NCERT Solutions for Class 11 Chemistry Chapter 6: Equilibrium

Concept. A liquid in a closed container at constant temperature is in dynamic liquid–vapour equilibrium: H₂O(l) ⇌ H₂O(vap), with rate of evaporation = rate of condensation, and the vapour exerts a constant equilibrium vapour pressure that depends only on the temperature and on the nature of the liquid. The question asks what happens when this equilibrium is disturbed by a sudden increase of volume and how Le Chatelier’s principle helps to restore it.

1. (a) Initial effect on vapour pressure. When the volume is suddenly increased, the same number of vapour molecules occupies a larger volume; the number of molecules per unit volume (concentration) falls, and since p ∝ concentration at constant T, the vapour pressure decreases suddenly (below the equilibrium value).
2. (b) Initial change of rates. The rate of evaporation depends on the surface area of the liquid and its temperature, which have not changed, so the rate of evaporation remains the same. The rate of condensation depends on the number of vapour molecules striking the liquid surface per second; as the vapour is now more dilute, the rate of condensation decreases. Hence initially rate of evaporation > rate of condensation.
3. (c) Final state. Because evaporation exceeds condensation, more liquid evaporates; the vapour pressure rises and the rate of condensation increases until the two rates become equal again — a new equilibrium. The final vapour pressure is the same as the original equilibrium vapour pressure, because the vapour pressure at a given temperature is a constant characteristic of the liquid, independent of the volume of the container or the amounts of liquid and vapour. Only the amounts change: more vapour (and less liquid) are present than before.

Extra depth. This is Le Chatelier’s principle for a physical equilibrium: the system opposes the pressure drop by evaporating more liquid. The result also explains why the vapour pressure is characterised as a constant at a given temperature (Table 6.1). It is possible to restore the original pressure only if enough liquid is present; if all the liquid evaporates before the pressure is regained, the container holds an unsaturated vapour and the pressure stays lower. The effect of temperature is different: raising the temperature raises the equilibrium vapour pressure.

Conclusion: The vapour pressure first falls and the rate of condensation decreases while evaporation continues unchanged; liquid then evaporates until equilibrium is restored at the same vapour pressure as before.

Concept. The equilibrium constant in terms of molar concentrations, K_c, is the product of the equilibrium concentrations of the products, each raised to the power of its stoichiometric coefficient, divided by the corresponding product for the reactants (law of chemical equilibrium). For a A + b B ⇌ c C + d D, K_c = [C]^c[D]^d/[A]^a[B]^b. The coefficients of the balanced equation become the exponents.

1. Write the expression: for 2SO₂ + O₂ ⇌ 2SO₃: K_c = [SO₃]²/([SO₂]²[O₂]).
2. Substitute the values: K_c = (1.90)²/[(0.60)² × 0.82] = 3.61/(0.36 × 0.82).
3. Calculate: denominator = 0.36 × 0.82 = 0.2952; K_c = 3.61/0.2952 = 12.23 L mol⁻¹ (about 12.2).
4. Unit: numerator (mol L⁻¹)², denominator (mol L⁻¹)³, so K_c has the unit L mol⁻¹. The value K_c > 1 shows products are favoured at this temperature.

Extra depth. The reaction is the key step in the contact process for sulphuric acid. Squaring the concentrations of SO₂ and SO₃ matters — the commonest error is to forget the exponents. If the equation were written for one mole of SO₃ (SO₂ + ½O₂ ⇌ SO₃), K_c would be the square root, 12.23^1/2 = 3.50, which shows why the balanced equation must be quoted with K (Table 6.4).

Conclusion: K_c = [SO₃]²/[SO₂]²[O₂] = (1.90)²/[(0.60)²(0.82)] = 12.2 L mol⁻¹.

Concept. For a gaseous equilibrium, K_p is written with partial pressures: for I₂(g) ⇌ 2I(g), K_p = (p_I)²/p_I2. The partial pressure of a gas in a mixture = mole fraction × total pressure, and for gases percentage by volume equals percentage by number of moles (Avogadro’s law).

1. Composition by volume (moles): I atoms 40 %, so I₂ = 100 − 40 = 60 %. Mole fractions: x_I = 0.40, x_I2 = 0.60.
2. Total pressure: 10⁵ Pa = 1 bar (1 bar = 10⁵ Pa). Partial pressures: p_I = 0.40 × 1 bar = 0.40 bar; p_I2 = 0.60 × 1 bar = 0.60 bar.
3. K_p = (p_I)²/p_I2 = (0.40)²/0.60 = 0.16/0.60 = 0.267 bar (≈ 2.67 × 10⁴ Pa if pressures are in Pa).

Extra depth. K_p carries the unit of pressure raised to Δn = 2 − 1 = 1, so 0.267 bar. The standard state for pressure is 1 bar, so pressures should be in bar when K_p is used with ΔG^⊖. Equivalent K_c can be found by K_c = K_p/(RT)^Δn if the temperature is known. Dissociation of iodine molecules into atoms is endothermic, so the 40 % dissociated state corresponds to a high temperature.

Conclusion: K_p = (0.40 bar)²/(0.60 bar) = 0.267 bar.

