This MCQ module is based on: NCERT Exercises and Solutions: Equilibrium
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NCERT Exercises and Solutions: Equilibrium
🎓 Class 11
Chemistry
CBSE
Theory
Ch 6 – Equilibrium
⏱ ~8 min
🌐 Language: [gtranslate]
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NCERT Exercises and Solutions: Equilibrium
Chapter 6 Summary — Equilibrium in One Page
| Concept | Key Equation / Idea |
|---|---|
| Dynamic equilibrium | Forward rate = reverse rate; concentrations constant |
| Law of Mass Action | K_c = [C]^c[D]^d / [A]^a[B]^b |
| K_p–K_c relation | K_p = K_c (RT)^Δn_g |
| Reaction quotient | Q < K → forward; Q > K → reverse; Q = K → equilibrium |
| Le Chatelier's principle | System shifts to counteract imposed change |
| Heterogeneous equilibrium | Pure solids & liquids excluded from K |
| Ionisation of water | Kw = [H⁺][OH⁻] = 10⁻¹⁴ at 298 K |
| pH | pH = −log[H⁺]; pH + pOH = 14 |
| Weak acid | Ka = cα² / (1−α); for α << 1: α ≈ √(Ka/c) |
| Conjugate pair | Ka × Kb = Kw; pKa + pKb = 14 |
| Buffer (Henderson) | pH = pKa + log([salt]/[acid]) |
| Salt hydrolysis | S+W base → acidic; W+S base → basic; S+S → neutral |
| Solubility product | Ksp = [M⁺][X⁻]; precipitate if Q_sp > Ksp |
| Common-ion effect | Adding a common ion shifts equilibrium backward |
Key Terms
Equilibrium • Dynamic equilibrium • Vapour pressure • Henry's law • Reversible reaction • Law of mass action • Equilibrium constant K_c • K_p • Reaction quotient Q • Homogeneous / heterogeneous equilibrium • Le Chatelier's principle • Catalyst • Inert gas • Haber process • Contact process • Arrhenius / Brønsted–Lowry / Lewis acids and bases • Conjugate acid–base pair • Ionic product of water Kw • pH, pOH • Ka, Kb • Ostwald's dilution law • Hydrolysis • Buffer solution • Henderson–Hasselbalch • Common-ion effect • Solubility • Solubility product Ksp
🎯 Interactive: Equilibrium Quick Reference
Pick a concept to recall its formula and a usage tip.
K_c = [Products]^n / [Reactants]^m
Constant at given T; pure solids/liquids excluded.
NCERT Exercises — Worked Solutions
6.1 A liquid is in equilibrium with its vapour in a closed vessel at fixed T. What happens to vapour pressure if V is suddenly increased?
Initially VP drops below saturation. More liquid evaporates until equilibrium is restored. Final VP = same saturated VP at that T (depends only on T).
6.2 What is K_c for: 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) given equilibrium [NOCl] = 1.6 M, [NO] = 1.4 × 10⁻³ M, [Cl₂] = 7.0 × 10⁻⁴ M.
K_c = [NO]² [Cl₂] / [NOCl]² = (1.4 × 10⁻³)² × 7.0 × 10⁻⁴ / (1.6)² = (1.96 × 10⁻⁶ × 7 × 10⁻⁴) / 2.56 = 5.36 × 10⁻¹⁰ M.
6.3 At 700 K, K_c = 1.7 × 10⁻³ for: 2HI(g) ⇌ H₂(g) + I₂(g). Find K_p. (R = 0.0831 L·bar K⁻¹ mol⁻¹)
Δn_g = (1 + 1) − 2 = 0 → K_p = K_c = 1.7 × 10⁻³.
6.4 K_p for 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) at 800 K is 1.8 × 10⁴ atm⁻¹. Calculate K_c.
Δn_g = 2 − 3 = −1; K_p = K_c (RT)⁻¹ → K_c = K_p × RT.
Using R = 0.0821 L·atm K⁻¹ mol⁻¹: K_c = (1.8 × 10⁴) × (0.0821) × (800) = 1.18 × 10⁶ M⁻¹.
Using R = 0.0821 L·atm K⁻¹ mol⁻¹: K_c = (1.8 × 10⁴) × (0.0821) × (800) = 1.18 × 10⁶ M⁻¹.
6.5 Predict the direction of shift on increasing pressure for: (a) PCl₅ ⇌ PCl₃ + Cl₂; (b) CaCO₃(s) ⇌ CaO(s) + CO₂(g); (c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g); (d) H₂(g) + I₂(g) ⇌ 2HI(g).
