ChemistryCBSE Class 11 · Equilibrium

Ionic Equilibrium: pH, Weak Electrolytes, Buffers and Solubility Product Class 11

Updated 17 October 2026 · Fact-checked

Strong electrolytes ionise completely, so pH follows directly from the concentration. For a weak acid HA, [H⁺] = √(Ka·C) when ionisation is small, and the degree of ionisation is α = √(Ka/C). A common ion suppresses ionisation, a mixture of a weak acid and its salt resists changes in pH (pH = pKa + log [salt]/[acid]), and a sparingly soluble salt obeys its solubility product Ksp, which decides whether a precipitate forms.

A⁻ absorbs added H⁺ and HA absorbs added OH⁻, so the ratio [A⁻]/[HA] and the pH change very little.

Illustration drawn for QuestionBuddy.

Understand Ionic Equilibrium: pH, Weak Electrolytes, Buffers and Solubility Product

Water and pH. $K_{w} = [H^{+}] [O H^{-}] = 1.0 \times 1 0^{- 14}$ at 298 K, so $p H + pO H = 14$ . $p H = - lo g [H^{+}]$ . Dilute strong acids below about $1 0^{- 6}$ M need the contribution of water included.

Weak acid and weak base. For $H A ⇌ H^{+} + A^{-}$ at concentration $C$ with degree of ionisation $α$ , $K_{a} = \frac{C α ^{2}}{1 - α} \approx C α^{2}$ , so $α = K_{a} / C$ (Ostwald dilution law) and $[H^{+}] = C α = K_{a} C$ . The approximation holds when $α$ is below about 5%; otherwise solve the quadratic. For a weak base, replace $K_{a}$ by $K_{b}$ and $[H^{+}]$ by $[O H^{-}]$ . For a conjugate pair, $K_{a} K_{b} = K_{w}$ .

Common ion effect. A salt that supplies an ion already present in a weak electrolyte pushes the ionisation back. For a weak acid with added salt of concentration $C_{s}$ , $[H^{+}] \approx K_{a} \frac{C _{a}}{C _{s}}$ .

Hydrolysis. Salt of a strong base and weak acid: basic, $K_{h} = K_{w} / K_{a}$ , $p H = 7 + \frac{1}{2} p K_{a} + \frac{1}{2} lo g C$ . Salt of a weak base and strong acid: acidic, $p H = 7 - \frac{1}{2} p K_{b} - \frac{1}{2} lo g C$ . Salt of a strong acid and strong base: neutral.

Buffers. A weak acid with its salt (or a weak base with its salt) resists a change in pH on adding small amounts of acid or base: $p H = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}$ (Henderson–Hasselbalch). The buffer works best when the ratio is near 1, that is $p H \approx p K_{a}$ . A strong acid added converts salt to acid; a strong base converts acid to salt.

Solubility product. For $A_{x} B_{y} (s) ⇌ x A^{y +} + y B^{x -}$ , $K_{s p} = [A^{y +}]^{x} [B^{x -}]^{y}$ at equilibrium. Precipitation occurs when the ionic product $Q > K_{s p}$ ; no precipitate if $Q < K_{s p}$ . A common ion lowers the solubility.

Ionic Equilibrium: pH, Weak Electrolytes, Buffers and Solubility Product: formulas and key facts

Ionic product of water: $K_{w} = [H^{+}] [O H^{-}] = 1.0 \times 1 0^{- 14}, p H + pO H = 14$; At 298 K.
Weak acid: $α = \frac{K _{a}}{C}, [H^{+}] = K_{a} C$; When α is small (below about 5%).
Conjugate pair: $K_{a} \times K_{b} = K_{w}$; For an acid and its conjugate base.
Hydrolysis of a salt of a weak acid: $p H = 7 + \frac{1}{2} p K_{a} + \frac{1}{2} lo g C$; Strong base with weak acid.
Henderson–Hasselbalch: $p H = p K_{a} + lo g \frac{[ salt ]}{[ acid ]}$; Acidic buffer.
Solubility product of A₂B or AB₂: $K_{s p} = 4 s^{3}$; s is the molar solubility in pure water.

