethanoate accept a proton from water? |
| ‘Kb of NH₄⁺ is 1.8 × 10⁻⁵.’ | Attached NH₃’s Kb to the wrong species | Which species is the base in the named equilibrium? |
| ‘pOH = 5.13 so pH = 5.13.’ | Forgot pH + pOH at the stated temperature | Does the predicted direction match the calculated pH? |
| ‘Every acidic solution is a buffer.’ | Conflated pH with buffer capacity | Where is the appreciable conjugate partner? |
Five changed questions to prove real understanding
- Predict whether aqueous sodium nitrate is acidic, neutral or alkaline in the simplified strong-acid/strong-base model. Explain both ions.
- Write the hydrolysis equation for NH₄⁺ and identify its conjugate base.
- At 25 °C, calculate Kb(CH₃COO⁻) if Ka(CH₃COOH) = 2.0 × 10⁻⁵.
- At 25 °C, estimate the pH of 0.200 mol dm⁻³ sodium ethanoate using Ka(CH₃COOH) = 1.8 × 10⁻⁵.
- Explain why aqueous NH₄CH₃COO should not be classified from the name ‘ammonium’ alone.
Answer key
1. Approximately neutral: Na⁺ and NO₃⁻ are treated as spectator ions for these hydrolysis calculations. 2. NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺; conjugate base NH₃. 3. Kb = 1.0 × 10⁻¹⁴ / 2.0 × 10⁻⁵ = 5.0 × 10⁻¹⁰. 4. [OH⁻] ≈ √(5.56 × 10⁻¹⁰ × 0.200) = 1.05 × 10⁻⁵, pOH ≈ 4.98, so pH ≈ 9.02. 5. Because acetate also hydrolyses; the two tendencies must be compared.
A practical study route for JC1 and JC2
- Before lectures: revise strong versus weak acids and bases and identify conjugate pairs.
- After lectures: write the ionisation or hydrolysis equation with states before any arithmetic.
- Before tutorials: classify ten salts by the behaviour of each ion; mark the unusual cases.
- After corrections: solve one numerical NH₄⁺ example and one ethanoate example from a blank page.
- Two days later: do a changed concentration and check the direction of pH before calculator work.
- At exam preparation: mix salt hydrolysis with a buffer and a titration-equivalence question so chapter headings no longer provide the answer.
In my preferred tutoring check, the student explains the ion first, chooses Ka or Kb second, and only then presses log on the calculator. Small-group work can make that explanation visible, but class size by itself does not guarantee the misconception has been repaired.
What a Hougang parent should ask
Ask for a marked aqueous-equilibrium question, a brief diagnosis and a new unaided check. Does your JC student mix up the parent acid with its conjugate ion, confuse pH and pOH, apply a formula to the wrong species, or simply rush the logarithm? Those are four different teaching jobs. Discuss H2 syllabus experience, teaching location, timetable, journey and fit before enrolling anywhere; a Hougang-focused article does not itself confirm a physical Hougang classroom.
Why this matters beyond the A Levels
Conjugate-pair reasoning appears again in chemistry laboratories, drug formulation, environmental monitoring, aqueous materials, biochemical systems and chemical engineering. Different courses require different entry criteria, and no tuition method can guarantee university admission. Still, a student who can explain why an apparently plain salt changes the chemistry of water is learning to reason like a practising chemist: identify the species, write its interaction, choose a model, test its assumptions and compare the result with reality.
Common questions
Why does NH₄Cl lower the pH of water?
NH₄⁺ acts as a weak acid and produces H₃O⁺ through hydrolysis. Cl⁻ is largely a spectator in this acid–base model.
Why is sodium ethanoate alkaline?
The ethanoate ion accepts a proton from water, producing some OH⁻ and ethanoic acid.
Do all salts dissolve to give pH 7?
No. It depends on the hydrolysis of the dissolved ions, and some salts also involve more complex metal-ion chemistry.
Does every H2 Chemistry question use pKw = 14?
Only under the familiar 25 °C treatment where Kw is approximately 1.0 × 10⁻¹⁴. Use the temperature and data given.
Can I use Henderson–Hasselbalch for a single salt solution?
Do not apply a buffer-ratio equation mechanically when there is no appreciable conjugate pair initially. Start from the relevant hydrolysis equilibrium.
Continue learning: How buffer solutions resist a pH change · Why equivalence-point pH may not be 7 · Hougang Chemistry Library · SEAB A-Level syllabus directory. For current H2 tuition enquiries, contact eduKate Singapore.