Ionic Equilibrium ISC Class 11: pH, Buffers and Solubility Product

What Ionic Equilibrium is about

Weightage: Ionic Equilibrium belongs to unit 6 of the ISC Class XI Chemistry syllabus. CISCE weights units 1-6 together (Physical Chemistry) at 41 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Chemistry syllabus, examination year 2028, Class XI unit weightage

Key concepts

Electrolytes and degree of ionisation

Electrolytes conduct electricity in solution or when molten because they produce ions; non-electrolytes (sugar, urea, ethanol) do not. Strong electrolytes (strong acids such as HCl, HNO₃, H₂SO₄; strong bases such as NaOH, KOH; nearly all salts) are almost completely ionised. Weak electrolytes (CH₃COOH, HCN, NH₄OH/NH₃(aq)) are partly ionised, and an equilibrium exists between ions and un-ionised molecules. The degree of ionisation α = (moles ionised)/(moles dissolved); it increases with dilution and with temperature. Acid strength in water is measured by Ka (or pKa = −log Ka): the larger Ka, the stronger the acid.

Worked example

Classify as strong electrolyte, weak electrolyte or non-electrolyte: KNO₃, CH₃COOH, glucose, NH₃(aq), HCl.

Try it first, then show the solutionSolution

1. Strong: KNO₃ (a salt), HCl (strong acid)
2. Weak: CH₃COOH, NH₃(aq)
3. Non-electrolyte: glucose

Answer: Strong: KNO₃, HCl; weak: CH₃COOH, NH₃(aq); non-electrolyte: glucose

Arrhenius, Brønsted-Lowry and Lewis concepts

Arrhenius: an acid gives H⁺ (H₃O⁺) in water, a base gives OH⁻; limited to aqueous solutions and cannot explain NH₃ as a base. Brønsted-Lowry: an acid is a proton donor, a base a proton acceptor. After donating H⁺, an acid becomes its conjugate base; a base becomes its conjugate acid (HCl/Cl⁻, NH₄⁺/NH₃, H₂O/OH⁻, H₃O⁺/H₂O). A strong acid has a weak conjugate base. Water is amphoteric (amphiprotic). Lewis: an acid is an electron-pair acceptor (BF₃, AlCl₃, H⁺, metal ions such as Cu²⁺), a base an electron-pair donor (NH₃, H₂O, OH⁻, F⁻); this covers reactions without protons.

Worked example

In NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, identify the Brønsted acids and bases and the conjugate pairs. Classify BF₃ and NH₃ in the Lewis sense.

Try it first, then show the solutionSolution

1. H₂O donates H⁺: acid; NH₃ accepts it: base
2. Conjugate pairs: NH₄⁺/NH₃ and H₂O/OH⁻
3. BF₃ (incomplete octet) accepts a pair: Lewis acid; NH₃ (lone pair) donates: Lewis base

Answer: Acids H₂O, NH₄⁺; bases NH₃, OH⁻; BF₃ Lewis acid, NH₃ Lewis base

Ostwald's dilution law

For a weak acid HA ⇌ H⁺ + A⁻ at concentration C with degree of ionisation α: at equilibrium [HA] = C(1 − α), [H⁺] = [A⁻] = Cα, so Ka = Cα²/(1 − α). For weak acids α ≪ 1, so Ka ≈ Cα², giving α = √(Ka/C) and [H⁺] = √(KaC). The law shows that α increases as the solution is diluted (C decreases). The same holds for weak bases with Kb and [OH⁻] = √(KbC). The approximation fails for fairly strong electrolytes or very dilute solutions.

Worked example

Find α, [H⁺] and pH of 0.10 M acetic acid (Ka = 1.8 × 10⁻⁵).