How to solve Ionic Equilibrium: pH, Weak Electrolytes, Buffers and Solubility Product questions

Decide first whether the species is strong or weak, and whether a common ion or a buffer is present. Then pick the right shortcut and check the approximation.

1Strong acid or base: take the concentration as [H⁺] or [OH⁻]; convert with pH + pOH = 14.
2Weak acid or base: use [H⁺] = √(Ka C) and check that α is below about 5%.
3Common ion present: use [H⁺] = Ka × [acid]/[salt] or the Henderson–Hasselbalch equation.
4Buffer plus added acid or base: update the moles of acid and salt first, then use the equation.
5Sparingly soluble salt: write Ksp in terms of s, include the common-ion concentration, and compare Q with Ksp for precipitation.

Ionic Equilibrium: pH, Weak Electrolytes, Buffers 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.

Antacid and stomach acid

Suppose a stomach holds 200 mL of a gastric fluid with [H⁺] = 0.050 mol/L and an antacid tablet neutralises 0.0060 mol of H⁺.

Find the pH before and after, ignoring any change in volume.

Try it first, then show the solutionSolution

Before: pH = −log 0.050 = 1.30.
Moles of H⁺ initially = 0.050 × 0.200 = 0.0100 mol; after = 0.0040 mol; [H⁺] = 0.0040/0.200 = 0.020 mol/L; pH = 1.70.

Answer: pH changes from 1.30 to 1.70.

Memory hook: pH moves only by 0.4 although most of the acid is gone, because pH is a logarithm.

A buffer in a blood-like fluid

Suppose a fluid is buffered by a weak acid of pKa = 6.10 with equal concentrations of the acid and its salt.

Find the pH, and the pH if the salt-to-acid ratio becomes 10 to 1.

Try it first, then show the solutionSolution

Equal concentrations: pH = pKa + log 1 = 6.10.
Ratio 10: pH = 6.10 + log 10 = 7.10.

Answer: 6.10, and 7.10 at a ratio of 10.

Memory hook: A tenfold change in the ratio moves the pH by exactly 1.

Tricks to remember Ionic Equilibrium: pH, Weak Electrolytes, Buffers and Solubility Product

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

Sum to 14
pH + pOH = 14 and pKa + pKb = 14 (conjugates).
Use it when: Switching between acid and base.
Salt on top
In Henderson–Hasselbalch the salt (base form) goes on top: pH = pKa + log(salt/acid).
Use it when: Buffer numericals.
Strong wins the hydrolysis
The salt's solution takes the nature of the stronger parent: weak acid with strong base is basic.
Use it when: Acidic, basic or neutral salts.
Q against Ksp
Q > Ksp precipitates; Q < Ksp stays dissolved.
Use it when: Precipitation questions.

Ionic Equilibrium: pH, Weak Electrolytes, Buffers and Solubility Product: solved examples

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

Example 1: pH of a strong base · 2 marks · Very short answerBoard level

Find the pH of 0.0050 M Ba(OH)₂ at 298 K, assuming complete ionisation.

Try it first, then show the solutionSolution

Each formula unit gives two hydroxide ions: [OH⁻] = 2 × 0.0050 = 0.010 M. (1)
pOH = 2, so pH = 14 − 2 = 12. (1)

Answer: pH = 12.

Example 2: Degree of ionisation of a weak acid · 3 marks · Short answerHarder board level

A weak monobasic acid has Ka = 1.8 × 10⁻⁵. Find [H⁺], the degree of ionisation and the pH of its 0.050 M solution. Check that the approximation is valid.

Try it first, then show the solutionSolution

[H⁺] = √(Ka C) = √(1.8 × 10⁻⁵ × 0.050) = √(9.0 × 10⁻⁷) = 9.5 × 10⁻⁴ M. (1)
α = 9.5 × 10⁻⁴/0.050 = 0.019, that is 1.9%, which is below 5%, so the approximation holds. (1)
pH = −log (9.49 × 10⁻⁴) = 3.02. (1)

Answer: [H⁺] = 9.5 × 10⁻⁴ M; α = 1.9%; pH = 3.02.

