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State Board · Class 12 · Chemistry

Lesson 3: Ionic Equilibria

7 concepts · 56 practice questions with answers · Free to read

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Electrolytes and theories of acids and bases

From daily life

Lemon is sour and soap is slippery — acids and bases are everywhere, we just need to recognise them.

Simple definition

An electrolyte gives ions in solution. Brønsted–Lowry: acid = H⁺ donor, base = H⁺ acceptor. Lewis: acid = electron-pair acceptor.

TheoriesAcid–base theoriesTheoryAcidBasee.g.Arrheniusgives H⁺gives OH⁻HCl, NaOHBrønsted–LowryH⁺ donorH⁺ acceptorNH₄⁺, NH₃Lewise⁻-pair acceptore⁻-pair donorBF₃, NH₃Conjugate pairs: HA / A⁻; stronger acid ⇒ weaker conjugate baseNH₃ + H₂O ⇌ NH₄⁺ + OH⁻NH₃ base, H₂O acid; NH₄⁺ conjugate acid, OH⁻ conjugate baseStrong electrolytes dissociate fully (HCl, NaOH); weak ones partially (CH₃COOH)

In the figure : Three theories.

Rule 1: Strong and weak — Strong: fully dissociated (HCl, NaOH, KCl); weak: partially (CH₃COOH, NH₃); non-electrolytes give no ions (glucose).

  • Easy : HCl ⇒ H⁺ + Cl⁻ (→).
  • Medium : CH₃COOH ⇌ CH₃COO⁻ + H⁺.
  • Exam-type : Sugar gives no ions.

Rule 2: Brønsted–Lowry — Conjugate pair HA / A⁻; stronger acid ⇒ weaker conjugate base; amphoteric: H₂O, HCO₃⁻.

  • Easy : NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.
  • Medium : Conjugate base of HCl is Cl⁻ (weak).
  • Exam-type : H₂O is both acid and base.

Rule 3: Arrhenius and Lewis — Arrhenius: acid gives H⁺, base gives OH⁻; Lewis: BF₃ is an acid, NH₃ a base (no H⁺ needed).

  • Easy : BF₃ + NH₃ ⇒ F₃B←NH₃.
  • Medium : AlCl₃ is a Lewis acid.
  • Exam-type : Lewis is the broadest.

Reasoning: why HCO₃⁻ is amphoteric

  • Step 1: Donating H⁺ gives CO₃²⁻, so it acts as an acid.
  • Step 2: Accepting H⁺ gives H₂CO₃, so it acts as a base.
  • Step 3: Both behaviours ⇒ amphoteric.

CET level (MCQ)

  • CET 1) Conjugate acid of NH₃? 1) NH₄⁺ 2) NH₂⁻ 3) OH⁻ 4) H₂O — Answer: 1).
  • CET 2) Lewis acid? 1) BF₃ 2) NH₃ 3) OH⁻ 4) Cl⁻ — Answer: 1).

Real-life use: Stomach acidity, buffers, drugs, catalysis.

Remember

Brønsted: H⁺ donor/acceptor; Lewis: e⁻-pair acceptor/donor; conjugate pairs.

Common mistake

Treating a conjugate pair as differing by OH⁻.

CET trap

A conjugate pair differs by one H⁺.

Shortcut

Memory clue: "acid gives, base takes".

Spot the difference

TheoryAcid
Arrheniusgives H⁺
Brønsted–LowryH⁺ donor
Lewise⁻ acceptor

Practice questions

Easy
  1. 1Strong vs weak electrolytes.
  2. 2Brønsted–Lowry acid and base.
  3. 3What is a conjugate pair?
Medium
  1. 4Conjugate acid of NH₃.
  2. 5Examples of Lewis acids.
  3. 6What is amphoteric?
Challenging
  1. 7Conjugate base of HCl.
  2. 8Limitation of Arrhenius theory.
Show answers (try them yourself first)
  1. 1Full vs partial dissociation.
  2. 2H⁺ donor and acceptor.
  3. 3Differ by one H⁺.
  4. 4NH₄⁺.
  5. 5BF₃, AlCl₃.
  6. 6Acts as acid and base.
  7. 7Cl⁻.
  8. 8Needs aqueous solution and H⁺.