Try it first, then show the solutionSolution

1. α = √(1.8 × 10⁻⁵/0.10) = √(1.8 × 10⁻⁴) = 0.0134 (1.34%)
2. [H⁺] = Cα = 1.34 × 10⁻³ M
3. pH = −log(1.34 × 10⁻³) = 2.87

Answer: α = 0.0134; [H⁺] = 1.34 × 10⁻³ M; pH = 2.87

Ionic product of water, pH and pOH

Water self-ionises: H₂O ⇌ H⁺ + OH⁻, with Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K (Kw rises with temperature). pH = −log[H⁺], pOH = −log[OH⁻], pH + pOH = pKw = 14 at 298 K. Neutral: [H⁺] = [OH⁻], pH 7; acidic: pH < 7; basic: pH > 7. For strong acids and bases take full ionisation, remembering the number of H⁺ or OH⁻ per formula unit. For very dilute acids (below about 10⁻⁶ M), the H⁺ from water must be included.

Worked example

Find the pH of 0.0050 M Ba(OH)₂ and of 0.0010 M HCl.

Try it first, then show the solutionSolution

1. Ba(OH)₂ gives 2 OH⁻: [OH⁻] = 0.010 M, pOH = 2.0, pH = 12.0
2. HCl: [H⁺] = 1.0 × 10⁻³ M, pH = 3.0

Answer: 12.0; 3.0

Polybasic acids and multistage ionisation

Acids with more than one ionisable H ionise in steps, each with its own constant: H₂S ⇌ H⁺ + HS⁻ (Ka₁), HS⁻ ⇌ H⁺ + S²⁻ (Ka₂); H₃PO₄ has three steps. Always Ka₁ > Ka₂ > Ka₃, because removing H⁺ from a negative ion is harder. So almost all the H⁺ comes from the first step, and [H⁺] ≈ √(Ka₁C). For a dibasic acid like H₂S, the concentration of the second anion is approximately Ka₂ when no other acid is present; adding a strong acid lowers [S²⁻] sharply.

Worked example

For 0.10 M H₂S (Ka₁ = 1.0 × 10⁻⁷, Ka₂ = 1.2 × 10⁻¹³), find [H⁺] and [S²⁻].

Try it first, then show the solutionSolution

1. First step dominates: [H⁺] = √(1.0 × 10⁻⁷ × 0.10) = 1.0 × 10⁻⁴ M; [HS⁻] ≈ [H⁺]
2. Second step: Ka₂ = [H⁺][S²⁻]/[HS⁻]; since [H⁺] ≈ [HS⁻], [S²⁻] ≈ Ka₂

Answer: [H⁺] = 1.0 × 10⁻⁴ M; [S²⁻] ≈ 1.2 × 10⁻¹³ M

Common ion effect and its use in salt analysis

Adding a strong electrolyte that shares an ion with a weak electrolyte suppresses the weak electrolyte's ionisation (Le Chatelier). Examples: CH₃COONa added to CH₃COOH lowers [H⁺]; NH₄Cl added to NH₄OH lowers [OH⁻]. In qualitative analysis: Group II: H₂S is passed in the presence of dilute HCl; the extra H⁺ lowers [S²⁻], so only sulphides with very low Ksp (CuS, PbS, CdS, As₂S₃) precipitate. Group III: NH₄OH with NH₄Cl gives a low [OH⁻], enough to precipitate Fe(OH)₃, Al(OH)₃ and Cr(OH)₃ (very low Ksp) but not Zn, Mn, Ni, Co or Mg hydroxides. Group IV: H₂S in ammoniacal solution: OH⁻ removes H⁺, [S²⁻] rises, and ZnS, MnS, NiS, CoS (higher Ksp) precipitate.

Worked example

Find the pH of 0.10 M acetic acid containing 0.10 M sodium acetate (Ka = 1.8 × 10⁻⁵), and compare with 0.10 M acetic acid alone (pH 2.87).