(a) Δn_g = +1 → ↑P shifts ←.
(b) Δn_g = +1 → ↑P shifts ←.
(c) Δn_g = 0 (4 vs 4 gases) → no shift.
(d) Δn_g = 0 → no shift.
(b) Δn_g = +1 → ↑P shifts ←.
(c) Δn_g = 0 (4 vs 4 gases) → no shift.
(d) Δn_g = 0 → no shift.
6.6 K_c for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 0.061 at 500 K. Predict direction of reaction with [N₂] = 3, [H₂] = 2, [NH₃] = 0.5 (all in M).
Q_c = (0.5)² / (3 × 2³) = 0.25 / 24 = 0.0104.
Q (0.0104) < K (0.061) → reaction proceeds forward (→).
Q (0.0104) < K (0.061) → reaction proceeds forward (→).
6.7 Conjugate base of: HF, H₂SO₄, HCO₃⁻, NH₄⁺.
Remove one proton: HF → F⁻; H₂SO₄ → HSO₄⁻; HCO₃⁻ → CO₃²⁻; NH₄⁺ → NH₃.
6.8 Calculate pH of 0.001 M HCl.
HCl is strong; complete dissociation → [H⁺] = 10⁻³ M → pH = 3.
6.9 Ka of acetic acid = 1.74 × 10⁻⁵. Find pKa, [H⁺] and pH of 0.05 M acetic acid.
pKa = −log(1.74 × 10⁻⁵) = 4.76.
[H⁺] ≈ √(Ka × c) = √(1.74 × 10⁻⁵ × 0.05) = √(8.7 × 10⁻⁷) = 9.33 × 10⁻⁴ M.
pH = −log(9.33 × 10⁻⁴) = 3.03.
[H⁺] ≈ √(Ka × c) = √(1.74 × 10⁻⁵ × 0.05) = √(8.7 × 10⁻⁷) = 9.33 × 10⁻⁴ M.
pH = −log(9.33 × 10⁻⁴) = 3.03.
6.10 Ksp of AgCl = 1.8 × 10⁻¹⁰. What is the molar solubility of AgCl in 0.01 M NaCl?
[Cl⁻] from NaCl = 0.01 M. Let solubility of AgCl = s. [Ag⁺] = s; [Cl⁻] ≈ 0.01 (s is tiny).
Ksp = s × 0.01 = 1.8 × 10⁻¹⁰ → s = 1.8 × 10⁻⁸ M.
Compare with pure water: s = 1.34 × 10⁻⁵ M. Common-ion effect reduces solubility ~750×.
Ksp = s × 0.01 = 1.8 × 10⁻¹⁰ → s = 1.8 × 10⁻⁸ M.
Compare with pure water: s = 1.34 × 10⁻⁵ M. Common-ion effect reduces solubility ~750×.
6.11 What is the pH of 0.1 M ammonia solution? Kb(NH₃) = 1.8 × 10⁻⁵.
[OH⁻] = √(Kb × c) = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M.
pOH = −log(1.34 × 10⁻³) = 2.87 → pH = 14 − 2.87 = 11.13.
pOH = −log(1.34 × 10⁻³) = 2.87 → pH = 14 − 2.87 = 11.13.
6.12 A buffer is 0.10 M NH₃ and 0.20 M NH₄Cl. Find pH. (Kb = 1.8 × 10⁻⁵)
For a basic buffer: pOH = pKb + log([salt]/[base]) = 4.74 + log(0.20/0.10) = 4.74 + 0.301 = 5.04.
pH = 14 − 5.04 = 8.96.
pH = 14 − 5.04 = 8.96.
6.13 The solubility of AgI in pure water at 25 °C is 9.22 × 10⁻⁹ M. Find Ksp.
AgI ⇌ Ag⁺ + I⁻; s = [Ag⁺] = [I⁻] = 9.22 × 10⁻⁹ M.
Ksp = s² = (9.22 × 10⁻⁹)² = 8.5 × 10⁻¹⁷.
Ksp = s² = (9.22 × 10⁻⁹)² = 8.5 × 10⁻¹⁷.
6.14 Will a precipitate form when 50 mL of 0.001 M AgNO₃ is mixed with 50 mL of 0.001 M NaCl? Ksp(AgCl) = 1.8 × 10⁻¹⁰.
After mixing (volume doubles): [Ag⁺] = 0.0005 M, [Cl⁻] = 0.0005 M.
