What value of R should I use in Kp = Kc(RT)^(delta n)?

Use R = 0.0831 L.bar.K-1.mol-1 when pressures are in bar, or R = 0.0821 L.atm.K-1.mol-1 when pressures are in atm. The choice must match the pressure units in the problem.

NCERT Solutions for Class 11 Chemistry Chapter 6: Equilibrium (NCERT 2026–27)

Q: What is the difference between Kc and Kp?

Kc is the equilibrium constant in terms of molar concentrations, while Kp is in terms of partial pressures of gases. They are related by Kp = Kc(RT)^(delta n), where delta n is the difference in the number of gaseous moles between products and reactants.

These Class 11 Chemistry Chapter 6 solutions cover Equilibrium with step-by-step answers to all 73 NCERT exercise questions. Every K_c, K_p, pH and solubility numerical is worked out fully with units and the final value cross-checked against the official NCERT answer key, updated for session 2026–27.

Class: 11 Subject: Chemistry Chapter: 6 Chapter Name: Equilibrium Exercises: 73 questions Session: 2026–27

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Chapter overview
Key concepts & definitions
Important formulas
NCERT Exercises (6.1–6.73) — solutions
Extra practice questions
MCQs & Assertion–Reason
Common mistakes
Exam tips
FAQs

Class 11 Chemistry Chapter 6 Solutions – Overview

Chapter 6, Equilibrium, is one of the most important chapters of Class 11 Chemistry and the foundation for Chemical Kinetics, Electrochemistry and Ionic Equilibrium in higher classes. It explains how reversible processes—both physical (melting, evaporation, dissolution) and chemical—reach a state of dynamic equilibrium where forward and reverse rates become equal and macroscopic properties stay constant. The chapter develops the law of mass action, the equilibrium constants K_c and K_p, the reaction quotient Q, and Le Chatelier’s principle for predicting how concentration, pressure, volume, temperature and catalysts shift equilibrium. Its second half covers ionic equilibrium—Arrhenius, Brönsted–Lowry and Lewis acids and bases, ionisation of weak acids/bases, the ionic product of water, the pH scale, buffer solutions, hydrolysis of salts, and the solubility product (K_sp) with the common-ion effect.

Key Concepts & Definitions

Dynamic equilibrium: a reversible reaction reaches equilibrium when the rate of the forward reaction equals the rate of the reverse reaction; concentrations of reactants and products then stay constant though both reactions continue.

Law of chemical equilibrium: for aA + bB ⇌ cC + dD, K_c = [C]^c[D]^d / [A]^a[B]^b at constant temperature.

K_p and K_c: for gaseous equilibria K_p = K_c(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants).

Reaction quotient Q: same expression as K but with non-equilibrium concentrations. If Q < K the reaction goes forward; if Q > K it goes backward; if Q = K it is at equilibrium.

Le Chatelier’s principle: if a system at equilibrium is disturbed, it shifts in the direction that partly opposes the change (in concentration, pressure, volume or temperature).

Acids and bases: Brönsted acid = proton donor, Brönsted base = proton acceptor; a conjugate acid–base pair differs by one H⁺. Lewis acid = electron-pair acceptor, Lewis base = electron-pair donor.

pH scale: pH = −log[H⁺]. At 298 K, K_w = [H⁺][OH⁻] = 10⁻¹⁴, so pH + pOH = 14; neutral pH = 7.

Solubility product (K_sp): the product of ionic concentrations (each raised to its stoichiometric power) in a saturated solution of a sparingly soluble salt. Precipitation occurs when the ionic product exceeds K_sp.

Important Formulas (Chapter 6)

Equilibrium constant: K_c = [products]^coeff / [reactants]^coeff.

K_p–K_c relation: K_p = K_c(RT)^Δn; R = 0.0831 L·bar·K⁻¹·mol⁻¹ (or 0.0821 L·atm·K⁻¹·mol⁻¹).

Reverse / multiplied reactions: K(reverse) = 1 / K(forward); for nA ⇌ products, K′ = Kⁿ.

Gibbs energy: ΔG° = −RT ln K, so K = e^−ΔG°/RT.

Weak acid (Ostwald): K_a = cα²/(1−α) ≈ cα²; [H⁺] = √(K_ac); α = √(K_a/c).

pH: pH = −log[H⁺]; pOH = −log[OH⁻]; pH + pOH = 14; K_a·K_b = K_w = 10⁻¹⁴.

Solubility: for A_xB_y, K_sp = (xS)^x(yS)^y.

NCERT Exercises (6.1–6.73) — Solutions

All questions are reproduced verbatim from the NCERT textbook (Reprint 2026–27). Answers are original, step-by-step, and the final numerical values agree with the official NCERT answer key.

6.1 A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. (a) What is the initial effect of the change on vapour pressure? (b) How do rates of evaporation and condensation change initially? (c) What happens when equilibrium is restored finally and what will be the final vapour pressure?

ANSWER (a) On suddenly increasing the volume, the same amount of vapour now occupies a larger space, so the vapour pressure decreases initially (it falls below the equilibrium value). (b) The rate of evaporation stays the same, because it depends only on temperature and surface area (both unchanged). The rate of condensation decreases, because there are now fewer vapour molecules per unit volume to strike the liquid surface. (c) Since evaporation now exceeds condensation, more liquid evaporates until the vapour pressure rises back to its original value. Equilibrium is restored when the two rates become equal again, and the final vapour pressure equals the original vapour pressure (vapour pressure depends only on temperature, which is unchanged).

6.2 What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂] = 0.60 M, [O₂] = 0.82 M and [SO₃] = 1.90 M ? 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

ANSWER K_c = [SO₃]² / ([SO₂]²[O₂]) = (1.90)² / ((0.60)² × 0.82) = 3.61 / (0.36 × 0.82) = 3.61 / 0.2952 K_c = 12.23 M⁻¹ (i.e. 12.23 mol⁻¹ L). Matches NCERT answer 12.229.

6.3 At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms. I₂(g) ⇌ 2I(g) Calculate K_p for the equilibrium.

ANSWER By volume, mole fraction of I = 0.40 and of I₂ = 0.60. Partial pressures: p_I = 0.40 × 10⁵ = 4 × 10⁴ Pa; p_I₂ = 0.60 × 10⁵ = 6 × 10⁴ Pa. K_p = (p_I)² / p_I₂ = (4 × 10⁴)² / (6 × 10⁴) = (16 × 10⁸) / (6 × 10⁴) K_p = 2.67 × 10⁴ Pa. Matches NCERT answer 2.67 × 10⁴.

6.4 Write the expression for the equilibrium constant, K_c for each of the following reactions: (i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g) (ii) 2Cu(NO₃)₂(s) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g) (iii) CH₃COOC₂H₅(aq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(aq) (iv) Fe³⁺(aq) + 3OH⁻(aq) ⇌ Fe(OH)₃(s) (v) I₂(s) + 5F₂ ⇌ 2IF₅

ANSWER Pure solids and pure liquids are omitted (their activity = 1). (i) K_c = [NO]²[Cl₂] / [NOCl]² (ii) K_c = [NO₂]⁴[O₂] (CuO and Cu(NO₃)₂ are solids) (iii) K_c = [CH₃COOH][C₂H₅OH] / [CH₃COOC₂H₅] (water is the solvent, omitted) (iv) K_c = 1 / ([Fe³⁺][OH⁻]³) (Fe(OH)₃ is a solid) (v) K_c = [IF₅]² / [F₂]⁵ (I₂ is a solid)

6.5 Find out the value of K_c for each of the following equilibria from the value of K_p: (i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 × 10⁻² at 500 K (ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K

ANSWER Use K_p = K_c(RT)^Δn ⇒ K_c = K_p / (RT)^Δn, with R = 0.0831 L·bar·K⁻¹·mol⁻¹. (i) Δn = (2 + 1) − 2 = 1. RT = 0.0831 × 500 = 41.55. K_c = (1.8 × 10⁻²) / 41.55 = 4.33 × 10⁻⁴ mol L⁻¹. (ii) Δn = 1 − 0 = 1. RT = 0.0831 × 1073 = 89.17. K_c = 167 / 89.17 = 1.90 mol L⁻¹. Matches NCERT key.

