ChemistryISC Class 11 · Ionic Equilibrium

Buffer Solutions, Hydrolysis of Salts and Solubility Product

Q: Why does a buffer resist a change in pH?

It contains both a species that consumes added acid and one that consumes added base, so [H⁺] changes only slightly.

Q: Is a high Ksp always a high solubility?

Only when comparing salts with the same ion ratio. AgCl and AgBr compare directly; AgCl and Ag₂CrO₄ do not, because the formulae differ.

Q: Why does the solution of Na₂CO₃ turn red litmus blue?

The carbonate ion hydrolyses to give OH⁻, making the solution basic.

Updated 17 October 2026 · Fact-checked

A buffer resists a change in pH and obeys pH = pKa + log([salt]/[acid]). Salts of weak acids or bases hydrolyse, giving basic or acidic solutions with Kh = Kw/Ka. A sparingly soluble salt has a solubility product Ksp = product of ion concentrations; precipitation occurs when the ionic product exceeds Ksp.

Notice that the salt anion A⁻ mops up added acid and the weak acid HA mops up added base, so the ratio, and pH, barely move.

Illustration drawn for QuestionBuddy.

Understand Buffers, Salt Hydrolysis and Solubility Product

Common ion effect. Adding an ion common to a weak electrolyte suppresses its ionisation (for example, CH₃COONa added to CH₃COOH lowers [H⁺]), and adding a common ion lowers the solubility of a sparingly soluble salt.

Buffer solutions. An acidic buffer is a weak acid with its salt (CH₃COOH + CH₃COONa); a basic buffer is a weak base with its salt (NH₄OH + NH₄Cl). Added acid is consumed by the salt anion (A⁻ + H⁺ → HA) and added base by the weak acid (HA + OH⁻ → A⁻ + H₂O), so the pH changes very little. The Henderson-Hasselbalch equation is pH = pKa + log([salt]/[acid]) for an acidic buffer, and pOH = pKb + log([salt]/[base]) for a basic one. A buffer works best when the ratio is close to 1, that is pH ≈ pKa.

Hydrolysis of salts. Strong acid + strong base (NaCl): no hydrolysis, pH 7. Weak acid + strong base (CH₃COONa): anion hydrolyses, solution basic, Kh = Kw/Ka, [OH⁻] = √(Kh c). Strong acid + weak base (NH₄Cl): cation hydrolyses, acidic, Kh = Kw/Kb, [H⁺] = √(Kh c). Weak acid + weak base: depends on Ka and Kb.

Solubility product. For a sparingly soluble salt AₓB_y ⇌ xAʸ⁺ + yBˣ⁻, Ksp = [Aʸ⁺]ˣ[Bˣ⁻]ʸ at saturation. For AB: Ksp = s²; for AB₂ or A₂B: Ksp = 4s³. If the ionic product Q > Ksp, the salt precipitates; if Q < Ksp, the solution is unsaturated. In a solution with a common ion, the solubility falls: for AgCl in c M Cl⁻, s = Ksp/c.

Indicators. The colour change of an indicator takes place in the pH range of about pKᵢₙ ± 1; the indicator for a titration is chosen so that the equivalence-point pH falls in its range.

Buffers, Salt Hydrolysis and Solubility Product: formulas and key facts

Henderson equation: $p H = p K a + l o g ([s a l t] / [a c i d])$; Basic buffer: pOH = pKb + log([salt]/[base]).
Hydrolysis constant: $K h = K w / K a (ani o n) ∣ K h = K w / K b (c a t i o n)$; [OH⁻] or [H⁺] = √(Kh c).
Solubility product: $A B : K s p = s^{2} ∣ A B_{2} : K s p = 4 s^{3}$; s = molar solubility.
Precipitation: Q > Ksp: precipitate | Q < Ksp: no precipitate; Q = ionic product at the given concentrations.

Buffers, Salt Hydrolysis and Solubility Product in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Keeping a solution at a steady pH

Suppose a lab prepares 1 L of a buffer from 0.10 mol acetic acid (Ka = 1.8 × 10⁻⁵, pKa = 4.74) and 0.10 mol sodium acetate.

Find its pH and the pH after adding 0.01 mol of HCl.

Try it first, then show the solutionSolution

pH = pKa + log(0.10/0.10) = 4.74.
HCl converts 0.01 mol acetate to acetic acid: acid = 0.11, salt = 0.09.
pH = 4.74 + log(0.09/0.11) = 4.74 − 0.087 = 4.65.