Q_sp = (5 × 10⁻⁴)² = 2.5 × 10⁻⁷.
Q_sp (2.5 × 10⁻⁷) > Ksp (1.8 × 10⁻¹⁰) → precipitate forms.
Q_sp = (5 × 10⁻⁴)² = 2.5 × 10⁻⁷.
Q_sp (2.5 × 10⁻⁷) > Ksp (1.8 × 10⁻¹⁰) → precipitate forms.
6.15 Why is BaSO₄ insoluble while BaCl₂ is soluble?
BaSO₄ has very low Ksp (1.1 × 10⁻¹⁰) — its lattice energy outweighs its hydration energy, so the salt does not appreciably dissolve. BaCl₂, by contrast, has high hydration energy of Ba²⁺ + 2Cl⁻ that overcomes its smaller lattice energy → soluble.
This is why BaSO₄ is used as a barium meal in X-ray imaging — it is so insoluble that the toxic Ba²⁺ never reaches the bloodstream.
This is why BaSO₄ is used as a barium meal in X-ray imaging — it is so insoluble that the toxic Ba²⁺ never reaches the bloodstream.
🧪 Concept Wrap-up — Universal Indicator Tour
Make universal indicator from red cabbage juice (boil chopped purple cabbage; the violet liquid is your indicator). Test these with a few drops:
- Lemon juice
- Vinegar
- Tap water
- Soap solution
- Baking soda solution
- Drain cleaner (CAUTION!)
1. Lemon juice (red) — pH ≈ 2 (citric acid). 2. Vinegar (red-pink) — pH ≈ 3. 3. Tap water (purple) — pH 6.5–7.5 (depends on dissolved CO₂). 4. Soap (blue) — pH ≈ 9. 5. Baking soda (blue) — pH ≈ 9. 6. Drain cleaner (yellow-green) — pH ≈ 13–14.
Why colour? Anthocyanin pigments in cabbage exist in different protonated forms at different pH values — each colour corresponds to a different equilibrium species.
🎯 Competency-Based Questions — Final Mix
Q1. The reaction quotient Q is equal to K when: L1 Remember
Answer: (c) At equilibrium, Q = K. Otherwise, Q ≠ K and reaction proceeds toward equilibrium.
Q2. State Le Chatelier's principle in your own words and give one example. L2 Understand
Answer: A system at equilibrium responds to a stress by shifting in the direction that partially offsets the stress. Example: in the synthesis of NH₃ (N₂ + 3H₂ ⇌ 2NH₃; ΔH < 0), increasing pressure shifts equilibrium toward NH₃ (fewer moles); cooling also shifts toward NH₃ (forward is exothermic).
Q3. Calculate the pH of a solution made by mixing 50 mL of 0.10 M HCl with 50 mL of 0.10 M NaOH. L3 Apply
Answer: Equal moles → complete neutralisation. Resulting solution is dilute NaCl (neutral salt). pH = 7.
Q4. Why does the equilibrium constant of an exothermic reaction decrease with rising temperature? L4 Analyse
Answer: Heat is a "product" of an exothermic reaction. Adding heat (raising T) shifts equilibrium backward (Le Chatelier), reducing [products]/[reactants] and hence K. Quantitatively, ln(K₂/K₁) = (−ΔH/R)(1/T₂ − 1/T₁); ΔH < 0 → K falls as T rises.
Q5. HOT (Create): Design a buffer solution of pH 5.00 using acetic acid (pKa 4.74). What ratio [salt]/[acid] do you need? L6 Create
Answer: pH = pKa + log([salt]/[acid]) → 5.00 = 4.74 + log r → log r = 0.26 → r = 10⁰·²⁶ = 1.82. So [acetate]/[acetic acid] ≈ 1.82. For example, 0.10 M acetic acid + 0.182 M sodium acetate would give pH ≈ 5.00. Buffer capacity is good as long as both species exceed ~0.05 M.
🧠 Assertion–Reason Questions — Wrap-up
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: The equilibrium constant K does not change when a catalyst is added.
R: A catalyst affects only the rate of the reaction, not the position of equilibrium.
Answer: (A). Both true; R explains A.
A: The pH of a buffer solution remains nearly constant on small additions of strong acid or base.
R: Buffers contain a weak acid–conjugate base pair that absorbs added H⁺ or OH⁻.
Answer: (A). Both true; R explains A. Buffers neutralise added strong acid/base by converting them to the weak conjugate.
A: AgCl is less soluble in 0.1 M HCl than in pure water.