Concept. The expression for K_c places the equilibrium concentrations of the products (raised to their coefficients) in the numerator and those of the reactants in the denominator. In heterogeneous equilibria the concentration of a pure solid or pure liquid is constant (independent of amount) and is omitted; only gases and species in solution appear. The coefficient of each species in the balanced equation is its exponent.

1. (i) all gases: K_c = [NO]²[Cl₂]/[NOCl]².
2. (ii) Cu(NO₃)₂ and CuO are solids and are omitted: K_c = [NO₂]⁴[O₂].
3. (iii) all species are in the homogeneous aqueous phase; as written (water included in the reaction): K_c = [CH₃COOH][C₂H₅OH]/([CH₃COOC₂H₅][H₂O]). (When water is the solvent in a dilute solution its concentration, about 55.5 M, is practically constant and can be absorbed into the constant, giving K′_c = [CH₃COOH][C₂H₅OH]/[CH₃COOC₂H₅]; the first form is the full expression.)
4. (iv) Fe(OH)₃ is a solid and is omitted: K_c = 1/([Fe³⁺][OH⁻]³).
5. (v) I₂ is a solid and is omitted; F₂ and IF₅ are gases: K_c = [IF₅]²/[F₂]⁵.

Extra depth. Reaction (ii) shows that at a given temperature the product [NO₂]⁴[O₂] is fixed — the pressure of the gaseous products is constant whatever the amount of solid. Reaction (iv) is the reverse of dissolution, so its K_c is the reciprocal of the solubility product of Fe(OH)₃. The word ‘K_c’ in (i) and (ii) could also be written as K_p with partial pressures. For reaction (iii) the question of water: NCERT treats all species of this ester hydrolysis as being in solution; in Exercise 6.18, where water is not in excess and is not a solvent, it is explicitly included in Q_c.

Conclusion: (i) [NO]²[Cl₂]/[NOCl]²; (ii) [NO₂]⁴[O₂]; (iii) [CH₃COOH][C₂H₅OH]/[CH₃COOC₂H₅][H₂O]; (iv) 1/[Fe³⁺][OH⁻]³; (v) [IF₅]²/[F₂]⁵.

Concept. K_p and K_c are related by K_p = K_c(RT)^Δn (eq. 6.15), where Δn is the number of moles of gaseous products minus the number of moles of gaseous reactants in the balanced equation, T is in kelvin, and R = 0.0831 L bar mol⁻¹ K⁻¹ if pressure is in bar. So K_c = K_p/(RT)^Δn.

1. (i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g): Δn = (2 + 1) − 2 = 1. RT = 0.0831 × 500 = 41.55. K_c = K_p/RT = 1.8 × 10⁻²/41.55 = 4.33 × 10⁻⁴ (mol L⁻¹).
2. (ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g): solids are ignored; Δn = 1 − 0 = 1. RT = 0.0831 × 1073 = 89.17. K_c = 167/89.17 = 1.87 (mol L⁻¹).

Extra depth. In (ii) only the gaseous CO₂ counts for Δn: solids do not contribute to pressure or concentration expressions. Since K_p = p_CO2 = 167 (in the units used by the textbook) and K_c = [CO₂] = p/RT, the result 1.87 is the equilibrium concentration of CO₂ in mol L⁻¹ — a useful consistency check. For reactions with Δn = 0 (such as H₂ + I₂ ⇌ 2HI) K_p = K_c. If R = 8.314 J K⁻¹ mol⁻¹ were used, pressure would have to be in Pa and volume in m³ — use R = 0.0831 with bar to avoid errors.

Conclusion: (i) K_c = 1.8 × 10⁻²/(0.0831 × 500) = 4.33 × 10⁻⁴; (ii) K_c = 167/(0.0831 × 1073) = 1.87.

Concept. For a reaction and its reverse, the equilibrium constant expressions are inverses of each other: if K_c = [NO₂][O₂]/([NO][O₃]) for the forward reaction, then for the reverse reaction NO₂ + O₂ ⇌ NO + O₃ the expression is [NO][O₃]/([NO₂][O₂]) = 1/K_c. NCERT states: the equilibrium constant for the reverse reaction is the inverse of that of the forward reaction (K′_c = 1/K_c, eq. 6.8).

1. Forward: NO + O₃ ⇌ NO₂ + O₂, K_c = 6.3 × 10¹⁴.
2. Reverse: NO₂ + O₂ ⇌ NO + O₃, K′_c = 1/K_c = 1/(6.3 × 10¹⁴) = 1.59 × 10⁻¹⁵.
3. Meaning: K′_c << 10⁻³, so the reverse reaction hardly proceeds at equilibrium — the forward reaction goes essentially to completion (K_c >> 10³).
4. Kinetic reading (extra data in the question): as both are elementary bimolecular reactions, K_c = k_f/k_r, so k_r = k_f/6.3 × 10¹⁴: the reverse rate constant is 6.3 × 10¹⁴ times smaller than the forward one.

Extra depth. The statement about elementary bimolecular reactions means that the rate laws are rate_f = k_f[NO][O₃] and rate_r = k_r[NO₂][O₂]; at equilibrium the rates are equal, so K_c = k_f/k_r — the kinetic origin of the law of mass action. This reaction is important in atmospheric chemistry: nitric oxide destroys ozone (NO + O₃ → NO₂ + O₂), and the huge K_c shows this depletion is thermodynamically very favourable.