Ostwald's dilution law, K_a and K_b

From daily life

Dilute the tea and the sugar spreads out — dilution makes a weak acid dissociate more.

Simple definition

K_a = Cα²/(1 – α); for α ≪ 1: K_a ≈ Cα², α = √(K_a/C), [H⁺] = √(K_a C).

DilutionOstwald's dilution lawdilutionαKa = Cα² / (1 – α)α ≪ 1: Ka ≈ Cα²α = √(Ka / C)[H⁺] = Cα = √(Ka C)0.1 M CH₃COOH, Ka = 1.8×10⁻⁵α = √(1.8×10⁻⁴) = 0.0134 ; [H⁺] = 1.34×10⁻³ ; pH = 2.87Dilution raises α; Ka depends only on T0.01 M ⇒ α = 0.0424 (10× dilution ⇒ α ≈ √10 times)

In the figure : α and dilution.

Live figure

Ostwald's dilution law, K_a and K_b

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Rule 1: K_a and K_b — HA ⇌ H⁺ + A⁻: K_a = [H⁺][A⁻]/[HA]; B + H₂O ⇌ BH⁺ + OH⁻: K_b; K_a K_b = K_w for a conjugate pair.

  • Easy : CH₃COOH K_a = 1.8×10⁻⁵.
  • Medium : pK_a = 4.74.
  • Exam-type : Larger K_a ⇒ stronger acid.

Rule 2: Finding α — α = √(K_a/C): 0.1 M ⇒ 0.0134; 0.01 M ⇒ 0.0424; [H⁺] = Cα.

  • Easy : 0.1 M: [H⁺] = 1.34×10⁻³, pH = 2.87.
  • Medium : Dilution ⇒ α↑.
  • Exam-type : 10× dilution ⇒ α about 3.16×.

Rule 3: Measuring K_a — 0.1 M HA with pH = 3 ⇒ [H⁺] = 10⁻³ ⇒ K_a = (10⁻³)²/0.099 = 1.0×10⁻⁵.

  • Easy : For bases NH₃: K_b = 1.8×10⁻⁵.
  • Medium : 0.1 M NH₃ ⇒ pH = 11.13.
  • Exam-type : pK = –log K.

Calculation: 0.1 M CH₃COOH, K_a = 1.8×10⁻⁵

  • Step 1: α = √(1.8×10⁻⁵/0.1) = 0.0134.
  • Step 2: [H⁺] = Cα = 1.34×10⁻³ M.
  • Step 3: pH = 2.87.

CET level (MCQ)

  • CET 1) On dilution α? 1) increases 2) decreases 3) same 4) zero — Answer: 1).
  • CET 2) K_a K_b = ? 1) K_w 2) 1 3) K_w² 4) 10⁻⁷ — Answer: 1).

Real-life use: Drugs, water treatment, acidity tests.

Remember

K_a = Cα²/(1–α); α = √(K_a/C); K_a K_b = K_w.

Common mistake

Using α ≪ 1 without checking.

CET trap

(1 – α) ≈ 1 only when α < 5 %.

Shortcut

Memory clue: "α = √(K/C)" — more dilute, bigger α.

Spot the difference

Cα
0.1 M0.0134
0.01 M0.0424

Practice questions

Easy
  1. 1Derive Ostwald's dilution law.
  2. 2α of 0.1 M CH₃COOH.
  3. 3Its pH.
Medium
  1. 4Relation of K_a and K_b.
  2. 5pH = 3, 0.1 M HA ⇒ K_a.
  3. 6pH of 0.1 M NH₃.
Challenging
  1. 7Effect of dilution on α.
  2. 8When is (1 – α) ≈ 1?
Show answers (try them yourself first)
  1. 1K = Cα²/(1–α).
  2. 20.0134.
  3. 32.87.
  4. 4K_w.
  5. 51.0×10⁻⁵.
  6. 611.13.
  7. 7α increases.
  8. 8When α is small.