Answer: 4.74; 4.65 after adding the acid.

Memory hook: The pH changed by only about 0.09 unit.

Tricks to remember Buffers, Salt Hydrolysis and Solubility Product

Short memory aids. Say each one aloud once, then use it on a question the same day.

Salt over acid
Henderson: pH = pKa + log(base form/acid form).
Use it when: Buffer calculations.
Strong decides
The strong parent of a salt decides the nature: strong base gives basic, strong acid gives acidic.
Use it when: Acidic or basic salt questions.

Buffers, Salt Hydrolysis and Solubility Product: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Composition of a buffer · 1 mark · MCQBoard level

Which pair of substances makes an acidic buffer?
(a) HCl and NaCl
(b) CH₃COOH and CH₃COONa
(c) NH₄OH and NH₄Cl
(d) NaOH and HCl

Try it first, then show the solutionSolution

A weak acid with its conjugate base (salt) can neutralise small additions of acid or base.

Answer: (b) CH₃COOH and CH₃COONa

Example 2: Henderson equation · 3 marks · Short answerBoard level

Calculate the pH of a buffer containing 0.10 M CH₃COOH and 0.10 M CH₃COONa (Ka = 1.8 × 10⁻⁵), and the pH after 0.01 mol of HCl is added to 1 L of it.

Try it first, then show the solutionSolution

pKa = −log(1.8 × 10⁻⁵) = 4.74. pH = 4.74 + log 1 = 4.74. (1 mark)
After HCl: salt 0.09 M, acid 0.11 M. (1 mark)
pH = 4.74 + log(0.09/0.11) = 4.74 − 0.09 = 4.65. (1 mark)

Answer: pH = 4.74; after the acid, pH = 4.65.

Example 3: Hydrolysis of sodium acetate · 4 marks · Short answerHarder board level

A bottle holds 0.050 M sodium acetate solution. Find its pH (Ka of acetic acid = 1.8 × 10⁻⁵) and explain why it is not neutral.

Try it first, then show the solutionSolution

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, so OH⁻ is produced. (1 mark)
Kh = Kw/Ka = 10⁻¹⁴/1.8 × 10⁻⁵ = 5.56 × 10⁻¹⁰. (1 mark)
[OH⁻] = √(Kh c) = √(2.78 × 10⁻¹¹) = 5.27 × 10⁻⁶ M; pOH = 5.28. (1 mark)
pH = 14 − 5.28 = 8.72 (basic). (1 mark)

Answer: pH = 8.72

Example 4: Solubility and common ion effect · 3 marks · Short answerHarder board level

The Ksp of AgCl is 1.8 × 10⁻¹⁰ at 298 K. Find its molar solubility in pure water and in 0.10 M NaCl.

Try it first, then show the solutionSolution

In water: s = √Ksp = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M. (1 mark)
In 0.10 M NaCl: [Cl⁻] ≈ 0.10, so s = Ksp/0.10 = 1.8 × 10⁻⁹ M. (1 mark)
The solubility falls about 7500 times because of the common ion Cl⁻. (1 mark)

Answer: 1.34 × 10⁻⁵ M in water; 1.8 × 10⁻⁹ M in 0.10 M NaCl.

Example 5: Magnesium hydroxide · 5 marks · Long answerHOTS / competency

The Ksp of Mg(OH)₂ is 1.2 × 10⁻¹¹. Find (i) its solubility in water, (ii) the pH of its saturated solution, and (iii) its solubility in 0.010 M MgCl₂.

Try it first, then show the solutionSolution

Mg(OH)₂ ⇌ Mg²⁺ + 2OH⁻: Ksp = s(2s)² = 4s³. (1 mark)
(i) s = (1.2 × 10⁻¹¹/4)^(1/3) = (3.0 × 10⁻¹²)^(1/3) = 1.44 × 10⁻⁴ M. (1 mark)
(ii) [OH⁻] = 2s = 2.88 × 10⁻⁴; pOH = 3.54; pH = 10.46. (1 mark)
(iii) With [Mg²⁺] = 0.010: [OH⁻]² = Ksp/0.010 = 1.2 × 10⁻⁹, so [OH⁻] = 3.46 × 10⁻⁵ M. (1 mark)
s = [OH⁻]/2 = 1.7 × 10⁻⁵ M, about eight times lower than in water. (1 mark)

Answer: s = 1.44 × 10⁻⁴ M; pH = 10.46; s = 1.7 × 10⁻⁵ M in 0.010 M MgCl₂.