Conclusion: The equilibrium constant for the reverse reaction is K′_c = 1/(6.3 × 10¹⁴) = 1.59 × 10⁻¹⁵.

Concept. The equilibrium constant expression contains concentrations (molar concentrations or partial pressures) of the species whose amounts can vary. For a pure solid or pure liquid, the ‘concentration’ is the amount per unit volume of that substance itself — and that quantity does not change however much of the substance is present. So it behaves as a constant and is merged into K.

1. Concentration of a pure solid or liquid = density/molar mass. Active mass (concentration) = (mass/volume) ÷ molar mass = d/M. Both density d and molar mass M are fixed at a given temperature, so the concentration is fixed.
2. It is independent of the amount. A 1 g lump and a 100 g lump of CaCO₃ have the same moles per litre of solid, because both the mass and the volume scale together. Changing the quantity changes how much reaction can occur, not the molar concentration.
3. It is absorbed into K. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), K_c = ([CaO][CO₂])/[CaCO₃]. Since [CaO] and [CaCO₃] are constants, K_c′ = K_c[CaCO₃]/[CaO] = [CO₂], and we simply write the new constant as K_c = [CO₂].
4. Same for a pure liquid. For water, 1000 g L⁻¹/18 g mol⁻¹ ≈ 55.5 mol L⁻¹; a dilute solution reaction changes this negligibly, so it is not written separately.

Extra depth. This argument is valid only for pure phases. If the solid is part of a solution or the liquid is a mixture (say, ethanol dissolved in water), its concentration can vary and must appear in the expression. It also explains why, at a fixed temperature, the pressure of CO₂ above CaCO₃ is the same whether we have a small or large sample of the solid — the solid only has to be present. If a solid is entirely consumed, equilibrium can no longer be established.

Conclusion: The concentrations of pure solids and pure liquids are constant at constant temperature (d/M), so they are included in the value of K and omitted from the expression.

Concept. When K_c is extremely small (here 10⁻³⁷), the reaction goes to a negligible extent, so the equilibrium amounts of the reactants are practically equal to their initial amounts. This lets us calculate the tiny concentration of the product directly without solving a complicated polynomial — the ‘small x’ approximation.

1. Initial concentrations: [N₂] = 0.482 mol/10 L = 0.0482 M; [O₂] = 0.933 mol/10 L = 0.0933 M; [N₂O] = 0.
2. Equilibrium (ICE): if 2x mol L⁻¹ of N₂O forms, then [N₂] = 0.0482 − 2x, [O₂] = 0.0933 − x. Since K_c is so small, x is negligible, so [N₂] ≈ 0.0482 M and [O₂] ≈ 0.0933 M.
3. Expression: K_c = [N₂O]²/([N₂]²[O₂]) = 2.0 × 10⁻³⁷.
4. Solve: [N₂O]² = 2.0 × 10⁻³⁷ × (0.0482)² × 0.0933 = 2.0 × 10⁻³⁷ × 2.168 × 10⁻⁴ = 4.34 × 10⁻⁴¹; so [N₂O] = 6.6 × 10⁻²¹ M.
5. Amount in the 10 L vessel: 6.6 × 10⁻²¹ × 10 = 6.6 × 10⁻²⁰ mol of N₂O.

Extra depth. A K_c of 10⁻³⁷ means that essentially no N₂O forms — an ideal illustration of the rule ‘K < 10⁻³ : products are hardly formed’. In practice N₂ and O₂ in air do not react to give N₂O at ordinary temperatures. The approximation is justified a posteriori: x = 3.3 × 10⁻²¹ is far below 5% of 0.0482 or 0.0933. In the textbook’s own words the equilibrium is reached when concentrations of the reactants and products show no further change, and here it is reached with practically all the reactants unchanged.

Conclusion: At equilibrium: [N₂] = 0.0482 M, [O₂] = 0.0933 M, [N₂O] = 6.6 × 10⁻²¹ M.

Concept. This is a pure stoichiometry (ICE-table) problem: the amount of product formed at equilibrium tells us how much reactant was used, through the mole ratio of the balanced equation: 2 mol NO + 1 mol Br₂ → 2 mol NOBr.

1. Set the ICE table (in mol): let NOBr formed = 0.0518 mol (this equals 2x, so x = 0.0259).
2. NO consumed: the ratio NO : NOBr = 2 : 2 = 1 : 1, so 0.0518 mol of NO is consumed. Equilibrium NO = 0.087 − 0.0518 = 0.0352 mol.
3. Br₂ consumed: Br₂ : NOBr = 1 : 2, so Br₂ used = 0.0518/2 = 0.0259 mol. Equilibrium Br₂ = 0.0437 − 0.0259 = 0.0178 mol.

Extra depth. A check by mass balance: the nitrogen atoms are 0.087 initially and 0.0352 + 0.0518 = 0.087 at equilibrium; the bromine atoms are 2 × 0.0437 = 0.0874 initially and 2 × 0.0178 + 0.0518 = 0.0874 at equilibrium. Both agree. The volume of the container is not given, so the question asks only for amounts, not concentrations; if the volume V were known, K_c = (0.0518/V)²/[(0.0352/V)²(0.0178/V)] = 121.6V.

Conclusion: At equilibrium NO = 0.0352 mol and Br₂ = 0.0178 mol.