Example 3: Common ion effect on a weak base · 3 marks · Short answerHarder board level

A weak base B has Kb = 2.0 × 10⁻⁵. Find the pH of (a) 0.10 M B and (b) a solution containing 0.10 M B and 0.050 M of its chloride salt BHCl.

Try it first, then show the solutionSolution

(a) [OH⁻] = √(Kb C) = √(2.0 × 10⁻⁶) = 1.41 × 10⁻³ M; pOH = 2.85; pH = 11.15. (1)
(b) The common ion BH⁺ suppresses ionisation: [OH⁻] = Kb × [B]/[BH⁺] = 2.0 × 10⁻⁵ × 0.10/0.050 = 4.0 × 10⁻⁵ M. (1)
pOH = 4.40 and pH = 9.60, so the pH falls by about 1.55 units. (1)

Answer: (a) pH = 11.15; (b) pH = 9.60.

Example 4: Buffer action on adding a strong acid · 4 marks · Long answerHarder board level

100 mL of a buffer contains 0.20 M CH₃COOH and 0.20 M CH₃COONa (pKa of the acid = 4.74). 10 mL of 0.10 M HCl is added. (a) Find the pH of the buffer before and after. (b) Find the pH if the same HCl were added to 100 mL of pure water.

Try it first, then show the solutionSolution

(a) Before: equal concentrations, so pH = pKa = 4.74. (1)
Moles: acid 0.020, salt 0.020; HCl adds 0.0010 mol H⁺, which converts 0.0010 mol of acetate to acid: acid 0.021 mol, salt 0.019 mol. (1)
After: pH = 4.74 + log (0.019/0.021) = 4.74 − 0.043 = 4.70. (1)
(b) In water: [H⁺] = 0.0010 mol/0.110 L = 9.1 × 10⁻³ M, so pH = 2.04, a drop of nearly 5 units. (1)

Answer: (a) 4.74 to 4.70; (b) 2.04.

Example 5: Solubility product and the common ion effect · 4 marks · Long answerHOTS / competency

Ag₂CrO₄ has Ksp = 4.0 × 10⁻¹² (data assumed for the problem). Find (a) its molar solubility in pure water, (b) its molar solubility in 0.010 M Na₂CrO₄, and (c) the [Ag⁺] needed to start precipitation from 0.010 M chromate. Explain the change from (a) to (b).

Try it first, then show the solutionSolution

(a) Ksp = [Ag⁺]²[CrO₄²⁻] = (2s)²(s) = 4s³, so s³ = 1.0 × 10⁻¹² and s = 1.0 × 10⁻⁴ M. (1)
(b) [CrO₄²⁻] ≈ 0.010 M, so (2s)²(0.010) = 4.0 × 10⁻¹², 4s² = 4.0 × 10⁻¹⁰, s = 1.0 × 10⁻⁵ M. (1)
(c) Precipitation begins when [Ag⁺]² × 0.010 = 4.0 × 10⁻¹², so [Ag⁺] = √(4.0 × 10⁻¹⁰) = 2.0 × 10⁻⁵ M. (1)
The chromate already present shifts the solubility equilibrium to the left, reducing the solubility tenfold. (1)

Answer: (a) 1.0 × 10⁻⁴ M; (b) 1.0 × 10⁻⁵ M; (c) 2.0 × 10⁻⁵ M; the common ion lowers solubility.

Where students lose marks in Ionic Equilibrium: pH, Weak Electrolytes, Buffers and Solubility Product

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

Mistake: Forgetting that Ba(OH)₂ or Ca(OH)₂ gives two OH⁻ per formula unit.
Using the formula concentration as [OH⁻].
Do this: Multiply by the number of hydroxide ions.
Usually costs: The pH value.
Mistake: Using [H⁺] = √(Ka C) when α is large.
Approximation not checked.
Do this: Check α < 5%; otherwise solve x² + Ka x − Ka C = 0.
Usually costs: Accuracy marks.
Mistake: Inverting the ratio in Henderson–Hasselbalch.
Salt and acid reversed.
Do this: pH = pKa + log (salt/acid).
Usually costs: A buffer numerical.
Mistake: Ignoring the common-ion concentration in a Ksp problem.
Using s alone for the ion also present in solution.
Do this: Add the common-ion concentration, then check that s is small enough to neglect.
Usually costs: The solubility.