R: The hydronium ions react with chloride ions reducing their concentration.
Answer: (C). A is TRUE — common ion (Cl⁻ from HCl) suppresses dissolution. R is FALSE — H₃O⁺ does not react with Cl⁻ (HCl is fully dissociated and a strong electrolyte).
Frequently Asked Questions — NCERT Exercises and Solutions: Equilibrium
What are the most important formulas in Class 11 Chemistry Chapter 6 exercises?
Key formulas for NCERT Class 11 Chemistry Chapter 6 equilibrium include: K_c = [products]^coeff / [reactants]^coeff; K_p = K_c (RT)^Δn_g; Q vs K for direction; pH = −log[H⁺]; pOH = −log[OH⁻]; pH + pOH = 14 at 25°C; for weak acid [H⁺] = √(K_a × C); Henderson-Hasselbalch pH = pK_a + log([salt]/[acid]); K_sp = product of ion concentrations; molar solubility s = (K_sp)^(1/n) for AB-type salts. Memorise these and the sign conventions for all CBSE board numerical problems.
How do you set up an ICE table for equilibrium problems?
An ICE table (Initial-Change-Equilibrium) is the systematic method for solving equilibrium problems in NCERT Class 11 Chemistry Chapter 6: (1) write balanced equation; (2) list initial concentrations; (3) define change in terms of x using stoichiometric ratios; (4) write equilibrium concentrations as initial + change; (5) substitute into K expression and solve for x. Example: for H₂ + I₂ ⇌ 2HI starting with 0.1 M each, at equilibrium [H₂] = [I₂] = 0.1 − x, [HI] = 2x. Set K_c = (2x)² / (0.1−x)² and solve. This systematic method handles all NCERT equilibrium problems.
How do you calculate pH of mixtures and buffers?
To calculate pH in NCERT Class 11 Chemistry Chapter 6: (1) for strong acid/base, pH = −log(concentration); (2) for weak acid, [H⁺] = √(K_a × C); (3) for weak base, [OH⁻] = √(K_b × C), then pOH and pH = 14 − pOH; (4) for buffer, apply Henderson-Hasselbalch pH = pK_a + log([salt]/[acid]); (5) for mixture of strong acid and base, find excess after neutralisation. Always check assumptions (e.g., x << C for weak acid). Example: 0.1 M CH₃COOH + 0.1 M CH₃COONa has pH = pK_a = 4.74. The MyAiSchool exercise set drills all these patterns.
How is solubility calculated from Ksp?
To calculate molar solubility s from K_sp in NCERT Class 11 Chemistry Chapter 6: (1) write dissolution equation; (2) express ion concentrations as multiples of s; (3) substitute into K_sp expression; (4) solve for s. For salt A_m B_n, K_sp = (ms)^m × (ns)^n = m^m × n^n × s^(m+n). Examples: AgCl (1:1) → K_sp = s², s = √K_sp; PbI₂ (1:2) → K_sp = s × (2s)² = 4s³, s = (K_sp/4)^(1/3); Ag₂CrO₄ (2:1) → K_sp = (2s)² × s = 4s³, s = (K_sp/4)^(1/3). Always state units (mol/L).
How does the common ion effect change solubility?
The common ion effect reduces the solubility of a sparingly soluble salt when a common ion is added. Example from NCERT Class 11 Chemistry Chapter 6: solubility of AgCl in 0.1 M NaCl is much less than in pure water. Calculation: K_sp(AgCl) = [Ag⁺][Cl⁻] = 1.8 × 10⁻¹⁰. In 0.1 M NaCl, [Cl⁻] ≈ 0.1 M (NaCl is strong electrolyte), so [Ag⁺] = K_sp/[Cl⁻] = 1.8 × 10⁻⁹ M, much lower than √K_sp = 1.34 × 10⁻⁵ M in pure water. This principle is used in qualitative analysis for selective precipitation.
What 5-mark questions are common in Chapter 6 board exams?
Common 5-mark CBSE Class 11 Chemistry Chapter 6 questions: (1) define equilibrium constant; derive relation between K_p and K_c; apply to a given reaction; (2) state Le Chatelier's principle; apply to changes in concentration, pressure, temperature with examples; (3) discuss acid-base theories (Arrhenius, Brønsted-Lowry, Lewis) with examples; (4) explain buffer action using Henderson-Hasselbalch equation and calculate pH; (5) calculate solubility and predict precipitation using K_sp. The MyAiSchool exercise set provides model answers, marking schemes and conceptual links.
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