Concept. The relation between the two constants is K_p = K_c(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants). Here pressure is in bar so R = 0.0831 L bar K⁻¹ mol⁻¹. The unit ‘/bar’ of K_p already tells us Δn = −1.

1. Δn: 2SO₂ + O₂ → 2SO₃: Δn = 2 − (2 + 1) = −1.
2. Rearrange: K_c = K_p/(RT)^Δn = K_p(RT)⁺¹ = K_p × RT.
3. Substitute: RT = 0.0831 × 450 = 37.4 L bar mol⁻¹. K_c = 2.0 × 10¹⁰ × 37.4 = 7.48 × 10¹¹ L mol⁻¹.

Extra depth. A very large K_c (10¹¹) shows that at 450 K the oxidation of SO₂ to SO₃ goes almost to completion thermodynamically; the contact process is run at about 700 K with a V₂O₅ catalyst because the reaction is too slow at 450 K — a nice example of equilibrium (how far) versus kinetics (how fast). Because the reaction is exothermic, K decreases when T is raised.

Conclusion: K_c = K_p(RT) = 2.0 × 10¹⁰ × 0.0831 × 450 = 7.5 × 10¹¹.

Concept. Since only HI is present initially, the H₂ and I₂ pressures at equilibrium arise solely from the decomposition of HI. At constant T and V, partial pressure is proportional to the number of moles, so we can run an ICE table directly in pressures.

1. ICE in atm: initial p(HI) = 0.2; if 2x atm of HI decomposes, p(HI) = 0.2 − 2x, p(H₂) = p(I₂) = x.
2. Use the given equilibrium pressure: 0.2 − 2x = 0.04, so 2x = 0.16 and x = 0.08 atm. Hence p(H₂) = p(I₂) = 0.08 atm.
3. K_p = p(H₂)p(I₂)/p(HI)² = (0.08 × 0.08)/(0.04)² = 0.0064/0.0016 = 4.0 (no unit, since Δn = 0).

Extra depth. Since Δn = 2 − 2 = 0, K_p = K_c = 4.0 and has no unit. Notice the useful consistency check: the total pressure at equilibrium (0.04 + 0.08 + 0.08 = 0.20 atm) equals the initial pressure, as it must when Δn = 0. The degree of dissociation of HI is 0.16/0.2 = 0.80 (80 %).

Conclusion: K_p = (0.08)(0.08)/(0.04)² = 4.0.

Concept. To decide whether a mixture is at equilibrium we compute the reaction quotient Q_c, which has the same form as K_c but uses the concentrations present now. Comparison rule: Q_c = K_c: at equilibrium; Q_c < K_c: net reaction goes forward (products form); Q_c > K_c: net reaction goes backward (reactants form).

1. Concentrations (divide by 20 L): [N₂] = 1.57/20 = 0.0785 M; [H₂] = 1.92/20 = 0.0960 M; [NH₃] = 8.13/20 = 0.4065 M.
2. Q_c = [NH₃]²/([N₂][H₂]³) = (0.4065)²/[(0.0785)(0.0960)³] = 0.1652/(0.0785 × 8.847 × 10⁻⁴) = 0.1652/6.945 × 10⁻⁵ = 2.38 × 10³.
3. Compare: Q_c = 2.38 × 10³ > K_c = 1.7 × 10². The mixture is not at equilibrium: it has too much NH₃, so the net reaction proceeds in the reverse direction — NH₃ decomposes into N₂ and H₂ until Q_c falls to 170.

Extra depth. The volume matters here because Δn = −2 ≠ 0: using moles instead of concentrations would give a wrong Q_c (it would be off by a factor of V² = 400). The reverse direction is the same as saying the system consumes the excess product, exactly as Le Chatelier’s principle predicts for a mixture that has more product than the equilibrium requires.

Conclusion: Q_c = 2.38 × 10³ > K_c (170), so the mixture is not at equilibrium and the net reaction is in the backward direction.

Concept. In the equilibrium constant expression the species in the numerator are the products and those in the denominator are the reactants; the exponents are the stoichiometric coefficients. So we can reverse-engineer the equation by reading the expression.

1. Numerator (products): NH₃ with power 4 and O₂ with power 5, i.e. 4NH₃ + 5O₂.
2. Denominator (reactants): NO with power 4 and H₂O with power 6, i.e. 4NO + 6H₂O.
3. Equation: 4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g).
4. Check the balance: N: 4 = 4; H: 12 = 12; O: 4 + 6 = 10 and 10 on the right. Balanced.

Extra depth. This is the reverse of the well-known catalytic oxidation of ammonia (4NH₃ + 5O₂ → 4NO + 6H₂O, the first step of Ostwald’s nitric acid process), whose K_c is the reciprocal of the given one. That explains the numerical subtlety: the given K_c must be extremely small, because ammonia oxidation is strongly forward. Without any information to the contrary the species are all gases, as the question says ‘gas reaction’.

Conclusion: 4NO(g) + 6H₂O(g) ⇌ 4NH₃(g) + 5O₂(g).

Concept. This is the water–gas shift reaction. Because 40 % of water by mass reacts, 40 % of its moles react too (same substance). With equal moles of H₂O and CO and 1:1:1:1 stoichiometry, each of the other species changes by the same amount.