Exam tips

State the approximation (α is small) and verify it in a line.
Write Ksp in terms of s before substituting.
For buffers update the moles of acid and salt first, then take the ratio.
Give pH to two decimal places and units to concentrations.
Explain hydrolysis by naming the ion that reacts with water.

Practice questions on Ionic Equilibrium: pH, Weak Electrolytes, Buffers 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
Which of the following mixtures forms an acidic buffer?
(a) NaOH + NaCl (b) HCl + NaCl (c) CH₃COOH + CH₃COONa (d) NH₄Cl + HCl
Hint
A weak acid with its salt.
Answer
(c).
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): An acetic acid and sodium acetate buffer has its greatest capacity when [salt] = [acid].
Reason (R): At that ratio pH = pKa, and the buffer can neutralise added acid and added base to the same extent.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Equal acid and salt means pH = pKa.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
Find the pH of 0.020 M NaOH at 298 K. What is the pH of the solution after it is diluted tenfold?
Hint
[OH⁻] = 0.020 M.
Answer
pOH = 1.70, so pH = 12.30. After tenfold dilution [OH⁻] = 0.0020 M, pOH = 2.70 and pH = 11.30.
Q4 · 3 marks · Short answerHarder board level
A weak acid with Ka = 1.6 × 10⁻⁵ is present at 0.040 M. (a) Find the degree of ionisation and the pH. (b) At what concentration would its degree of ionisation be 4.0%?
Hint
α = √(Ka/C), so C = Ka/α².
Answer
(a) [H⁺] = √(1.6 × 10⁻⁵ × 0.040) = 8.0 × 10⁻⁴ M, α = 0.020 (2.0%), pH = 3.10. (b) C = Ka/α² = 1.6 × 10⁻⁵/(0.04)² = 0.010 M.
Q5 · 4 marks · Long answerHOTS / competency
Ka of acetic acid is 1.8 × 10⁻⁵ and the ionic product of water is 1.0 × 10⁻¹⁴. (a) Find the pH of 0.10 M sodium acetate. (b) Find the pH of 0.10 M ammonium chloride, taking Kb of ammonia as 1.8 × 10⁻⁵. (c) Explain why the two pH values are on opposite sides of 7.
Hint
Kh = Kw/Ka or Kw/Kb, then h = √(Kh/C).
Answer
(a) Kh = Kw/Ka = 5.6 × 10⁻¹⁰, h = √(5.6 × 10⁻¹⁰/0.10) = 7.5 × 10⁻⁵, [OH⁻] = 7.5 × 10⁻⁶ M, pOH = 5.13, pH = 8.87. (b) By the same working with Kb, [H⁺] = 7.5 × 10⁻⁶ M, pH = 5.13. (c) Acetate is the conjugate base of a weak acid and takes H⁺ from water (basic); NH₄⁺ is the conjugate acid of a weak base and donates H⁺ (acidic).

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.Kw at 298 K?Show answer
1.0 × 10⁻¹⁴.
2.Degree of ionisation of a weak acid?Show answer
α = √(Ka/C).
3.Henderson–Hasselbalch equation?Show answer
pH = pKa + log (salt/acid).
4.Relation between Ka and Kb of a conjugate pair?Show answer
Ka × Kb = Kw.
5.Condition for precipitation?Show answer
Ionic product Q > Ksp.
6.Ksp of A₂B in terms of s?Show answer
4s³.

Ionic Equilibrium: pH, Weak Electrolytes, Buffers and Solubility Product: frequently asked questions

Why does a buffer resist change in pH?

The acid component neutralises added base and the salt component neutralises added acid, so the ratio changes only slightly.

When is a buffer most effective?

When [salt] and [acid] are about equal, so that pH is close to pKa.

Does diluting a buffer change its pH?

Very little, because the ratio is unchanged; the buffer capacity falls.

Why is a solution of Na₂CO₃ basic?

The carbonate ion, the conjugate base of a weak acid, takes H⁺ from water and leaves excess OH⁻.

Can two salts with the same Ksp have the same solubility?

Only if their formulas have the same ion ratio; otherwise the power of s in Ksp differs.