6.6 For the following equilibrium, K_c = 6.3 × 10¹⁴ at 1000 K. NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g). Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?

ANSWER For the reverse reaction, K_c(reverse) = 1 / K_c(forward). = 1 / (6.3 × 10¹⁴) = 1.59 × 10⁻¹⁵.

6.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

ANSWER The active mass (molar concentration) of a pure solid or pure liquid is fixed, because it equals density ÷ molar mass, both of which are constant at a given temperature and do not change as the reaction proceeds. Since these terms are constants, they are merged into the equilibrium constant itself. Equivalently, the activity of a pure solid or pure liquid is taken as 1, so they do not appear in the K_c (or K_p) expression.

6.8 Reaction between N₂ and O₂ takes place as follows: 2N₂(g) + O₂(g) ⇌ 2N₂O(g). If a mixture of 0.482 mol N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 × 10⁻³⁷, determine the composition of equilibrium mixture.

ANSWER Initial concentrations: [N₂] = 0.482/10 = 0.0482 M; [O₂] = 0.933/10 = 0.0933 M. Let 2x mol L⁻¹ of N₂O form. Because K_c is extremely small (2.0 × 10⁻³⁷), x is negligible, so [N₂] and [O₂] stay essentially unchanged. K_c = [N₂O]² / ([N₂]²[O₂]) ⇒ [N₂O]² = K_c × [N₂]² × [O₂]. = 2.0 × 10⁻³⁷ × (0.0482)² × 0.0933 = 2.0 × 10⁻³⁷ × 2.168 × 10⁻⁴ = 4.336 × 10⁻⁴¹. [N₂O] = √(4.336 × 10⁻⁴¹) = 6.6 × 10⁻²¹ mol L⁻¹. Equilibrium mixture: [N₂] = 0.0482 M, [O₂] = 0.0933 M, [N₂O] = 6.6 × 10⁻²¹ M.

6.9 Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below: 2NO(g) + Br₂(g) ⇌ 2NOBr(g). When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂.

ANSWER From stoichiometry, forming 0.0518 mol NOBr consumes 0.0518 mol NO and 0.0518/2 = 0.0259 mol Br₂. NO remaining = 0.087 − 0.0518 = 0.0352 mol. Br₂ remaining = 0.0437 − 0.0259 = 0.0178 mol. Matches NCERT key.

6.10 At 450 K, K_p = 2.0 × 10¹⁰/bar for the given reaction at equilibrium. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). What is K_c at this temperature ?

ANSWER Δn = 2 − 3 = −1, so K_p = K_c(RT)⁻¹ ⇒ K_c = K_p × RT. RT = 0.0831 × 450 = 37.395 L·bar·mol⁻¹. K_c = (2.0 × 10¹⁰) × 37.395 = 7.48 × 10¹¹ M⁻¹ (NCERT key: 7.47 × 10¹¹).

6.11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium ? 2HI(g) ⇌ H₂(g) + I₂(g)

ANSWER HI decomposed = 0.2 − 0.04 = 0.16 atm. By stoichiometry, p_H₂ = p_I₂ = 0.16/2 = 0.08 atm each. K_p = (p_H₂ · p_I₂) / (p_HI)² = (0.08 × 0.08) / (0.04)² = 0.0064 / 0.0016 K_p = 4.0 (dimensionless, since Δn = 0). Matches NCERT key.

6.12 A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1.7 × 10². Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

ANSWER Concentrations: [N₂] = 1.57/20 = 0.0785 M; [H₂] = 1.92/20 = 0.096 M; [NH₃] = 8.13/20 = 0.4065 M. Q_c = [NH₃]² / ([N₂][H₂]³) = (0.4065)² / (0.0785 × (0.096)³). = 0.1652 / (0.0785 × 8.847 × 10⁻⁴) = 0.1652 / (6.945 × 10⁻⁵) = 2.38 × 10³. Since Q_c (2.38 × 10³) > K_c (1.7 × 10²), the mixture is not at equilibrium; the net reaction proceeds in the reverse direction (NH₃ decomposes into N₂ and H₂) until Q = K.

6.13 The equilibrium constant expression for a gas reaction is, K_c = [NH₃]⁴[O₂]⁵ / ([NO]⁴[H₂O]⁶). Write the balanced chemical equation corresponding to this expression.

ANSWER Species in the numerator are products and those in the denominator are reactants, each power giving its coefficient. Balanced equation: 4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g).

6.14 One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g). Calculate the equilibrium constant for the reaction.

ANSWER 40% of 1 mol of water reacts = 0.4 mol. So 0.4 mol each of H₂ and CO₂ form, and 0.4 mol of CO is used. At equilibrium: H₂O = 0.6, CO = 0.6, H₂ = 0.4, CO₂ = 0.4 mol (in 10 L). Since Δn = 0, the volume cancels and we may use moles directly. K_c = ([H₂][CO₂]) / ([H₂O][CO]) = (0.4 × 0.4) / (0.6 × 0.6) = 0.16 / 0.36 K_c = 0.44 (dimensionless). Matches NCERT key.

6.15 At 700 K, equilibrium constant for the reaction: H₂(g) + I₂(g) ⇌ 2HI(g) is 54.8. If 0.5 mol L⁻¹ of HI(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

ANSWER For the reverse decomposition, K_c(rev) = 1/54.8. By symmetry [H₂] = [I₂] = x. For the given (formation) direction: K_c = [HI]² / ([H₂][I₂]) = (0.5)² / x² = 54.8. x² = 0.25 / 54.8 = 4.562 × 10⁻³ ⇒ x = √(4.562 × 10⁻³) = 0.068 mol L⁻¹. So [H₂] = [I₂] = 0.068 mol L⁻¹. Matches NCERT key.

6.16 What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl(g) ⇌ I₂(g) + Cl₂(g); K_c = 0.14

ANSWER Let x M of I₂ (and x M of Cl₂) form; then ICl used = 2x. At equilibrium: [ICl] = 0.78 − 2x, [I₂] = [Cl₂] = x. K_c = ([I₂][Cl₂]) / [ICl]² = x² / (0.78 − 2x)² = 0.14. Take square root: x / (0.78 − 2x) = √0.14 = 0.374. x = 0.374(0.78 − 2x) = 0.2917 − 0.748x ⇒ 1.748x = 0.2917 ⇒ x = 0.167 M. [I₂] = [Cl₂] = 0.167 M; [ICl] = 0.78 − 2(0.167) = 0.446 M. Matches NCERT key.

6.17 K_p = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? C₂H₆(g) ⇌ C₂H₄(g) + H₂(g)

ANSWER Let p atm of C₂H₆ dissociate. At equilibrium: p_C₂H₆ = 4.0 − p; p_C₂H₄ = p_H₂ = p. K_p = (p × p) / (4.0 − p) = 0.04 ⇒ p² + 0.04p − 0.16 = 0. p = [−0.04 + √(0.0016 + 0.64)] / 2 = (−0.04 + 0.801) / 2 = 0.38 atm. Equilibrium p_C₂H₆ = 4.0 − 0.38 = 3.62 atm. Matches NCERT key.

6.18 Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l) (i) Write the concentration ratio (reaction quotient), Q_c, for this reaction (note: water is not in excess and is not a solvent in this reaction) (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?