1. Moles reacted: 40 % of 1 mol = 0.40 mol of H₂O (and 0.40 mol CO) react, forming 0.40 mol H₂ and 0.40 mol CO₂.
2. Equilibrium moles: H₂O = 0.60, CO = 0.60, H₂ = 0.40, CO₂ = 0.40. Concentrations in 10 L: 0.060, 0.060, 0.040, 0.040 M.
3. K_c = [H₂][CO₂]/([H₂O][CO]) = (0.040 × 0.040)/(0.060 × 0.060) = 0.0016/0.0036 = 0.444 (no unit, Δn = 0).

Extra depth. Because Δn = 0, the volume (10 L) cancels: K_c = (0.4 × 0.4)/(0.6 × 0.6) = 0.444 using moles directly gives the same result — a quick check. The reaction is used industrially to increase hydrogen yield from synthesis gas; being exothermic, K falls as T rises, so it is run in two stages (high-temperature and low-temperature).

Conclusion: K_c = (0.04 × 0.04)/(0.06 × 0.06) = 0.444.

Concept. Since we started with only HI, the decomposition 2HI ⇌ H₂ + I₂ produces equal amounts of H₂ and I₂, so [H₂] = [I₂] at equilibrium. The equilibrium constant for HI decomposition is the reciprocal of that for formation: K′ = 1/54.8.

1. Let [H₂] = [I₂] = x at equilibrium; [HI] = 0.5 M.
2. Use the forward expression: K_c = [HI]²/([H₂][I₂]) = (0.5)²/x² = 54.8.
3. Solve: x² = 0.25/54.8 = 4.56 × 10⁻³, so x = 0.0675 mol L⁻¹.
4. Result: [H₂] = [I₂] = 0.068 M (0.0675).

Extra depth. The same final composition is reached whether we start from H₂ + I₂ or from HI — the hallmark of a true chemical equilibrium (Fig. 6.4 of the chapter). The initial concentration of HI was 0.5 + 2(0.0675) = 0.635 M, so HI is dissociated to only 21 %; this is consistent with a large K_c for the formation of HI.

Conclusion: [H₂] = [I₂] = √(0.25/54.8) = 0.068 mol L⁻¹.

Concept. Starting with only ICl, the decomposition gives equal amounts of I₂ and Cl₂. Let x mol L⁻¹ of I₂ be formed. Then 2x mol L⁻¹ of ICl is used (coefficient 2). The perfect-square form of the expression lets us take a square root rather than solve a quadratic.

1. ICE: [ICl] = 0.78 − 2x, [I₂] = x, [Cl₂] = x.
2. Expression: K_c = [I₂][Cl₂]/[ICl]² = x²/(0.78 − 2x)² = 0.14.
3. Square root of both sides: x/(0.78 − 2x) = √0.14 = 0.374.
4. Solve: x = 0.374(0.78 − 2x) = 0.292 − 0.748x, so 1.748x = 0.292 and x = 0.167 M.
5. Equilibrium concentrations: [I₂] = [Cl₂] = 0.167 M; [ICl] = 0.78 − 2(0.167) = 0.446 M.

Extra depth. Check: (0.167)²/(0.446)² = 0.0279/0.199 = 0.140 — matches K_c. The degree of dissociation of ICl is 0.334/0.78 = 43 %, which is large because K_c is only a little less than 1. The ‘small x’ approximation could not be used here, so the square-root method is the right tool. Whenever the equilibrium expression is a perfect square (equal numbers of moles on both sides) the square root trick avoids a quadratic.

Conclusion: At equilibrium [I₂] = [Cl₂] = 0.167 M and [ICl] = 0.446 M.

Concept. Here the equilibrium constant is in terms of partial pressures. We first find the equilibrium partial pressure of C₂H₆ by an ICE table in atm, and then convert it into molar concentration using c = p/RT (ideal gas law, with R = 0.0821 L atm K⁻¹ mol⁻¹ because the pressure is in atm).

1. ICE (atm): let x atm of C₂H₆ dissociate: p(C₂H₆) = 4.0 − x, p(C₂H₄) = p(H₂) = x.
2. K_p = x²/(4.0 − x) = 0.04, so x² + 0.04x − 0.16 = 0.
3. Quadratic formula: x = [−0.04 + √(0.0016 + 0.64)]/2 = (−0.04 + 0.801)/2 = 0.38 atm.
4. Equilibrium pressure of ethane: p(C₂H₆) = 4.0 − 0.38 = 3.62 atm.
5. Convert to concentration: [C₂H₆] = p/RT = 3.62/(0.0821 × 899) = 3.62/73.8 = 0.049 mol L⁻¹.

Extra depth. Check: (0.38)²/3.62 = 0.0399 ≈ 0.04. The small-x approximation would give x = √(0.04 × 4) = 0.40, about 5 % off — just at the limit of acceptability, so the quadratic is safer. Only 0.38/4.0 = 9.5 % of ethane has dissociated at 899 K. This is the cracking-type dehydrogenation that gives ethene industrially; since it increases the number of gas moles (Δn = +1), low pressure favours it.

Conclusion: [C₂H₆] = 3.62 atm/(0.0821 × 899) = 0.049 mol L⁻¹.