Try it first, then show the solutionSolution

1. Ka = [H⁺][CH₃COO⁻]/[CH₃COOH]; acetate comes almost entirely from the salt: [CH₃COO⁻] ≈ 0.10
2. [H⁺] = Ka × 0.10/0.10 = 1.8 × 10⁻⁵ M
3. pH = 4.74

Answer: pH 4.74: [H⁺] falls about 75-fold

Buffer solutions and the Henderson equation

A buffer resists change in pH when a little acid or base is added or when it is diluted. Acidic buffer: weak acid + its salt with a strong base (CH₃COOH + CH₃COONa): added H⁺ is taken up by CH₃COO⁻, added OH⁻ by CH₃COOH. Basic buffer: weak base + its salt with a strong acid (NH₄OH + NH₄Cl). Henderson equation: pH = pKa + log([salt]/[acid]); for basic buffers pOH = pKb + log([salt]/[base]). A buffer works best when [salt] ≈ [acid], i.e. pH ≈ pKa ± 1.

Worked example

Find the pH of a buffer that is 0.20 M in NH₃ and 0.10 M in NH₄Cl (Kb = 1.8 × 10⁻⁵).

Try it first, then show the solutionSolution

1. pKb = −log(1.8 × 10⁻⁵) = 4.74
2. pOH = 4.74 + log(0.10/0.20) = 4.74 − 0.30 = 4.44
3. pH = 14 − 4.44 = 9.56

Answer: pH ≈ 9.56

Hydrolysis of salts and the pH of salt solutions

Salt hydrolysis is the reaction of a salt's ions with water that changes the pH. Strong acid + strong base (NaCl, KNO₃): no hydrolysis, pH 7. Strong acid + weak base (NH₄Cl): the cation hydrolyses, NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, so acidic: pH = 7 − ½pKb − ½log C. Weak acid + strong base (CH₃COONa, Na₂CO₃): the anion hydrolyses, CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, so basic: pH = 7 + ½pKa + ½log C. Weak acid + weak base (CH₃COONH₄): both hydrolyse: pH = 7 + ½pKa − ½pKb, independent of concentration (pH 7 when Ka = Kb).

Worked example

Find the pH of 0.10 M CH₃COONa (pKa = 4.74) and of 0.10 M NH₄Cl (pKb = 4.74).

Try it first, then show the solutionSolution

1. CH₃COONa: pH = 7 + 2.37 + ½(−1) = 8.87
2. NH₄Cl: pH = 7 − 2.37 − ½(−1) = 5.13

Answer: 8.87 (basic); 5.13 (acidic)

pH indicators and their choice in titration

An indicator is a weak acid (or base) whose un-ionised and ionised forms have different colours: HIn ⇌ H⁺ + In⁻. It changes colour over about pK_In ± 1. Methyl orange: pH 3.1–4.4 (red to yellow). Phenolphthalein: pH 8.3–10.0 (colourless to pink). The indicator's range must lie within the steep part of the titration curve at the equivalence point: strong acid–strong base: either; weak acid–strong base (equivalence pH > 7): phenolphthalein; strong acid–weak base (equivalence pH < 7): methyl orange; weak acid–weak base: no sharp change, so no simple indicator is suitable.

Worked example

Choose indicators for titrating (i) ethanoic acid with NaOH (ii) ammonia solution with HCl.

Try it first, then show the solutionSolution

1. (i) Equivalence point is basic (acetate hydrolyses): phenolphthalein
2. (ii) Equivalence point is acidic (NH₄⁺ hydrolyses): methyl orange

Answer: Phenolphthalein; methyl orange

Solubility and solubility product

For a sparingly soluble salt in equilibrium with its saturated solution, AₓBᵧ(s) ⇌ xA^{y+} + yB^{x−}, the solubility product Ksp = [A]ˣ[B]ʸ. With molar solubility s: AB: Ksp = s²; AB₂ or A₂B: Ksp = 4s³; AB₃: Ksp = 27s⁴. A precipitate forms when the ionic product (same expression with actual concentrations) exceeds Ksp; none forms if it is less. A common ion reduces solubility (AgCl is much less soluble in NaCl solution). This principle underlies selective precipitation in salt analysis.

Worked example

Find the solubility of AgCl (Ksp = 1.8 × 10⁻¹⁰, M = 143.5) in mol L⁻¹ and g L⁻¹, and of CaF₂ (Ksp = 3.9 × 10⁻¹¹) in mol L⁻¹.