ANSWER (i) Q_c = [CH₃COOC₂H₅][H₂O] / ([CH₃COOH][C₂H₅OH]). (ii) Forming 0.171 mol ester uses 0.171 mol acid and 0.171 mol ethanol, and makes 0.171 mol water. At equilibrium: acid = 1.00 − 0.171 = 0.829, ethanol = 0.18 − 0.171 = 0.009, ester = 0.171, water = 0.171 mol. Volume cancels (Δn = 0), so use moles. K_c = (0.171 × 0.171) / (0.829 × 0.009) = 0.02924 / 0.007461 = 3.92. Matches NCERT key. (iii) Now acid = 1.0 − 0.214 = 0.786, ethanol = 0.5 − 0.214 = 0.286, ester = water = 0.214 mol. Q_c = (0.214 × 0.214) / (0.786 × 0.286) = 0.0458 / 0.2248 = 0.204. Since Q_c (0.204) < K_c (3.92), equilibrium has not been reached; more ester will form.

6.19 A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 × 10⁻¹ mol L⁻¹. If value of K_c is 8.3 × 10⁻³, what are the concentrations of PCl₃ and Cl₂ at equilibrium? PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

ANSWER Since PCl₅ gives equal amounts of PCl₃ and Cl₂, let [PCl₃] = [Cl₂] = x. K_c = ([PCl₃][Cl₂]) / [PCl₅] = x² / (0.05) = 8.3 × 10⁻³. x² = 8.3 × 10⁻³ × 0.05 = 4.15 × 10⁻⁴ ⇒ x = √(4.15 × 10⁻⁴) = 0.0204 ≈ 0.02 mol L⁻¹. So [PCl₃] = [Cl₂] = 0.02 mol L⁻¹. Matches NCERT key.

6.20 One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO₂. FeO(s) + CO(g) ⇌ Fe(s) + CO₂(g); K_p = 0.265 atm at 1050 K. What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are: p_CO = 1.4 atm and p_CO₂ = 0.80 atm?

ANSWER Q_p = p_CO₂ / p_CO = 0.80 / 1.4 = 0.571. Since Q_p (0.571) > K_p (0.265), reaction goes backward (CO₂ → CO). Let p atm of CO₂ convert to CO. At equilibrium: p_CO = 1.4 + p, p_CO₂ = 0.80 − p. K_p = (0.80 − p) / (1.4 + p) = 0.265 ⇒ 0.80 − p = 0.371 + 0.265p ⇒ 0.429 = 1.265p ⇒ p = 0.339. p_CO = 1.4 + 0.339 = 1.74 atm; p_CO₂ = 0.80 − 0.339 = 0.46 atm. Matches NCERT key.

6.21 Equilibrium constant, K_c for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K is 0.061. At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L⁻¹ N₂, 2.0 mol L⁻¹ H₂ and 0.5 mol L⁻¹ NH₃. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

ANSWER Q_c = [NH₃]² / ([N₂][H₂]³) = (0.5)² / (3.0 × (2.0)³) = 0.25 / (3.0 × 8) = 0.25 / 24 = 0.0104. Since Q_c (0.0104) < K_c (0.061), the reaction is not at equilibrium; it proceeds in the forward direction (forming more NH₃) until Q = K.

6.22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl(g) ⇌ Br₂(g) + Cl₂(g) for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10⁻³ mol L⁻¹, what is its molar concentration in the mixture at equilibrium?

ANSWER Let 2x decompose: [Br₂] = [Cl₂] = x, [BrCl] = 3.3 × 10⁻³ − 2x. K_c = x² / (3.3 × 10⁻³ − 2x)² = 32. Square root: x / (3.3 × 10⁻³ − 2x) = √32 = 5.657. x = 5.657(3.3 × 10⁻³) − 11.314x ⇒ 12.314x = 0.01867 ⇒ x = 1.516 × 10⁻³. [BrCl] = 3.3 × 10⁻³ − 2(1.516 × 10⁻³) = 3.3 × 10⁻³ − 3.03 × 10⁻³ = 3 × 10⁻⁴ mol L⁻¹. Matches NCERT key.

6.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass. C(s) + CO₂(g) ⇌ 2CO(g). Calculate K_c for this reaction at the above temperature.

ANSWER In 100 g of gas: CO = 90.55 g → 90.55/28 = 3.234 mol; CO₂ = 9.45 g → 9.45/44 = 0.215 mol. Total = 3.449 mol. Mole fractions: x_CO = 3.234/3.449 = 0.9377; x_CO₂ = 0.0623. At 1 atm, p_CO = 0.9377 atm, p_CO₂ = 0.0623 atm. K_p = (p_CO)² / p_CO₂ = (0.9377)² / 0.0623 = 0.8793 / 0.0623 = 14.11 atm. K_c = K_p / (RT)^Δn, Δn = 2 − 1 = 1; RT = 0.0821 × 1127 = 92.53. K_c = 14.11 / 92.53 = 0.153 mol L⁻¹ (NCERT key: 0.149; small rounding difference).

6.24 Calculate a) ΔG° and b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K. NO(g) + ½ O₂(g) ⇌ NO₂(g) where Δ_fG°(NO₂) = 52.0 kJ/mol, Δ_fG°(NO) = 87.0 kJ/mol, Δ_fG°(O₂) = 0 kJ/mol.

ANSWER (a) Δ_rG° = Δ_fG°(NO₂) − [Δ_fG°(NO) + ½Δ_fG°(O₂)] = 52.0 − (87.0 + 0) = −35.0 kJ/mol. (b) ΔG° = −RT ln K ⇒ ln K = −ΔG°/RT = −(−35000) / (8.314 × 298) = 35000 / 2477.6 = 14.126. K = e^14.126 = 1.36 × 10⁶ (NCERT key: 1.365 × 10⁶).

6.25 Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume? (a) PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) (b) CaO(s) + CO₂(g) ⇌ CaCO₃(s) (c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)

ANSWER A decrease in pressure (volume increase) shifts equilibrium toward the side with more moles of gas. (a) Products have 2 mol gas vs 1 mol on the left, so equilibrium shifts right: moles of products increase. (b) Left has 1 mol gas (CO₂), right has 0 mol gas, so equilibrium shifts left: moles of product (CaCO₃) decrease. (c) Gaseous moles are equal on both sides (4 H₂O vs 4 H₂); pressure change has no effect, so moles of products remain the same.

6.26 Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction. (i) COCl₂(g) ⇌ CO(g) + Cl₂(g) (ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g) (iii) CO₂(g) + C(s) ⇌ 2CO(g) (iv) 2H₂(g) + CO(g) ⇌ CH₃OH(g) (v) CaCO₃(s) ⇌ CaO(s) + CO₂(g) (vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)

ANSWER Increasing pressure shifts equilibrium toward the side with fewer gaseous moles. Compare Δn_gas = (products − reactants). (i) 2 mol vs 1 mol ⇒ shifts backward (toward COCl₂). (ii) 3 mol vs 3 mol ⇒ no effect. (iii) 2 mol vs 1 mol ⇒ shifts backward (toward CO₂). (iv) 1 mol vs 3 mol ⇒ shifts forward (toward CH₃OH). (v) 1 mol vs 0 mol gas ⇒ shifts backward (toward CaCO₃). (vi) 10 mol vs 9 mol ⇒ shifts backward (toward NH₃ + O₂).