Ionic product of water, pH and pOH

From daily life

Pure water ionises very slightly — a tiny fraction forms H⁺ and OH⁻.

Simple definition

K_w = [H⁺][OH⁻] = 1.0×10⁻¹⁴ at 298 K; pH = –log[H⁺], pOH = –log[OH⁻], pH + pOH = 14.

pHKw, pH and pOH01234567891011121314acidicneutral 7basicKw = [H⁺][OH⁻] = 1.0×10⁻¹⁴ (298 K)pH = –log[H⁺] ; pOH = –log[OH⁻] ; pH + pOH = 140.01 M HCl ⇒ pH = 20.001 M NaOH ⇒ pOH = 3 ⇒ pH = 11T↑ ⇒ Kw↑ (about 5.5×10⁻¹³ at 100 °C, neutral pH ≈ 6.1)A change of 1 pH unit = a factor of 10 in [H⁺]

In the figure : The pH scale.

Rule 1: K_w — H₂O ⇌ H⁺ + OH⁻; pure water [H⁺] = [OH⁻] = 10⁻⁷; T↑ ⇒ K_w↑.

  • Easy : 100 °C: K_w ≈ 5.5×10⁻¹³.
  • Medium : Neutral pH ≈ 6.1 then.
  • Exam-type : K_w depends only on T.

Rule 2: Calculating pH — Strong acid: [H⁺] = C; strong base: [OH⁻] = C; 0.01 M HCl ⇒ 2; 0.001 M NaOH ⇒ 11.

  • Easy : pH 3 ⇒ [H⁺] = 10⁻³.
  • Medium : pH 2 vs 3: 10× difference.
  • Exam-type : pOH = 14 – pH.

Rule 3: pH in daily life — Blood 7.4; lemon juice 2.2; milk 6.5; soap 10; stomach acid 1.5–2.

  • Easy : Tooth decay below pH 5.5.
  • Medium : Rainwater pH ≈ 5.6.
  • Exam-type : Soil pH for crops.

Calculation: pH of 0.001 M NaOH

  • Step 1: [OH⁻] = 10⁻³ ⇒ pOH = 3.
  • Step 2: pH = 14 – 3.
  • Step 3: pH = 11.

CET level (MCQ)

  • CET 1) pH of 0.01 M HCl? 1) 2 2) 1 3) 12 4) 7 — Answer: 1).
  • CET 2) pH + pOH = ? 1) 14 2) 7 3) 1 4) 0 — Answer: 1).

Real-life use: Health, farming, water treatment, food industry.

Remember

K_w = [H⁺][OH⁻]; pH + pOH = 14 (298 K); each pH unit = 10×.

Common mistake

Assuming neutral pH is 7 at all temperatures.

CET trap

T↑ ⇒ K_w↑ ⇒ neutral pH < 7.

Shortcut

Memory clue: "pH + pOH = 14".

Spot the difference

SolutionpH
0.01 M HCl2
Pure water7
0.001 M NaOH11

Practice questions

Easy
  1. 1Define K_w.
  2. 2Define pH.
  3. 3pH of 0.01 M HCl.
Medium
  1. 4pH of 0.001 M NaOH.
  2. 5[H⁺] at pH 3.
  3. 6Effect of T on K_w.
Challenging
  1. 7pH of blood.
  2. 8Difference between pH 2 and 4.
Show answers (try them yourself first)
  1. 1[H⁺][OH⁻].
  2. 2–log[H⁺].
  3. 32.
  4. 411.
  5. 510⁻³.
  6. 6T↑ ⇒ K_w↑.
  7. 77.4.
  8. 8100× in [H⁺].

Hydrolysis of salts

From daily life

Washing soda feels slippery in water because it reacts with water and raises the pH.

Simple definition

Hydrolysis is the reaction of a salt's ions with water, changing the pH; K_h = K_w/K_a.