Concept. This esterification is a homogeneous liquid-phase equilibrium. Here all four species are liquids that are mixed, so none is a pure phase on its own; and, as the question tells us, water is a product, not the solvent, so its concentration must be included. Because Δn = 0 the volume cancels and mole numbers can be used directly.

1. (i) Reaction quotient: Q_c = [CH₃COOC₂H₅][H₂O]/([CH₃COOH][C₂H₅OH]).
2. (ii) Equilibrium amounts: ester formed = 0.171 mol, so water formed = 0.171 mol; acetic acid left = 1.00 − 0.171 = 0.829 mol; ethanol left = 0.18 − 0.171 = 0.009 mol.
3. K_c = (0.171 × 0.171)/(0.829 × 0.009) = 0.02924/0.007461 = 3.92 (volume cancels because the number of moles of reactants equals that of products).
4. (iii) New composition: ester = water = 0.214 mol; acetic acid = 1.0 − 0.214 = 0.786 mol; ethanol = 0.5 − 0.214 = 0.286 mol.
5. 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; the net reaction continues in the forward direction, forming more ester and water.

Extra depth. Notice how mixing equal volumes of reactants (rather than a large excess of ethanol) gives a much smaller conversion: only a modest fraction is esterified. In part (ii) ethanol is the limiting reagent and 95 % of it is used up. In practice, to push the esterification forward one removes water or uses excess of one reactant — Le Chatelier’s principle. Part (iii) is an excellent worked example of using Q_c to judge the direction of change: the mixture has far too little ester and water compared to equilibrium.

Conclusion: (i) Q_c = [ester][H₂O]/[acid][alcohol]; (ii) K_c = 3.92; (iii) Q_c = 0.204 < 3.92, equilibrium not reached.

Concept. Only pure PCl₅ was introduced, so the two products PCl₃ and Cl₂ are formed in equal amounts by dissociation: [PCl₃] = [Cl₂]. The equilibrium concentration of PCl₅ is given, so no ICE table is required — only substitution into K_c.

1. Let [PCl₃] = [Cl₂] = x at equilibrium; [PCl₅] = 0.05 M.
2. K_c = [PCl₃][Cl₂]/[PCl₅] = x²/0.05 = 8.3 × 10⁻³.
3. x² = 8.3 × 10⁻³ × 0.05 = 4.15 × 10⁻⁴, hence x = 2.04 × 10⁻² mol L⁻¹.
4. Result: [PCl₃] = [Cl₂] = 2.04 × 10⁻² M (about 0.02 M).

Extra depth. The initial concentration of PCl₅ was 0.050 + 0.0204 = 0.0704 M, so about 29 % has dissociated at 473 K. PCl₅ dissociation is endothermic (ΔH > 0), so more dissociation occurs at higher temperature; with Δn = +1, adding an inert gas at constant volume has no effect, but compressing the mixture would push the equilibrium back toward PCl₅ (Le Chatelier). This equilibrium is a classical textbook case of a gaseous dissociation with equal amounts of the two products.

Conclusion: [PCl₃] = [Cl₂] = √(8.3 × 10⁻³ × 0.05) = 2.04 × 10⁻² mol L⁻¹.

Concept. For this heterogeneous equilibrium the solids FeO and Fe are omitted, so K_p = p_CO2/p_CO = 0.265 (a pure ratio; Δn = 0). First we compare Q_p to K_p to see in which direction the reaction must go, and then do the ICE table.

1. Reaction quotient: Q_p = 0.80/1.4 = 0.571, which is greater than K_p = 0.265, so the reaction goes in the reverse direction: CO₂ decreases and CO increases.
2. ICE (atm): let p(CO₂) fall by x: p(CO₂) = 0.80 − x; p(CO) = 1.4 + x.
3. Equation: (0.80 − x)/(1.4 + x) = 0.265, so 0.80 − x = 0.371 + 0.265x, giving 0.429 = 1.265x and x = 0.339 atm.
4. Equilibrium pressures: p(CO) = 1.4 + 0.339 = 1.739 atm; p(CO₂) = 0.80 − 0.339 = 0.461 atm.

Extra depth. Notice that the pressure ratio is the only thing fixed by K_p, and the total pressure stays the same (2.20 atm) because Δn = 0. This reaction is the heart of the blast furnace: CO removes oxygen from iron ore. At 1050 K the equilibrium ratio of CO₂ : CO is only 0.265, which is why a large excess of CO (coke) has to be supplied to keep the reduction going and why the exit gas contains much unused CO.

Conclusion: p(CO) = 1.739 atm and p(CO₂) = 0.461 atm at equilibrium.

Concept. The composition is already given in concentrations, so we directly evaluate the reaction quotient and compare it with K_c: Q_c < K_c (forward), Q_c = K_c (equilibrium), Q_c > K_c (backward).

1. Q_c = [NH₃]²/([N₂][H₂]³) = (0.5)²/[(3.0)(2.0)³] = 0.25/(3.0 × 8) = 0.25/24.
2. Q_c = 0.0104.
3. Compare: Q_c = 0.0104 < K_c = 0.061, so the mixture is not at equilibrium; the reaction will proceed in the forward direction, forming more NH₃ from N₂ and H₂, until Q_c increases to 0.061.