6.27 The equilibrium constant for the following reaction is 1.6 × 10⁵ at 1024 K. H₂(g) + Br₂(g) ⇌ 2HBr(g). Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

ANSWER Start with pure HBr; it decomposes. For 2HBr ⇌ H₂ + Br₂, K′ = 1/(1.6 × 10⁵) = 6.25 × 10⁻⁶. Let p_H₂ = p_Br₂ = p; then p_HBr = 10.0 − 2p ≈ 10.0 (decomposition tiny). K′ = p² / (p_HBr)² ⇒ p² = 6.25 × 10⁻⁶ × (10.0)² = 6.25 × 10⁻⁴ ⇒ p = 2.5 × 10⁻² bar. So p_H₂ = p_Br₂ = 2.5 × 10⁻² bar; p_HBr ≈ 10.0 bar. Matches NCERT key.

6.28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g) (a) Write as expression for K_p for the above reaction. (b) How will the values of K_p and composition of equilibrium mixture be affected by (i) increasing the pressure (ii) increasing the temperature (iii) using a catalyst?

ANSWER (a) K_p = (p_CO · (p_H₂)³) / (p_CH₄ · p_H₂O). (b)(i) Increasing pressure: gaseous moles are 4 (right) vs 2 (left), so equilibrium shifts backward (less product); K_p is unchanged (it depends only on temperature). (b)(ii) Increasing temperature: the reaction is endothermic, so equilibrium shifts forward (more product) and the value of K_p increases. (b)(iii) Using a catalyst: it speeds up both forward and reverse reactions equally, so it does not change K_p or the equilibrium composition; it only helps equilibrium to be reached faster.

6.29 Describe the effect of: a) addition of H₂, b) addition of CH₃OH, c) removal of CO, d) removal of CH₃OH on the equilibrium of the reaction: 2H₂(g) + CO(g) ⇌ CH₃OH(g)

ANSWER By Le Chatelier’s principle: (a) Adding H₂ increases a reactant, so equilibrium shifts forward (more CH₃OH). (b) Adding CH₃OH increases a product, so equilibrium shifts backward (toward H₂ and CO). (c) Removing CO lowers a reactant, so equilibrium shifts backward (CH₃OH decomposes to replace CO). (d) Removing CH₃OH lowers a product, so equilibrium shifts forward (more CH₃OH formed).

6.30 At 473 K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 × 10⁻³. If decomposition is depicted as, PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) Δ_rH° = 124.0 kJ mol⁻¹ a) write an expression for K_c for the reaction. b) what is the value of K_c for the reverse reaction at the same temperature? c) what would be the effect on K_c if (i) more PCl₅ is added (ii) pressure is increased (iii) the temperature is increased ?

ANSWER (a) K_c = ([PCl₃][Cl₂]) / [PCl₅]. (b) K_c(reverse) = 1 / K_c = 1 / (8.3 × 10⁻³) = 120.48. Matches NCERT key. (c)(i) Adding more PCl₅: K_c is unchanged (only the position of equilibrium shifts, not the constant). (c)(ii) Increasing pressure: K_c is unchanged (K_c depends only on temperature). (c)(iii) Increasing temperature: the reaction is endothermic (ΔH positive), so K_c increases.

6.31 Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g). If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and steam such that p_CO = p_H₂O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_p = 10.1 at 400 °C

ANSWER Let p bar of CO react. At equilibrium: p_CO = p_H₂O = 4.0 − p; p_CO₂ = p_H₂ = p. K_p = (p × p) / ((4.0 − p)(4.0 − p)) = p² / (4.0 − p)² = 10.1. Square root: p / (4.0 − p) = √10.1 = 3.178 ⇒ p = 12.71 − 3.178p ⇒ 4.178p = 12.71 ⇒ p = 3.04 bar. So p_H₂ = p = 3.04 bar, and equilibrium p_CO = p_H₂O = 4.0 − 3.04 = 0.96 bar. (The NCERT key reports the residual reactant pressure, 0.96 bar.)

6.32 Predict which of the following reaction will have appreciable concentration of reactants and products: a) Cl₂(g) ⇌ 2Cl(g) K_c = 5 × 10⁻³⁹ b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g) K_c = 3.7 × 10⁸ c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g) K_c = 1.8

ANSWER A reaction has appreciable amounts of both reactants and products only when K_c is neither very large nor very small (roughly 10⁻³ to 10³). (a) K_c = 5 × 10⁻³⁹ (extremely small) ⇒ mainly reactants. (b) K_c = 3.7 × 10⁸ (very large) ⇒ mainly products. (c) K_c = 1.8 (moderate) ⇒ appreciable concentrations of both reactants and products.

6.33 The value of K_c for the reaction 3O₂(g) ⇌ 2O₃(g) is 2.0 × 10⁻⁵⁰ at 25 °C. If the equilibrium concentration of O₂ in air at 25 °C is 1.6 × 10⁻², what is the concentration of O₃?

ANSWER K_c = [O₃]² / [O₂]³ ⇒ [O₃]² = K_c × [O₂]³. [O₂]³ = (1.6 × 10⁻²)³ = 4.096 × 10⁻⁶. [O₃]² = 2.0 × 10⁻⁵⁰ × 4.096 × 10⁻⁶ = 8.192 × 10⁻⁵⁶. [O₃] = √(8.192 × 10⁻⁵⁶) = 2.86 × 10⁻²⁸ M. Matches NCERT key.

6.34 The reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1 L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.

ANSWER Volume = 1 L, so concentrations equal moles: [CO] = 0.30, [H₂] = 0.10, [H₂O] = 0.02 M. K_c = ([CH₄][H₂O]) / ([CO][H₂]³) ⇒ [CH₄] = K_c × [CO][H₂]³ / [H₂O]. = 3.90 × (0.30 × (0.10)³) / 0.02 = 3.90 × (0.30 × 10⁻³) / 0.02 = 3.90 × (3.0 × 10⁻⁴) / 0.02. = 3.90 × 0.015 = 5.85 × 10⁻² M. Matches NCERT key.

6.35 What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO₂, CN⁻, HClO₄, F⁻, OH⁻, CO₃²⁻, and S²⁻

ANSWER A conjugate acid–base pair is a pair of species that differ by one proton (H⁺): the acid has one more H⁺ than its conjugate base.

Species	Conjugate acid/base
HNO₂ (acid)	NO₂⁻ (base)
CN⁻ (base)	HCN (acid)
HClO₄ (acid)	ClO₄⁻ (base)
F⁻ (base)	HF (acid)
OH⁻ (base)	H₂O (acid)
CO₃²⁻ (base)	HCO₃⁻ (acid)
S²⁻ (base)	HS⁻ (acid)

6.36 Which of the followings are Lewis acids? H₂O, BF₃, H⁺, and NH₄⁺

ANSWER Lewis acids accept an electron pair. BF₃ (electron-deficient B) and H⁺ (empty orbital) are Lewis acids. NH₄⁺ is also counted as a Lewis acid as per the NCERT key (it can supply H⁺). H₂O is a Lewis base. Lewis acids: BF₃, H⁺, NH₄⁺.

6.37 What will be the conjugate bases for the Brönsted acids: HF, H₂SO₄ and HCO₃⁻?

ANSWER Remove one H⁺ from each acid: HF → F⁻; H₂SO₄ → HSO₄⁻; HCO₃⁻ → CO₃²⁻.

6.38 Write the conjugate acids for the following Brönsted bases: NH₂⁻, NH₃ and HCOO⁻.

ANSWER Add one H⁺ to each base: NH₂⁻ → NH₃; NH₃ → NH₄⁺; HCOO⁻ → HCOOH.

6.39 The species: H₂O, HCO₃⁻, HSO₄⁻ and NH₃ can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base.

ANSWER

Species	Conjugate acid (add H⁺)	Conjugate base (remove H⁺)
H₂O	H₃O⁺	OH⁻
HCO₃⁻	H₂CO₃	CO₃²⁻
HSO₄⁻	H₂SO₄	SO₄²⁻
NH₃	NH₄⁺	NH₂⁻

6.40 Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH⁻ (b) F⁻ (c) H⁺ (d) BCl₃.