HydrolysisHydrolysis of saltsSalt ofe.g.HydrolysispHStrong acid + strong baseNaClnonepH = 7Weak acid + strong baseCH₃COONaA⁻ + H₂O ⇌ HA + OH⁻pH > 7Strong acid + weak baseNH₄ClBH⁺ + H₂O ⇌ B + H₃O⁺pH < 7Weak + weakCH₃COONH₄depends on Ka, Kb≈ 7Kh = Kw / Ka ; h = √(Kh / C)0.1 M CH₃COONa: Kh = 5.56×10⁻¹⁰, h = 7.45×10⁻⁵[OH⁻] = 7.45×10⁻⁶ ⇒ pOH 5.13 ⇒ pH 8.870.1 M NH₄Cl: pH = 5.13

In the figure : Four kinds of salts.

Live figure

Hydrolysis of salts

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Rule 1: Weak acid + strong base — A⁻ + H₂O ⇌ HA + OH⁻ ⇒ pH > 7; K_h = K_w/K_a; h = √(K_h/C).

  • Easy : 0.1 M CH₃COONa: K_h = 5.56×10⁻¹⁰.
  • Medium : h = 7.45×10⁻⁵.
  • Exam-type : pH = 8.87.

Rule 2: Strong acid + weak base — BH⁺ + H₂O ⇌ B + H₃O⁺ ⇒ pH < 7; 0.1 M NH₄Cl ⇒ pH = 5.13.

  • Easy : K_h = K_w/K_b.
  • Medium : [H⁺] = Ch.
  • Exam-type : CuSO₄ is acidic.

Rule 3: Strong–strong and weak–weak — NaCl: pH 7, no hydrolysis; CH₃COONH₄: both ions hydrolyse, K_a ≈ K_b ⇒ pH ≈ 7.

  • Easy : KNO₃ is neutral.
  • Medium : Na₂CO₃ is basic.
  • Exam-type : NH₄Cl is acidic.

Calculation: 0.1 M CH₃COONa (K_a = 1.8×10⁻⁵)

  • Step 1: K_h = 10⁻¹⁴/1.8×10⁻⁵ = 5.56×10⁻¹⁰.
  • Step 2: h = √(5.56×10⁻¹⁰/0.1) = 7.45×10⁻⁵; [OH⁻] = 7.45×10⁻⁶.
  • Step 3: pOH = 5.13 ⇒ pH = 8.87.

CET level (MCQ)

  • CET 1) A solution of NH₄Cl is? 1) acidic 2) basic 3) neutral 4) buffer — Answer: 1).
  • CET 2) K_h = ? 1) K_w/K_a 2) K_a/K_w 3) K_a K_w 4) K_w² — Answer: 1).

Real-life use: Soap, washing soda, fertilisers, water pH.

Remember

Salt of weak acid ⇒ pH > 7; of weak base ⇒ pH < 7; K_h = K_w/K.

Common mistake

Treating ions of strong acids/bases as hydrolysing.

CET trap

Na⁺, Cl⁻, K⁺ and NO₃⁻ do not hydrolyse.

Shortcut

Memory clue: "the stronger parent decides the pH".

Spot the difference

SaltpH
NaCl7
CH₃COONa> 7
NH₄Cl< 7

Practice questions

Easy
  1. 1What is hydrolysis?
  2. 2Formula of K_h.
  3. 3pH of CH₃COONa.
Medium
  1. 4pH of NH₄Cl.
  2. 5Why is NaCl neutral?
  3. 6Why is Na₂CO₃ basic?
Challenging
  1. 7How is h found?
  2. 8Weak–weak salts.
Show answers (try them yourself first)
  1. 1Ion + water ⇒ pH change.
  2. 2K_w/K_a.
  3. 38.87.
  4. 45.13.
  5. 5Ions of strong parents.
  6. 6CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻.
  7. 7√(K_h/C).
  8. 8pH ≈ 7.

Buffer solutions

From daily life

A shock absorber soaks up bumps — a buffer soaks up pH shocks.

Simple definition

A buffer keeps pH nearly constant on adding a little acid or base; Henderson: pH = pK_a + log([salt]/[acid]).