Where students lose marks in Buffers, Salt Hydrolysis and Solubility Product

The slips examiners see most often, and what to write instead.

Mistake: Reversing the Henderson ratio
Writing log([acid]/[salt]).
Do this: It is log of salt (base form) over acid.
Usually costs: 2 marks.
Mistake: Wrong expression of Ksp for AB₂
Using Ksp = s³.
Do this: Ksp = s × (2s)² = 4s³.
Usually costs: 2 marks.
Mistake: Ignoring the common ion
Using s = √Ksp in a solution containing the common ion.
Do this: Use the total concentration of the common ion.
Usually costs: 1 to 2 marks.

Exam tips

Write the equilibrium or hydrolysis equation first.
State the Henderson equation and the assumption that the salt and acid concentrations are the equilibrium ones.
Compare Q with Ksp in precipitation questions.
Use Kh = Kw/Ka for salts of weak acids.

Practice questions on Buffers, Salt Hydrolysis and Solubility Product

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The aqueous solution of NH₄Cl is:
(a) acidic
(b) basic
(c) neutral
(d) amphoteric
Hint
It is a salt of a strong acid and a weak base.
Answer
(a) acidic
Q2 · 2 marks · Short answerBoard level
State whether the solutions of NaCl, Na₂CO₃ and NH₄Cl are acidic, basic or neutral.
Hint
Identify the parent acid and base.
Answer
NaCl: neutral; Na₂CO₃: basic (weak acid H₂CO₃, strong base NaOH); NH₄Cl: acidic.
Q3 · 3 marks · Short answerHarder board level
What ratio of [CH₃COONa]/[CH₃COOH] gives a buffer of pH 5.0? (pKa = 4.74; antilog 0.26 = 1.82)
Hint
Rearrange the Henderson equation.
Answer
log(salt/acid) = 5.0 − 4.74 = 0.26, so the ratio = 1.82.
Q4 · 3 marks · Short answerHarder board level
The Ksp of PbI₂ is 7.1 × 10⁻⁹. Find its molar solubility and the concentration of I⁻ in the saturated solution.
Hint
PbI₂ ⇌ Pb²⁺ + 2I⁻; Ksp = 4s³.
Answer
s = (7.1 × 10⁻⁹/4)^(1/3) = (1.775 × 10⁻⁹)^(1/3) = 1.21 × 10⁻³ M; [I⁻] = 2s = 2.4 × 10⁻³ M.
Q5 · 3 marks · Short answerHarder board level
Find the pH of 0.10 M NH₄Cl (Kb of ammonia = 1.8 × 10⁻⁵).
Hint
Kh = Kw/Kb; [H⁺] = √(Kh c).
Answer
Kh = 5.56 × 10⁻¹⁰; [H⁺] = 7.45 × 10⁻⁶; pH = 5.13.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a model body-fluid buffer is H₂CO₃/HCO₃⁻ with pKa = 6.1.
(i) Find the ratio [HCO₃⁻]/[H₂CO₃] for pH 7.4. (1 mark)
(ii) Find the pH if the ratio falls to 10. (1 mark)
(iii) Explain how this buffer handles added acid. (1 mark)
(iv) Would it work as a buffer at pH 3? (1 mark)
Hint
pH = pKa + log(ratio); antilog 1.3 = 20.
Answer
(i) log ratio = 7.4 − 6.1 = 1.3, ratio = 20. (ii) pH = 6.1 + log 10 = 7.1. (iii) Added H⁺ is consumed by HCO₃⁻ to form H₂CO₃, so the ratio changes only slightly. (iv) No: pH 3 is far from the pKa (6.1) and the ratio would be about 10⁻³·¹, so it would have almost no base form to neutralise acid.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Henderson equation?Show answer
pH = pKa + log([salt]/[acid]).
2.Kh for CH₃COONa?Show answer
Kw/Ka.
3.Ksp of AB₂?Show answer
4s³.
4.Condition for precipitation?Show answer
Ionic product exceeds Ksp.

Buffers, Salt Hydrolysis and Solubility Product: frequently asked questions

Why does a buffer resist a change in pH?

It contains both a species that consumes added acid and one that consumes added base, so [H⁺] changes only slightly.

Is a high Ksp always a high solubility?

Only when comparing salts with the same ion ratio. AgCl and AgBr compare directly; AgCl and Ag₂CrO₄ do not, because the formulae differ.

Why does the solution of Na₂CO₃ turn red litmus blue?

The carbonate ion hydrolyses to give OH⁻, making the solution basic.