ANSWER (a) OH⁻ — Lewis base; it donates a lone pair (e.g. OH⁻ + H⁺ → H₂O). (b) F⁻ — Lewis base; it donates a lone pair (e.g. F⁻ + BF₃ → BF₄⁻). (c) H⁺ — Lewis acid; it accepts a lone pair into its empty orbital (e.g. H⁺ + :NH₃ → NH₄⁺). (d) BCl₃ — Lewis acid; the electron-deficient boron accepts a lone pair (e.g. BCl₃ + :NH₃ → Cl₃B←NH₃).

6.41 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10⁻³ M. What is its pH?

ANSWER pH = −log[H⁺] = −log(3.8 × 10⁻³) = −(log 3.8 + log 10⁻³) = −(0.5798 − 3) = 3 − 0.58. pH = 2.42. Matches NCERT key.

6.42 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.

ANSWER [H⁺] = 10^−pH = 10^−3.76 = 10^0.24 × 10⁻⁴. antilog(0.24) = 1.74, so [H⁺] = 1.74 × 10⁻⁴ M (NCERT key: 1.7 × 10⁻⁴ M).

6.43 The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10⁻⁴, 1.8 × 10⁻⁴ and 4.8 × 10⁻⁹ respectively. Calculate the ionization constants of the corresponding conjugate base.

ANSWER For a conjugate base, K_b = K_w / K_a = 10⁻¹⁴ / K_a. F⁻: K_b = 10⁻¹⁴ / (6.8 × 10⁻⁴) = 1.47 × 10⁻¹¹ (key: 1.5 × 10⁻¹¹). HCOO⁻: K_b = 10⁻¹⁴ / (1.8 × 10⁻⁴) = 5.6 × 10⁻¹¹. CN⁻: K_b = 10⁻¹⁴ / (4.8 × 10⁻⁹) = 2.08 × 10⁻⁶. Matches NCERT key.

6.44 The ionization constant of phenol is 1.0 × 10⁻¹⁰. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?

ANSWER For pure phenol: [C₆H₅O⁻] = √(K_ac) = √(1.0 × 10⁻¹⁰ × 0.05) = √(5.0 × 10⁻¹²) = 2.24 × 10⁻⁶ M (key: 2.2 × 10⁻⁶). Degree of ionization (pure phenol) α = 2.24 × 10⁻⁶ / 0.05 = 4.47 × 10⁻⁵. With 0.01 M sodium phenolate (common ion C₆H₅O⁻): [H⁺] = K_a × c / [C₆H₅O⁻] = (1.0 × 10⁻¹⁰ × 0.05) / 0.01 = 5.0 × 10⁻¹⁰ M. Then α = [H⁺]/c = 5.0 × 10⁻¹⁰ / 0.05 = 1.0 × 10⁻⁸. Matches NCERT key.

6.45 The first ionization constant of H₂S is 9.1 × 10⁻⁸. Calculate the concentration of HS⁻ ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H₂S is 1.2 × 10⁻¹³, calculate the concentration of S²⁻ under both conditions.

ANSWER In 0.1 M H₂S alone: [HS⁻] = [H⁺] = √(K_a1c) = √(9.1 × 10⁻⁸ × 0.1) = √(9.1 × 10⁻⁹) = 9.54 × 10⁻⁵ M. With 0.1 M HCl, [H⁺] ≈ 0.1 M (from HCl). Then [HS⁻] = K_a1 × [H₂S] / [H⁺] = (9.1 × 10⁻⁸ × 0.1) / 0.1 = 9.1 × 10⁻⁸ M (much smaller, common-ion effect). [S²⁻] = K_a2 × [HS⁻] / [H⁺]. In pure H₂S, [HS⁻] = [H⁺], so [S²⁻] = K_a2 = 1.2 × 10⁻¹³ M. In 0.1 M HCl: [S²⁻] = (1.2 × 10⁻¹³ × 9.1 × 10⁻⁸) / 0.1 = 1.09 × 10⁻¹⁹ M. Matches NCERT key.

6.46 The ionization constant of acetic acid is 1.74 × 10⁻⁵. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

ANSWER α = √(K_a/c) = √(1.74 × 10⁻⁵ / 0.05) = √(3.48 × 10⁻⁴) = 0.01865 (≈ 1.87%). [CH₃COO⁻] = cα = 0.05 × 0.01865 = 9.3 × 10⁻⁴ M (key: 0.00093). [H⁺] = 9.3 × 10⁻⁴ M ⇒ pH = −log(9.3 × 10⁻⁴) = 3.03. Matches NCERT key.

6.47 It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.

ANSWER [H⁺] = 10^−4.15 = antilog(−4.15) = 7.08 × 10⁻⁵ M = [anion A⁻] = 7.08 × 10⁻⁵ M. K_a = [H⁺][A⁻] / [HA] = (7.08 × 10⁻⁵)² / (0.01 − 7.08 × 10⁻⁵) ≈ (5.01 × 10⁻⁹) / 0.00993 = 5.08 × 10⁻⁷. pK_a = −log(5.08 × 10⁻⁷) = 6.29. Matches NCERT key.

6.48 Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

ANSWER (a) [H⁺] = 0.003 M; pH = −log(3 × 10⁻³) = 2.52. (b) [OH⁻] = 0.005 M; pOH = −log(5 × 10⁻³) = 2.30; pH = 14 − 2.30 = 11.70. (c) [H⁺] = 0.002 M; pH = −log(2 × 10⁻³) = 2.70. (d) [OH⁻] = 0.002 M; pOH = 2.70; pH = 14 − 2.70 = 11.30. Matches NCERT key.

6.49 Calculate the pH of the following solutions: a) 2 g of TlOH dissolved in water to give 2 litre of solution. b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution. c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

ANSWER a) M(TlOH) = 204 + 17 = 221 g/mol. Moles = 2/221 = 9.05 × 10⁻³; [OH⁻] = 9.05 × 10⁻³/2 = 4.52 × 10⁻³ M. pOH = 2.34; pH = 11.65. b) M(Ca(OH)₂) = 74 g/mol. Moles = 0.3/74 = 4.05 × 10⁻³ in 0.5 L → 8.1 × 10⁻³ M; [OH⁻] = 2 × 8.1 × 10⁻³ = 1.62 × 10⁻² M. pOH = 1.79; pH = 12.21. c) M(NaOH) = 40 g/mol. Moles = 0.3/40 = 7.5 × 10⁻³ in 0.2 L → [OH⁻] = 3.75 × 10⁻² M. pOH = 1.43; pH = 12.57. d) After dilution [HCl] = (13.6 × 1)/1000 = 1.36 × 10⁻² M = [H⁺]. pH = −log(1.36 × 10⁻²) = 1.87. Matches NCERT key.

6.50 The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.

ANSWER [H⁺] = cα = 0.1 × 0.132 = 0.0132 M. pH = −log(0.0132) = 1.88. K_a = cα²/(1−α) = 0.1 × (0.132)²/(1 − 0.132) = 0.1 × 0.01742 / 0.868 = 2.01 × 10⁻³. pK_a = −log(2.01 × 10⁻³) = 2.70. Matches NCERT key.

6.51 The pH of 0.005 M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant and pK_b.

ANSWER pOH = 14 − 9.95 = 4.05; [OH⁻] = 10^−4.05 = 8.91 × 10⁻⁵ M. K_b = [OH⁻]²/(c) = (8.91 × 10⁻⁵)² / 0.005 = (7.94 × 10⁻⁹) / 0.005 = 1.6 × 10⁻⁶. pK_b = −log(1.6 × 10⁻⁶) = 5.8. Matches NCERT key.