BufferBuffer solutionsAcidic: CH₃COOH + CH₃COONapH = pKa + log [salt]/[acid]Basic: NH₃ + NH₄ClpOH = pKb + log [salt]/[base]Henderson: pH = pKa + log([A⁻]/[HA])1 L buffer 0.1 M / 0.1 M ⇒ pH = 4.740.2 M salt / 0.1 M acid ⇒ pH = 5.05Add 0.01 mol HCl: buffer 4.74 ⇒ 4.66; pure water 7 ⇒ 2Added acid is consumed by A⁻; added base by HABest buffering: pH ≈ pKa ± 1; blood pH ≈ 7.4 (H₂CO₃/HCO₃⁻)

In the figure : Buffers and the Henderson equation.

Live figure

Buffer solutions

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Rule 1: Types — Acidic: weak acid + its salt (CH₃COOH + CH₃COONa); basic: weak base + its salt (NH₃ + NH₄Cl).

  • Easy : Around pH 4.74.
  • Medium : NH₃/NH₄Cl pH ≈ 9.25.
  • Exam-type : Blood has H₂CO₃/HCO₃⁻.

Rule 2: Henderson — pH = pK_a + log([A⁻]/[HA]): 0.1/0.1 ⇒ 4.74; 0.2/0.1 ⇒ 5.05.

  • Easy : Equal amounts ⇒ pH = pK_a.
  • Medium : Ten times more salt ⇒ +1.
  • Exam-type : pOH = pK_b + log([BH⁺]/[B]).

Rule 3: Buffer action — Added acid is consumed by A⁻, added base by HA; 0.01 mol HCl: buffer 4.74 ⇒ 4.66, pure water 7 ⇒ 2.

  • Easy : Capacity: pK_a ± 1.
  • Medium : Larger amounts ⇒ greater capacity.
  • Exam-type : Injectable drugs are buffered.

Calculation: 1 L buffer (0.1 M CH₃COOH + 0.1 M CH₃COONa) with 0.01 mol HCl added

  • Step 1: HCl + A⁻ ⇒ HA: A⁻ = 0.09, HA = 0.11.
  • Step 2: pH = 4.74 + log(0.09/0.11).
  • Step 3: pH = 4.74 – 0.09 = 4.66.

CET level (MCQ)

  • CET 1) Buffer components? 1) CH₃COOH + CH₃COONa 2) HCl + NaCl 3) NaOH + NaCl 4) H₂SO₄ + Na₂SO₄ — Answer: 1).
  • CET 2) When [A⁻] = [HA], pH = ? 1) pK_a 2) 7 3) 14 4) 0 — Answer: 1).

Real-life use: Blood, drugs, biological experiments, food preservation.

Remember

pH = pK_a + log(salt/acid); best range pK_a ± 1.

Common mistake

Using a strong acid in a buffer.

CET trap

A buffer is a weak acid/base plus its salt.

Shortcut

Memory clue: "more salt, higher pH".

Spot the difference

A⁻/HApH
1 : 1pK_a
10 : 1pK_a + 1
1 : 10pK_a – 1

Practice questions

Easy
  1. 1What is a buffer?
  2. 2Two types.
  3. 3Henderson equation.
Medium
  1. 40.2/0.1 ⇒ pH.
  2. 5pH after adding HCl.
  3. 6Blood buffer.
Challenging
  1. 7What is buffer capacity?
  2. 8pH of the NH₃/NH₄Cl buffer.
Show answers (try them yourself first)
  1. 1Keeps pH steady.
  2. 2Acidic and basic.
  3. 3pK_a + log(salt/acid).
  4. 45.05.
  5. 54.66.
  6. 6H₂CO₃/HCO₃⁻.
  7. 7Resistance to pH change.
  8. 8≈ 9.25.

Solubility product and condition for precipitation

From daily life

Too much sugar in tea settles at the bottom — saturation exceeded.

Simple definition

K_sp is the product of ion concentrations in a saturated solution; Q > K_sp ⇒ precipitation.