Try it first, then show the solutionSolution

1. AgCl: s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M; × 143.5 = 1.93 × 10⁻³ g L⁻¹
2. CaF₂: Ksp = s(2s)² = 4s³ ⇒ s = (3.9 × 10⁻¹¹/4)^(1/3)
3. = 2.1 × 10⁻⁴ M

Answer: 1.34 × 10⁻⁵ M (1.93 mg L⁻¹); 2.1 × 10⁻⁴ M

Chemistry in real life: Ionic Equilibrium

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.

Topics in this chapter

• Buffers, Salt Hydrolysis and Solubility Product
• pH, pOH and Ionisation of Weak Acids and Bases

Ionic Equilibrium formula sheet

Ostwald's dilution law

$K_a=\frac{C{\alpha }^2}{1-\alpha }\approx C{\alpha }^2;\alpha =\sqrt{\frac{K_a}{C}};[H^{+}]=\sqrt{K_{a}C}$

Same form for weak bases with Kb.

Water and pH

$K_w=[H^{+}][O{H}^{-}]=10^{-14};pH=-\log [H^{+}];pH+pOH=14$

At 298 K.

Henderson equation

$pH=p{K}_a+\log \frac{[\text{salt}]}{[\text{acid}]};pOH=p{K}_b+\log \frac{[\text{salt}]}{[\text{base}]}$

Best near pH = pKa.

Salt of weak acid + strong base

$pH=7+\frac{1}{2}p{K}_a+\frac{1}{2}\log C$

Basic solution.

Salt of strong acid + weak base

$pH=7-\frac{1}{2}p{K}_b-\frac{1}{2}\log C$

Acidic solution.

Salt of weak acid + weak base

$pH=7+\frac{1}{2}p{K}_a-\frac{1}{2}p{K}_b$

Independent of concentration.

Solubility product

$AB:K_{sp}=s^2;A{B}_2:K_{sp}=4s^3;A{B}_3:K_{sp}=27s^4$

Precipitate if ionic product > Ksp.

Ionic Equilibrium: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Ionic Equilibrium

The question patterns to expect, and the method that works for each.

Tricks to remember Ionic Equilibrium

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

• Conjugates differ by one H⁺

  Acid minus H⁺ = conjugate base; base plus H⁺ = conjugate acid.

  Use it when: Brønsted-Lowry pairs.

• Dilute to ionise

  α = √(Ka/C): smaller C, larger α.

  Use it when: Ostwald's dilution law.

• Salt over acid

  pH = pKa + log(salt/acid): equal amounts give pH = pKa.

  Use it when: Buffers.

• Strong parent wins

  Salt of strong acid → acidic; of strong base → basic; both strong → neutral.

  Use it when: Salt hydrolysis.

• Phenol for weak acids, orange for weak bases

  Weak acid titration: phenolphthalein; weak base: methyl orange.

  Use it when: Choosing indicators.

Ionic Equilibrium: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Ionic Equilibrium

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

• ✗Mistake: Forgetting the number of OH⁻ or H⁺ per formula unit

  Students treat Ba(OH)₂ like NaOH.

  Wrong working

  0.005 M Ba(OH)₂: pOH = 2.3, pH = 11.7.

  Right working

  [OH⁻] = 0.010 M: pH = 12.0.

  ✓Do this: Multiply by the number of ions released.

  Usually costs: 1 mark.

• ✗Mistake: Using Ksp = s² for every salt

  The stoichiometry is ignored.

  Wrong working

  CaF₂: s = √Ksp.

  Right working

  CaF₂: Ksp = 4s³.

  ✓Do this: Write the dissolution equation and Ksp in terms of s.

  Usually costs: 1-2 marks.

• ✗Mistake: Not diluting on mixing

  Students use stock concentrations in the ionic product.

  Wrong working

  Mixing equal volumes of 10⁻⁴ M solutions: IP = 10⁻⁸.