6.52 What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 6.7 (K_b = 4.27 × 10⁻¹⁰). Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

ANSWER [OH⁻] = √(K_bc) = √(4.27 × 10⁻¹⁰ × 0.001) = √(4.27 × 10⁻¹³) = 6.53 × 10⁻⁷ M. pOH = −log(6.53 × 10⁻⁷) = 6.19; pH = 14 − 6.19 = 7.81. α = [OH⁻]/c = 6.53 × 10⁻⁷ / 0.001 = 6.53 × 10⁻⁴. For the conjugate acid (anilinium ion): K_a = K_w/K_b = 10⁻¹⁴ / (4.27 × 10⁻¹⁰) = 2.35 × 10⁻⁵. Matches NCERT key.

6.53 Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl ?

ANSWER K_a = 10^−4.74 = 1.82 × 10⁻⁵. For pure acid, α = √(K_a/c) = √(1.82 × 10⁻⁵/0.05) = √(3.64 × 10⁻⁴) = 0.019 (≈ 1.9%). (a) With 0.01 M HCl, [H⁺] ≈ 0.01 M. Then α = K_a/[H⁺] = 1.82 × 10⁻⁵/0.01 = 1.8 × 10⁻³. (b) With 0.1 M HCl, α = K_a/[H⁺] = 1.82 × 10⁻⁵/0.1 = 1.8 × 10⁻⁴. The common-ion effect strongly suppresses ionization. Matches NCERT key (0.0018, 0.00018).

6.54 The ionization constant of dimethylamine is 5.4 × 10⁻⁴. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M in NaOH?

ANSWER α = √(K_b/c) = √(5.4 × 10⁻⁴/0.02) = √(0.027) = 0.164 (≈ 16.4%). With 0.1 M NaOH, [OH⁻] ≈ 0.1 M (common ion). Then α = K_b/[OH⁻] = 5.4 × 10⁻⁴/0.1 = 5.4 × 10⁻³ = 0.54%. Matches NCERT key (α = 0.0054).

6.55 Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below: (a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2 (c) Human blood, 7.38 (d) Human saliva, 6.4.

ANSWER [H⁺] = 10^−pH: (a) 10^−6.83 = 1.48 × 10⁻⁷ M. (b) 10^−1.2 = 0.063 M (6.3 × 10⁻²). (c) 10^−7.38 = 4.17 × 10⁻⁸ M. (d) 10^−6.4 = 3.98 × 10⁻⁷ M. Matches NCERT key.

6.56 The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

ANSWER [H⁺] = 10^−pH: Milk (6.8): 1.58 × 10⁻⁷ M; Black coffee (5.0): 1.0 × 10⁻⁵ M; Tomato juice (4.2): 6.31 × 10⁻⁵ M; Lemon juice (2.2): 6.31 × 10⁻³ M; Egg white (7.8): 1.58 × 10⁻⁸ M. Matches NCERT key.

6.57 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?

ANSWER M(KOH) = 56.1 g/mol. Moles = 0.561/56.1 = 0.01 mol in 0.2 L → [KOH] = 0.05 M. KOH is a strong base: [K⁺] = [OH⁻] = 0.05 M. [H⁺] = K_w/[OH⁻] = 10⁻¹⁴/0.05 = 2.0 × 10⁻¹³ M. pH = −log(2.0 × 10⁻¹³) = 12.70. Matches NCERT key.

6.58 The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

ANSWER M(Sr(OH)₂) = 87.6 + 34 = 121.6 g/mol. Molar solubility = 19.23/121.6 = 0.1581 M. [Sr²⁺] = 0.1581 M; [OH⁻] = 2 × 0.1581 = 0.3162 M. pOH = −log(0.3162) = 0.50; pH = 14 − 0.50 = 13.50. Matches NCERT key.

6.59 The ionization constant of propanoic acid is 1.32 × 10⁻⁵. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also?

ANSWER α = √(K_a/c) = √(1.32 × 10⁻⁵/0.05) = √(2.64 × 10⁻⁴) = 1.63 × 10⁻². [H⁺] = cα = 0.05 × 1.63 × 10⁻² = 8.13 × 10⁻⁴ M; pH = −log(8.13 × 10⁻⁴) = 3.09. With 0.01 M HCl, [H⁺] ≈ 0.01 M. α = K_a/[H⁺] = 1.32 × 10⁻⁵/0.01 = 1.32 × 10⁻³. Matches NCERT key.

6.60 The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

ANSWER [H⁺] = 10^−2.34 = 4.57 × 10⁻³ M. α = [H⁺]/c = 4.57 × 10⁻³/0.1 = 0.0457. K_a = cα² = 0.1 × (0.0457)² = 0.1 × 2.09 × 10⁻³ = 2.09 × 10⁻⁴. Matches NCERT key.

6.61 The ionization constant of nitrous acid is 4.5 × 10⁻⁴. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

ANSWER NaNO₂ is a salt of strong base + weak acid. K_h = K_w/K_a = 10⁻¹⁴/(4.5 × 10⁻⁴) = 2.22 × 10⁻¹¹. [OH⁻] = √(K_hc) = √(2.22 × 10⁻¹¹ × 0.04) = √(8.89 × 10⁻¹³) = 9.43 × 10⁻⁷ M. pOH = 6.03; pH = 14 − 6.03 = 7.97. Degree of hydrolysis h = [OH⁻]/c = 9.43 × 10⁻⁷/0.04 = 2.36 × 10⁻⁵. Matches NCERT key.

6.62 A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.

ANSWER [H⁺] = 10^−3.44 = 3.63 × 10⁻⁴ M (salt of weak base + strong acid, acidic hydrolysis). K_h = [H⁺]²/c = (3.63 × 10⁻⁴)²/0.02 = (1.32 × 10⁻⁷)/0.02 = 6.6 × 10⁻⁶. K_b(pyridine) = K_w/K_h = 10⁻¹⁴/(6.6 × 10⁻⁶) = 1.5 × 10⁻⁹. Matches NCERT key.

6.63 Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF

ANSWER NaCl (strong acid + strong base) → neutral. KBr (strong acid + strong base) → neutral. NaCN (weak acid HCN + strong base) → basic. NH₄NO₃ (strong acid + weak base NH₃) → acidic. NaNO₂ (weak acid HNO₂ + strong base) → basic. KF (weak acid HF + strong base) → basic. Matches NCERT key.

6.64 The ionization constant of chloroacetic acid is 1.35 × 10⁻³. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?

ANSWER Acid: [H⁺] = √(K_ac) = √(1.35 × 10⁻³ × 0.1) = √(1.35 × 10⁻⁴) = 1.16 × 10⁻² M; pH = −log(1.16 × 10⁻²) = 1.94 ≈ 1.9. Salt (sodium chloroacetate, weak acid + strong base): K_h = K_w/K_a = 10⁻¹⁴/(1.35 × 10⁻³) = 7.41 × 10⁻¹². [OH⁻] = √(K_hc) = √(7.41 × 10⁻¹² × 0.1) = 8.61 × 10⁻⁷ M; pOH = 6.06; pH = 7.9. Matches NCERT key.

6.65 Ionic product of water at 310 K is 2.7 × 10⁻¹⁴. What is the pH of neutral water at this temperature?

ANSWER For neutral water, [H⁺] = [OH⁻] = √K_w = √(2.7 × 10⁻¹⁴) = 1.64 × 10⁻⁷ M. pH = −log(1.64 × 10⁻⁷) = 6.78. (Slightly less than 7 because K_w is larger at 310 K.) Matches NCERT key.