Extra depth. The mixture is too rich in reactants and too poor in NH₃ compared with equilibrium, so the system makes ammonia. Note that this K_c (0.061 at 500 K) is a typical value for the Haber process: even at equilibrium much N₂ and H₂ remain, so industry uses high pressure (about 200 atm), an iron catalyst and recycles the unreacted gases. The reaction quotient tool applies to any starting mixture, not only to mixtures prepared from pure reactants.

Conclusion: Q_c = 0.0104 < K_c = 0.061; not at equilibrium; net reaction goes in the forward direction.

Concept. Starting with pure BrCl, Br₂ and Cl₂ form in equal amounts. The expression K_c = [Br₂][Cl₂]/[BrCl]² is a perfect square, so we can solve it by taking the square root of both sides.

1. ICE: let [Br₂] = [Cl₂] = x; [BrCl] = 3.3 × 10⁻³ − 2x.
2. Expression: x²/(3.3 × 10⁻³ − 2x)² = 32, so x/(3.3 × 10⁻³ − 2x) = √32 = 5.657.
3. Solve: x = 5.657(3.3 × 10⁻³) − 11.314x ⇒ 12.314x = 1.867 × 10⁻² ⇒ x = 1.516 × 10⁻³ M.
4. [BrCl] at equilibrium = 3.3 × 10⁻³ − 2(1.516 × 10⁻³) = 3.3 × 10⁻³ − 3.03 × 10⁻³ = 2.7 × 10⁻⁴ mol L⁻¹.

Extra depth. About 92 % of BrCl dissociates at 500 K, consistent with K_c = 32 > 1. Because the answer is a small difference of two numbers (3.3 − 3.03), keep at least three significant figures in x or the result will be inaccurate — rounding x to 1.5 × 10⁻³ would give BrCl = 3 × 10⁻⁴, a visible error. A neat alternative is to solve directly for the BrCl concentration y: (3.3 × 10⁻³ − y)²/4y² = 32.

Conclusion: [BrCl]_eq = 2.7 × 10⁻⁴ mol L⁻¹.

Concept. The mass percentage must be converted into mole fractions, then partial pressures, then K_p, and finally K_c through K_p = K_c(RT)^Δn with Δn = 2 − 1 = +1 (only gases are counted).

1. Take 100 g of the gas mixture: CO = 90.55 g = 90.55/28 = 3.234 mol; CO₂ = 9.45 g = 9.45/44 = 0.2148 mol; total = 3.449 mol.
2. Mole fractions: x(CO) = 3.234/3.449 = 0.9377; x(CO₂) = 0.0623.
3. Partial pressures (P = 1 atm): p(CO) = 0.9377 atm; p(CO₂) = 0.0623 atm.
4. K_p = p(CO)²/p(CO₂) = (0.9377)²/0.0623 = 0.8793/0.0623 = 14.1.
5. K_c = K_p/(RT)^Δn = 14.1/(0.0821 × 1127) = 14.1/92.5 = 0.152 (mol L⁻¹).

Extra depth. This is the Boudouard (carbon gasification) equilibrium, important in the blast furnace: at about 1100 K and above the equilibrium lies on the CO side, as the data show (94 mol % CO). Solid carbon is excluded from K, so K_p depends only on the two gases. Using R = 0.0831 L bar mol⁻¹ K⁻¹ and pressure in bar would change the numerical value slightly (1 atm = 1.013 bar), a difference within the accuracy of the data.

Conclusion: K_p = (0.9377)²/0.0623 = 14.1; K_c = 14.1/(0.0821 × 1127) = 0.152.

Concept. The standard Gibbs energy change of a reaction is Δ_rG⊖ = ΣΔ_fG⊖(products) − ΣΔ_fG⊖(reactants), and it is linked to the equilibrium constant by Δ_rG⊖ = −RT ln K = −2.303 RT log K (eq. 6.26 of the chapter). A negative ΔG⊖ means K > 1 and a spontaneous reaction under standard conditions.

1. (a) Δ_rG⊖ = Δ_fG(NO₂) − [Δ_fG(NO) + ½Δ_fG(O₂)] = 52.0 − (87.0 + 0) = −35.0 kJ mol⁻¹.
2. (b) log K = −ΔG⊖/(2.303RT) = 35.0 × 1000 J/(2.303 × 8.314 J K⁻¹ mol⁻¹ × 298 K) = 35000/5705.8 = 6.134.
3. K = antilog 6.134 = 1.36 × 10⁶ (K is for the equation as written with ½O₂).

Extra depth. The large K shows that at 298 K the formation of NO₂ from NO and O₂ is thermodynamically very favourable, which is why colourless NO turns brown in air. Remember the units: ΔG⊖ must be converted to J when R is in J K⁻¹ mol⁻¹. Because the stoichiometry uses ½ O₂, doubling the equation (2NO + O₂ → 2NO₂) would give ΔG⊖ = −70 kJ and K = (1.36 × 10⁶)² = 1.8 × 10¹². Also, Δ_fG⊖ of an element in its standard state (O₂) is zero.

Conclusion: Δ_rG⊖ = −35 kJ mol⁻¹ and K = 1.36 × 10⁶ at 298 K.

Concept. By Le Chatelier’s principle, when the pressure on a gaseous equilibrium is lowered (by increasing the volume), the system shifts toward the side with the greater number of moles of gas to oppose the fall of pressure. Only gaseous species are counted; solids have negligible volume and do not matter. If the number of gas moles is the same on both sides, there is no shift.