K_spSolubility product KspSaltFormulaKsps (mol/L)AgClKsp = s²1.8×10⁻¹⁰1.34×10⁻⁵Ag₂CrO₄Ksp = 4s³1.1×10⁻¹²6.5×10⁻⁵PbI₂Ksp = 4s³6.9×10⁻⁹1.2×10⁻³AₓBᵧ ⇌ xA + yB: Ksp = [A]ˣ[B]ʸQ > Ksp ⇒ precipitate; Q = Ksp ⇒ saturated; Q < Ksp ⇒ stays dissolved0.001 M Ag⁺ + 0.001 M Cl⁻: Q = 1×10⁻⁶ > 1.8×10⁻¹⁰ ⇒ AgCl precipitatesCommon-ion effect: in 0.1 M NaCl, AgCl s = 1.8×10⁻⁹Solubility falls from 1.34×10⁻⁵ to 1.8×10⁻⁹: about 7000 times lower

In the figure : K_sp and solubility.

Live figure

Solubility product and condition for precipitation

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Rule 1: K_sp and s — AgCl: K_sp = s² ⇒ s = 1.34×10⁻⁵; A₂B: K_sp = 4s³; Ag₂CrO₄: s = (1.1×10⁻¹²/4)^(1/3) = 6.5×10⁻⁵.

  • Easy : PbI₂ s = 1.2×10⁻³ ⇒ K_sp = 6.9×10⁻⁹.
  • Medium : AB₃: 27s⁴.
  • Exam-type : K_sp depends on T.

Rule 2: Precipitation condition — Q > K_sp ⇒ precipitate; Q = K_sp ⇒ saturated; Q < K_sp ⇒ dissolved; concentrations halve on mixing equal volumes.

  • Easy : 0.002 M + 0.002 M, equal volumes ⇒ Q = 10⁻⁶.
  • Medium : 10⁻⁶ > 1.8×10⁻¹⁰ ⇒ AgCl precipitates.
  • Exam-type : Group separation in analysis.

Rule 3: Applications — Qualitative analysis; removing hardness; kidney stones (calcium oxalate); treating effluents.

  • Easy : BaSO₄ in barium meals.
  • Medium : Bone (hydroxyapatite).
  • Exam-type : Precipitating Ca²⁺.

Calculation: 0.001 M Ag⁺ and 0.001 M Cl⁻ mixed

  • Step 1: Q = [Ag⁺][Cl⁻] = 10⁻³ × 10⁻³ = 10⁻⁶.
  • Step 2: K_sp = 1.8×10⁻¹⁰.
  • Step 3: Q > K_sp ⇒ AgCl precipitates.

CET level (MCQ)

  • CET 1) K_sp for AgCl? 1) s² 2) 4s³ 3) 27s⁴ 4) s³ — Answer: 1).
  • CET 2) Precipitation occurs when? 1) Q > K_sp 2) Q < K_sp 3) Q = 0 4) Q = 1 — Answer: 1).

Real-life use: Analysis, water treatment, medicine, kidney stones.

Remember

AB: s²; A₂B/AB₂: 4s³; Q > K_sp ⇒ precipitate.

Common mistake

Forgetting to halve the concentrations on mixing.

CET trap

Volume increases on mixing ⇒ concentrations fall.

Shortcut

Memory clue: "big Q drops out".

Spot the difference

SaltK_sp
AgCls²
Ag₂CrO₄4s³
PbI₂4s³

Practice questions

Easy
  1. 1What is K_sp?
  2. 2s of AgCl.
  3. 3s of Ag₂CrO₄.
Medium
  1. 4K_sp of PbI₂ (s = 1.2×10⁻³).
  2. 5Condition of precipitation.
  3. 60.002 M + 0.002 M Ag⁺/Cl⁻ ⇒ ?
Challenging
  1. 7K_sp vs Q.
  2. 8K_sp of AB₃.
Show answers (try them yourself first)
  1. 1Ion product.
  2. 21.34×10⁻⁵.
  3. 36.5×10⁻⁵.
  4. 46.9×10⁻⁹.
  5. 5Q > K_sp.
  6. 6Q = 10⁻⁶ ⇒ precipitate.
  7. 7K_sp at equilibrium; Q at any state.
  8. 827s⁴.

Common-ion effect

From daily life

In a crowded room one more person pushes someone out — adding a common ion pushes the equilibrium back.

Simple definition

The common-ion effect: adding a salt with an ion already in the equilibrium reduces dissociation/solubility (Le Chatelier).