  Right working

  Each halves to 5 × 10⁻⁵ M: IP = 2.5 × 10⁻⁹.

  ✓Do this: Recalculate concentrations in the total volume.

  Usually costs: 1 mark.

• ✗Mistake: Inverting the Henderson ratio

  Students write acid/salt.

  Wrong working

  pH = 4.74 + log(0.10/0.20) = 4.44.

  Right working

  pH = 4.74 + log(0.20/0.10) = 5.05.

  ✓Do this: More salt (conjugate base) makes the buffer more basic.

  Usually costs: 1 mark.

• ✗Mistake: Wrong indicator choice

  Students use methyl orange for every titration.

  Wrong working

  Ethanoic acid vs NaOH with methyl orange.

  Right working

  Equivalence pH ≈ 8.7: phenolphthalein.

  ✓Do this: Estimate the equivalence pH first.

  Usually costs: 1 mark.

Practice questions on Ionic Equilibrium

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

1. Q1 · 1 mark · MCQExam level

   Which is a Lewis acid?
   (a) NH₃
   (b) H₂O
   (c) BF₃
   (d) Cl⁻

   Hint

   Electron-pair acceptor.

   Answer

   (c)

2. Q2 · 1 mark · MCQExam level

   The pH of pure water at 298 K is:
   (a) 0
   (b) 1
   (c) 7
   (d) 14

   Hint

   [H⁺] = 10⁻⁷.

   Answer

   (c)

3. Q3 · 1 mark · Very short answerExam level

   Write the conjugate acid of NH₃.

   Hint

   Add H⁺.

   Answer

   NH₄⁺

4. Q4 · 1 mark · Very short answerExam level

   What is the value of pKw at 298 K?

   Hint

   −log(10⁻¹⁴).

   Answer

   14

5. Q5 · 2 marks · Short answerExam level

   Predict whether aqueous solutions of NaCl, Na₂CO₃ and NH₄NO₃ are acidic, basic or neutral.

   Hint

   Parent acid and base.

   Answer

   NaCl neutral (strong/strong); Na₂CO₃ basic (weak acid, strong base); NH₄NO₃ acidic (strong acid, weak base)

6. Q6 · 2 marks · Short answerHarder exam level

   Find the pH of 0.010 M H₂SO₄, assuming complete ionisation.

   Hint

   Two H⁺ per molecule.

   Answer

   [H⁺] = 0.020 M; pH = 1.70

7. Q7 · 2 marks · Short answerHarder exam level

   Find [OH⁻] in a solution of pH 4.5.

   Hint

   pOH = 9.5.

   Answer

   10⁻⁹·⁵ = 3.2 × 10⁻¹⁰ M

8. Q8 · 3 marks · Short answerHarder exam level

   0.010 M HCN is 0.020% ionised. Find Ka and the pH.

   Hint

   α = 2.0 × 10⁻⁴.

   Answer

   Ka = Cα² = 0.010 × 4.0 × 10⁻⁸ = 4.0 × 10⁻¹⁰; [H⁺] = Cα = 2.0 × 10⁻⁶ M; pH = 5.70

9. Q9 · 3 marks · Short answerHarder exam level

   Find the pH of 0.050 M sodium benzoate (Ka of benzoic acid = 6.3 × 10⁻⁵).

   Hint

   Salt of a weak acid and a strong base: pH = 7 + ½pKa + ½log C.

   Answer

   pKa = −log(6.3 × 10⁻⁵) = 4.20; pH = 7 + 2.10 + ½log 0.050 = 7 + 2.10 − 0.65 = 8.45 (basic, because the benzoate ion hydrolyses)

10. Q10 · 2 marks · Short answerHarder exam level

    In what ratio should sodium acetate and acetic acid be mixed to make a buffer of pH 5.00 (pKa = 4.74)?

    Hint

    Henderson equation.