6.66 Calculate the pH of the resultant mixtures: a) 10 mL of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HCl b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂ c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH

ANSWER a) OH⁻ = 10 × 0.2 × 2 = 4 mmol; H⁺ = 25 × 0.1 = 2.5 mmol. Excess OH⁻ = 1.5 mmol in 35 mL → [OH⁻] = 1.5/35 = 0.0429 M. pOH = 1.37; pH = 12.6. b) H⁺ = 10 × 0.01 × 2 = 0.2 mmol; OH⁻ = 10 × 0.01 × 2 = 0.2 mmol. Exactly neutralised → pH = 7.00. c) H⁺ = 10 × 0.1 × 2 = 2 mmol; OH⁻ = 10 × 0.1 = 1 mmol. Excess H⁺ = 1 mmol in 20 mL → [H⁺] = 0.05 M. pH = −log(0.05) = 1.3. Matches NCERT key.

6.67 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions.

ANSWER Using K_sp values from Table 6.9: Ag₂CrO₄ (K_sp = 1.1 × 10⁻¹²): K_sp = (2S)²(S) = 4S³ ⇒ S = (1.1 × 10⁻¹²/4)^1/3 = 0.65 × 10⁻⁴ M; [Ag⁺] = 1.30 × 10⁻⁴ M, [CrO₄²⁻] = 0.65 × 10⁻⁴ M. BaCrO₄ (K_sp = 1.2 × 10⁻¹⁰): K_sp = S² ⇒ S = 1.1 × 10⁻⁵ M; [Ba²⁺] = [CrO₄²⁻] = 1.1 × 10⁻⁵ M. Fe(OH)₃ (K_sp = 1.0 × 10⁻³⁸): K_sp = S(3S)³ = 27S⁴ ⇒ S = 1.39 × 10⁻¹⁰ M; [Fe³⁺] = 1.39 × 10⁻¹⁰ M, [OH⁻] = 4.17 × 10⁻¹⁰ M. PbCl₂ (K_sp = 1.6 × 10⁻⁵): K_sp = S(2S)² = 4S³ ⇒ S = 1.59 × 10⁻² M; [Pb²⁺] = 1.59 × 10⁻² M, [Cl⁻] = 3.18 × 10⁻² M. Hg₂I₂ (K_sp = 4.5 × 10⁻²⁹): K_sp = S(2S)² = 4S³ ⇒ S = 2.24 × 10⁻¹⁰ M; [Hg₂²⁺] = 2.24 × 10⁻¹⁰ M, [I⁻] = 4.48 × 10⁻¹⁰ M. Matches NCERT key.

6.68 The solubility product constant of Ag₂CrO₄ and AgBr are 1.1 × 10⁻¹² and 5.0 × 10⁻¹³ respectively. Calculate the ratio of the molarities of their saturated solutions.

ANSWER Ag₂CrO₄: S₁ = (K_sp/4)^1/3 = (1.1 × 10⁻¹²/4)^1/3 = (2.75 × 10⁻¹³)^1/3 = 6.5 × 10⁻⁵ M. AgBr: S₂ = √K_sp = √(5.0 × 10⁻¹³) = 7.07 × 10⁻⁷ M. Ratio S₁/S₂ = (6.5 × 10⁻⁵)/(7.07 × 10⁻⁷) = 91.9 (Ag₂CrO₄ is about 92 times more soluble). Matches NCERT key.

6.69 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_sp = 7.4 × 10⁻⁸).

ANSWER On mixing equal volumes, each concentration is halved: [Cu²⁺] = [IO₃⁻] = 0.001 M. Ionic product = [Cu²⁺][IO₃⁻]² = (0.001)(0.001)² = 1.0 × 10⁻⁹. Since ionic product (1.0 × 10⁻⁹) < K_sp (7.4 × 10⁻⁸), the solution is unsaturated → no precipitation. Matches NCERT key.

6.70 The ionization constant of benzoic acid is 6.46 × 10⁻⁵ and K_sp for silver benzoate is 2.5 × 10⁻¹³. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

ANSWER At pH 3.19, [H⁺] = 10^−3.19 = 6.46 × 10⁻⁴ M. The ratio [C₆H₅COOH]/[C₆H₅COO⁻] = [H⁺]/K_a = (6.46 × 10⁻⁴)/(6.46 × 10⁻⁵) = 10. So total benzoate = [C₆H₅COO⁻](1 + 10) = 11 × [C₆H₅COO⁻]. In pure water, solubility S_w = √K_sp = √(2.5 × 10⁻¹³) = 5.0 × 10⁻⁷ M. In the buffer, K_sp = [Ag⁺][C₆H₅COO⁻], with [Ag⁺] = S and [C₆H₅COO⁻] = S/11. So S²/11 = K_sp ⇒ S = √(11 K_sp) = √(11 × 2.5 × 10⁻¹³) = 1.66 × 10⁻⁶ M. Ratio = S/S_w = (1.66 × 10⁻⁶)/(5.0 × 10⁻⁷) = 3.32 times more soluble. Matches NCERT key (3.317).

6.71 What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, K_sp = 6.3 × 10⁻¹⁸).

ANSWER On mixing equal volumes, each concentration halves. Let the original concentration be C; then [Fe²⁺] = [S²⁻] = C/2. At the threshold of precipitation, ionic product = K_sp: (C/2)(C/2) = 6.3 × 10⁻¹⁸ ⇒ C²/4 = 6.3 × 10⁻¹⁸. C² = 2.52 × 10⁻¹⁷ ⇒ C = 5.0 × 10⁻⁹ M. (Equilibrium ion conc. = C/2 = 2.5 × 10⁻⁹ M, matching the NCERT key.)

6.72 What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, K_sp is 9.1 × 10⁻⁶).

ANSWER Molar solubility S = √K_sp = √(9.1 × 10⁻⁶) = 3.02 × 10⁻³ M. M(CaSO₄) = 136 g/mol. Moles in 1 g = 1/136 = 7.35 × 10⁻³ mol. Volume = moles / S = (7.35 × 10⁻³) / (3.02 × 10⁻³) = 2.43 L of water. Matches NCERT key.

6.73 The concentration of sulphide ion in 0.1 M HCl solution saturated with hydrogen sulphide is 1.0 × 10⁻¹⁹ M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO₄, MnCl₂, ZnCl₂ and CdCl₂. in which of these solutions precipitation will take place?

ANSWER After mixing 10 mL + 5 mL = 15 mL: [S²⁻] = (1.0 × 10⁻¹⁹ × 10)/15 = 6.67 × 10⁻²⁰ M; [M²⁺] = (0.04 × 5)/15 = 1.33 × 10⁻² M. Ionic product = [M²⁺][S²⁻] = (1.33 × 10⁻²)(6.67 × 10⁻²⁰) = 8.87 × 10⁻²² for each metal. Compare with K_sp (Table 6.9): FeS = 6.3 × 10⁻¹⁸, MnS = 2.5 × 10⁻¹³, ZnS = 1.6 × 10⁻²⁴, CdS = 8.0 × 10⁻²⁷. Precipitation occurs only when ionic product > K_sp. Here 8.87 × 10⁻²² exceeds K_sp only for CdS (8.0 × 10⁻²⁷). So precipitation takes place only in the CdCl₂ solution. Matches NCERT key.

Extra Practice Questions

Short Answer Type Questions

Q1. State the law of chemical equilibrium.

ANSWERAt constant temperature, for a reversible reaction at equilibrium the ratio of the product of the molar concentrations of the products to that of the reactants (each raised to its stoichiometric coefficient) is a constant, called the equilibrium constant K_c.

Q2. Why is equilibrium said to be “dynamic”?

ANSWERBecause at equilibrium the forward and reverse reactions do not stop—they continue at equal rates. The concentrations remain constant only because both processes balance each other, not because the reactions have ceased.

Q3. What is the common-ion effect? Give one example.

ANSWERThe suppression of the ionization of a weak electrolyte by adding a strong electrolyte that has an ion in common. For example, adding CH₃COONa to acetic acid increases [CH₃COO⁻], which shifts the ionization equilibrium backward and lowers the degree of ionization of acetic acid.