Common ionCommon-ion effectCH₃COOH ⇌ CH₃COO⁻ + H⁺+ CH₃COONa ⇒ shifts leftAgCl ⇌ Ag⁺ + Cl⁻+ NaCl ⇒ solubility ↓Le Chatelier: adding a common ion shifts equilibrium back0.1 M CH₃COOH: pH = 2.87, α = 0.0134+ 0.1 M CH₃COONa: [H⁺] = Ka = 1.8×10⁻⁵ ⇒ pH = 4.74Uses: group precipitation in qualitative analysis (NH₄Cl + NH₄OH)Salting-out of soap: NaCl lowers soap's solubility

In the figure : Common ion.

Rule 1: In a weak acid — CH₃COOH + CH₃COONa lowers α; 0.1 M + 0.1 M ⇒ [H⁺] = K_a = 1.8×10⁻⁵ ⇒ pH 4.74 (acid alone: 2.87).

  • Easy : α 0.0134 ⇒ 1.8×10⁻⁴.
  • Medium : Basis of buffers.
  • Exam-type : NH₄Cl in NH₃.

Rule 2: On solubility — AgCl in 0.1 M NaCl: s = K_sp/[Cl⁻] = 1.8×10⁻⁹ (1.34×10⁻⁵ in pure water).

  • Easy : About 7000× lower.
  • Medium : PbCl₂ less soluble in NaCl.
  • Exam-type : The common ion is Cl⁻.

Rule 3: Applications — NH₄Cl + NH₄OH for group III in qualitative analysis; salting-out of soap; near-complete precipitation.

  • Easy : NaCl in soap making.
  • Medium : Excess reagent completes precipitation.
  • Exam-type : Making buffers.

Calculation: AgCl in 0.1 M NaCl

  • Step 1: [Cl⁻] ≈ 0.1 M.
  • Step 2: s = K_sp/[Cl⁻] = 1.8×10⁻¹⁰/0.1.
  • Step 3: s = 1.8×10⁻⁹ mol/L.

CET level (MCQ)

  • CET 1) Adding NaCl to AgCl solubility? 1) decreases 2) increases 3) same 4) infinite — Answer: 1).
  • CET 2) Adding CH₃COONa changes α of CH₃COOH to? 1) lower 2) higher 3) same 4) 1 — Answer: 1).

Real-life use: Qualitative analysis, buffers, soap industry.

Remember

Common ion ⇒ equilibrium shifts back; lower solubility/α.

Common mistake

Thinking K_sp changes.

CET trap

K_sp depends on T only; s changes.

Shortcut

Memory clue: "add the same ion, dissolve less".

Spot the difference

CaseEffect
Salt in acidα↓
AgCl + NaCls↓
K_spunchanged

Practice questions

Easy
  1. 1Explain the common-ion effect.
  2. 2s of AgCl in 0.1 M NaCl.
  3. 3Effect of CH₃COONa on CH₃COOH.
Medium
  1. 4How does pH 2.87 become 4.74?
  2. 5Salting-out of soap.
  3. 6Does K_sp change?
Challenging
  1. 7Link with buffers.
  2. 8Use in qualitative analysis.
Show answers (try them yourself first)
  1. 1Added ion reverses the equilibrium.
  2. 21.8×10⁻⁹.
  3. 3α decreases.
  4. 4[H⁺] = K_a.
  5. 5NaCl separates soap.
  6. 6No.
  7. 7Buffers rely on it.
  8. 8NH₄Cl + NH₄OH.

Quick revision of the whole lesson

ConceptKey formula / fact
K_a (dilution)K_a = Cα²/(1 – α)
K_w[H⁺][OH⁻] = 10⁻¹⁴
pHpH = –log[H⁺]
K_hK_w / K_a
HendersonpH = pK_a + log(A⁻/HA)
K_sp (AB)s²
K_sp (A₂B)4s³
PrecipitationQ > K_sp
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🎯 Most important points for the exam

  • Acid–base theories.
  • Ostwald.
  • K_w, pH.
  • Hydrolysis.
  • Buffer.
  • K_sp.
  • Common ion.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.