    Answer

    log(salt/acid) = 0.26 ⇒ ratio = 10^0.26 = 1.8 : 1

11. Q11 · 2 marks · Short answerHarder exam level

    The solubility of Ag₂CrO₄ is 6.5 × 10⁻⁵ mol L⁻¹. Find its Ksp.

    Hint

    A₂B type.

    Answer

    Ksp = (2s)²s = 4s³ = 4 × (6.5 × 10⁻⁵)³ = 1.1 × 10⁻¹²

12. Q12 · 3 marks · Short answerHarder exam level

    Explain the role of dilute HCl when H₂S is passed through a solution in group II analysis.

    Hint

    Common ion H⁺.

    Answer

    HCl supplies H⁺, which suppresses H₂S ⇌ 2H⁺ + S²⁻ and lowers [S²⁻]. Only sulphides with very small Ksp (CuS, PbS, CdS) exceed it and precipitate; group IV sulphides (ZnS, MnS) with larger Ksp stay in solution.

13. Q13 · 5 marks · Long answerHarder exam level

    (a) Derive Ostwald's dilution law for a weak acid. [2]
    (b) Find the degree of ionisation of 0.010 M acetic acid (Ka = 1.8 × 10⁻⁵). [1]
    (c) Compare it with 0.10 M acetic acid (α = 0.0134) and explain. [2]

    Hint

    [HA] = C(1 − α), [H⁺] = [A⁻] = Cα.

    Answer

    (a) Ka = (Cα)(Cα)/[C(1 − α)] = Cα²/(1 − α) ≈ Cα² for small α. (b) α = √(1.8 × 10⁻³) = 0.042. (c) About 3.2 times larger (√10): dilution shifts the equilibrium towards ions, as the dilution law predicts.

14. Q14 · 3 marks · Short answerHOTS / competency

    Find the pH after mixing 50 mL of 0.10 M HCl with 50 mL of 0.080 M NaOH.

    Hint

    Excess acid in the total volume.

    Answer

    H⁺: 5.0 mmol; OH⁻: 4.0 mmol; excess H⁺ 1.0 mmol in 100 mL = 0.010 M; pH = 2.0

15. Q15 · 3 marks · Short answerHOTS / competency

    Equal volumes of 0.020 M MgCl₂ and 0.020 M NH₃ are mixed. Will Mg(OH)₂ precipitate? (Kb = 1.8 × 10⁻⁵, Ksp = 1.8 × 10⁻¹¹)

    Hint

    Halve both concentrations first.

    Answer

    After mixing: [Mg²⁺] = 0.010 M, [NH₃] = 0.010 M; [OH⁻] = √(1.8 × 10⁻⁷) = 4.2 × 10⁻⁴ M; ionic product = 0.010 × 1.8 × 10⁻⁷ = 1.8 × 10⁻⁹ > Ksp, so yes.

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. 1.Brønsted-Lowry acid?Show answer

   A proton donor.

2. 2.Lewis base?Show answer

   An electron-pair donor.

3. 3.Ostwald's dilution law (approximate)?Show answer

   Ka ≈ Cα².

4. 4.Kw at 298 K?Show answer

   1.0 × 10⁻¹⁴.

5. 5.Henderson equation?Show answer

   pH = pKa + log([salt]/[acid]).

6. 6.pH of a salt of a weak acid and a strong base?Show answer

   7 + ½pKa + ½log C.

7. 7.Ksp of an AB₂ salt in terms of s?Show answer

   4s³.

8. 8.When does a precipitate form?Show answer

   When the ionic product exceeds Ksp.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you identify conjugate pairs and Lewis acids and bases?
• ☐Can you derive Ostwald's dilution law?
• ☐Can you find the pH of strong and weak acids and bases?
• ☐Can you find the pH of 10⁻⁸ M HCl correctly?
• ☐Can you calculate the pH of an acidic and a basic buffer?
• ☐Can you find the pH of a hydrolysing salt?
• ☐Can you choose an indicator for any titration?
• ☐Can you predict precipitation from Ksp and explain group II and III separations?

Ionic Equilibrium: frequently asked questions