Q4. Define buffer solution and give one example of an acidic buffer.

ANSWERA buffer is a solution that resists a change in pH on adding small amounts of acid or base. An acidic buffer is a mixture of a weak acid and its salt with a strong base, e.g. CH₃COOH + CH₃COONa (pH ≈ 4.75).

Q5. How does a catalyst affect a reaction at equilibrium?

ANSWERA catalyst lowers the activation energy of both forward and reverse reactions equally, so it speeds up the attainment of equilibrium but does not change the equilibrium constant or the equilibrium concentrations.

Long Answer Type Questions

Q1. State Le Chatelier’s principle and apply it to the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −ve, to give optimum conditions for the manufacture of ammonia.

ANSWERLe Chatelier’s principle states that when a system at equilibrium is disturbed, it adjusts so as to partly oppose the change. In the Haber process: (i) the forward reaction reduces gaseous moles (4 → 2), so high pressure (about 200 atm) favours NH₃; (ii) the reaction is exothermic, so a low temperature favours NH₃, but too low a temperature makes the rate impractically slow, hence a moderate temperature (about 700 K) is used as a compromise; (iii) continuously removing NH₃ and adding fresh N₂ and H₂ shifts equilibrium forward; (iv) a catalyst (finely divided iron with molybdenum) speeds up attainment of equilibrium without altering its position. These optimum conditions maximise both yield and rate.

Q2. Derive Ostwald’s dilution law for a weak acid and state its limitation.

ANSWERFor a weak acid HA ⇌ H⁺ + A⁻ with initial concentration c and degree of ionization α: at equilibrium [HA] = c(1−α), [H⁺] = [A⁻] = cα. So K_a = (cα)(cα)/c(1−α) = cα²/(1−α). For a weak acid α « 1, so (1−α) ≈ 1 and K_a ≈ cα², giving α = √(K_a/c). This shows the degree of ionization increases on dilution (as c decreases). Limitation: the approximation fails for moderately strong electrolytes (where α is not small) and for strong electrolytes, which are essentially completely ionized and do not obey the law.

Q3. Explain the relation K_p = K_c(RT)^Δn and discuss the three cases of Δn.

ANSWERFor a gaseous equilibrium, partial pressure p = (n/V)RT = (concentration)RT. Substituting concentrations expressed as p/RT into K_c and rearranging gives K_p = K_c(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants). Three cases arise: (i) if Δn = 0 (e.g. H₂ + I₂ ⇌ 2HI), then K_p = K_c and both are dimensionless; (ii) if Δn > 0 (e.g. PCl₅ ⇌ PCl₃ + Cl₂, Δn = +1), then K_p > K_c; (iii) if Δn < 0 (e.g. N₂ + 3H₂ ⇌ 2NH₃, Δn = −2), then K_p < K_c. The units of K depend on Δn.

MCQs & Assertion–Reason

1. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the value of Δn for K_p = K_c(RT)^Δn is:

(a) +2 (b) −2 (c) +1 (d) 0

2. If Q_c > K_c, the reaction will:

(a) proceed forward (b) proceed backward (c) be at equilibrium (d) stop completely

3. The pH of a 0.001 M HCl solution is:

(a) 1 (b) 2 (c) 3 (d) 11

4. Which one is a Lewis acid?

(a) NH₃ (b) H₂O (c) BF₃ (d) OH⁻

5. The conjugate base of HSO₄⁻ is:

(a) H₂SO₄ (b) SO₄²⁻ (c) SO₃²⁻ (d) HSO₃⁻

6. Adding an inert gas at constant volume to a gaseous equilibrium will:

(a) shift it forward (b) shift it backward (c) have no effect on the equilibrium (d) double K_c

7. For a sparingly soluble salt AB₂ with solubility S, K_sp equals:

(a) S² (b) 2S³ (c) 4S³ (d) 27S⁴

8. An aqueous solution of NH₄Cl is:

(a) neutral (b) acidic (c) basic (d) amphoteric

9. Increasing temperature increases K for a reaction that is:

(a) exothermic (b) endothermic (c) at ΔH = 0 (d) catalysed

10. The relation between K_a of an acid and K_b of its conjugate base is:

(a) K_a + K_b = K_w (b) K_a · K_b = K_w (c) K_a/K_b = K_w (d) K_a = K_b

Answer key: 1-(b), 2-(b), 3-(c), 4-(c), 5-(b), 6-(c), 7-(c), 8-(b), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A catalyst does not change the value of the equilibrium constant.

Reason: A catalyst speeds up the forward and reverse reactions equally.

A-R 2. Assertion: The pH of pure water decreases below 7 as temperature rises above 298 K.

Reason: The ionic product of water K_w increases with temperature.

A-R 3. Assertion: Pure solids and liquids are omitted from equilibrium constant expressions.

Reason: Their molar concentrations (active masses) are constant at a given temperature.

A-R 4. Assertion: Adding NaCl to a saturated solution of AgCl increases the solubility of AgCl.

Reason: The common-ion effect shifts the dissolution equilibrium toward more dissolution.

A-R 5. Assertion: For an exothermic reaction, the equilibrium constant decreases as temperature increases.

Reason: Increasing temperature shifts an exothermic equilibrium in the backward direction.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Common Mistakes to Avoid

Watch out for these

Including pure solids and liquids in K_c/K_p expressions—they must be omitted.
Using the wrong value of R: use 0.0831 L·bar·K⁻¹·mol⁻¹ with bar, or 0.0821 L·atm·K⁻¹·mol⁻¹ with atm—not 8.314.
Forgetting the correct sign and value of Δn (gaseous products minus gaseous reactants) in K_p = K_c(RT)^Δn.
Confusing Q with K: Q < K means forward, Q > K means backward.
Treating a weak acid as fully ionized—use [H⁺] = √(K_ac), not [H⁺] = c.
Forgetting to halve concentrations when two solutions of equal volume are mixed (K_sp problems).
Ignoring the stoichiometric power on each ion in the K_sp expression (e.g. (2S)² for a 1:2 salt).

How to score full marks in this chapter

Always write the balanced equation first, then the K expression, and clearly omit pure solids/liquids. For numericals, draw an ICE (Initial–Change–Equilibrium) table, substitute, and simplify with the small-x approximation only when K is very small or very large—then check it is valid. Carry units through every step (M, atm, bar) and state the final answer with units. For pH problems remember pH + pOH = 14 and K_aK_b = K_w. For Le Chatelier questions, state the change, the direction of shift, and the reason in one clean line.

Frequently Asked Questions

How many exercise questions are there in Class 11 Chemistry Chapter 6 Equilibrium?

There are 73 numbered exercise questions (6.1 to 6.73), covering physical and chemical equilibrium, K_c/K_p calculations, the reaction quotient, Le Chatelier’s principle, acids and bases, pH, hydrolysis and solubility product. All 73 are solved step by step on this page with verified answers.

What is the difference between K_c and K_p?

K_c is the equilibrium constant in terms of molar concentrations, while K_p is in terms of partial pressures of gases. They are related by K_p = K_c(RT)^Δn, where Δn is the difference in the number of gaseous moles between products and reactants.

What value of R should I use in K_p = K_c(RT)^Δn?

Use R = 0.0831 L·bar·K⁻¹·mol⁻¹ when pressures are in bar, or R = 0.0821 L·atm·K⁻¹·mol⁻¹ when pressures are in atm. The choice must match the pressure units in the problem.

Are these Class 11 Chemistry Chapter 6 solutions free?

Yes. All solutions are free and follow the official NCERT Chemistry textbook (Part I) for session 2026–27, with every numerical cross-checked against the NCERT answer key.

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