1. (a) PCl₅ ⇌ PCl₃ + Cl₂: 1 mol gas on the left, 2 mol gas on the right. Lower pressure favours the right side, so the number of moles of products increases.
2. (b) CaO(s) + CO₂(g) ⇌ CaCO₃(s): the only gas is CO₂ on the left (1 mol); the right has no gas (0). Lower pressure favours the left, so CaCO₃ decomposes and the moles of product (CaCO₃) decrease.
3. (c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g): 4 mol gas on each side, so Δn_g = 0 and the pressure change has no effect: the number of moles of products remains the same.

Extra depth. In (b) the equilibrium can be put into one sentence: K_p = 1/p(CO₂), so at a fixed temperature the equilibrium pressure of CO₂ is fixed. If the container is expanded the pressure of CO₂ momentarily falls, and CaCO₃ must decompose to restore it — until it is used up if the volume is large enough. K itself is unchanged in all three cases, since pressure does not alter K; only temperature does.

Conclusion: (a) increases; (b) decreases; (c) remains the same.

Concept. Pressure affects a gaseous equilibrium only if Δn_g (moles of gaseous products − moles of gaseous reactants) is not zero. Increasing the pressure shifts the equilibrium to the side with fewer moles of gas; for Δn_g = 0 there is no effect. Solids and liquids are not counted.

1. (i) COCl₂ ⇌ CO + Cl₂: 1 → 2 (Δn_g = +1). Affected; higher pressure shifts it backward.
2. (ii) CH₄ + 2S₂ ⇌ CS₂ + 2H₂S: 3 → 3 (Δn_g = 0). Not affected.
3. (iii) CO₂ + C(s) ⇌ 2CO: gas moles 1 → 2 (Δn_g = +1). Affected; shifts backward.
4. (iv) 2H₂ + CO ⇌ CH₃OH: 3 → 1 (Δn_g = −2). Affected; shifts forward.
5. (v) CaCO₃(s) ⇌ CaO(s) + CO₂(g): 0 → 1 (Δn_g = +1). Affected; shifts backward.
6. (vi) 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O: 9 → 10 (Δn_g = +1). Affected; shifts backward.

Extra depth. Only (iv) is favoured by high pressure: this is why methanol is manufactured at 50–100 atm. Reaction (vi), the oxidation of ammonia in Ostwald’s process, is only slightly affected (9 → 10 mol) — it is carried out at modest pressure and a catalyst. In all cases K does not change with pressure; only the position of equilibrium does.

Conclusion: Affected: (i), (iii), (v), (vi) backward; (iv) forward; (ii) unaffected.

Concept. The reaction written for the formation of HBr has a huge K_p; if we start from HBr, the decomposition 2HBr ⇌ H₂ + Br₂ has the reciprocal constant K′_p = 1/(1.6 × 10⁵) = 6.25 × 10⁻⁶, so very little HBr will decompose and the small-x approximation is justified.

1. ICE (bar): if x bar of H₂ forms, p(Br₂) = x too, and p(HBr) = 10.0 − 2x ≈ 10.0 (x is tiny).
2. K_p for formation: p(HBr)²/(p(H₂)p(Br₂)) = (10.0)²/x² = 1.6 × 10⁵.
3. Solve: x² = 100/(1.6 × 10⁵) = 6.25 × 10⁻⁴, so x = 2.5 × 10⁻² bar.
4. Result: p(H₂) = p(Br₂) = 2.5 × 10⁻² bar; p(HBr) = 10.0 − 2(0.025) = 9.95 bar.

Extra depth. The approximation is valid: 2x = 0.05 bar is only 0.5 % of 10 bar (< 5 %). Since Δn = 0, K_p = K_c and the total pressure remains 10.0 bar (9.95 + 0.025 + 0.025). The result agrees with the fact that HBr is a very stable molecule at 1024 K.

Conclusion: p(H₂) = p(Br₂) = 2.5 × 10⁻² bar; p(HBr) = 9.95 bar.

Concept. Part (a) needs the standard K_p expression in terms of partial pressures. Part (b) needs Le Chatelier’s principle: pressure changes the composition (when Δn_g ≠ 0) but not K; temperature changes K (the direction depends on the sign of ΔH); a catalyst changes neither K nor the composition.

1. (a) K_p = p(CO) × p(H₂)³/(p(CH₄) × p(H₂O)).
2. (b)(i) Increasing the pressure: gas moles 2 → 4 (Δn_g = +2), so the equilibrium shifts backward (toward CH₄ and H₂O); the amount of H₂ and CO falls. K_p is unchanged.
3. (ii) Increasing the temperature: the reaction is endothermic, so heat is absorbed in the forward direction; the equilibrium shifts forward, the yield of H₂ and CO increases, and K_p increases.
4. (iii) Catalyst: speeds up both forward and reverse reactions equally; it only helps to reach equilibrium faster. Neither K_p nor the composition changes.

Extra depth. This is why industrial steam reforming is carried out at high temperature (about 1100 K) and moderate pressure, with a nickel catalyst. The pressure compromise: high pressure lowers the equilibrium yield, but compressing gas is cheaper for downstream processes (Haber, methanol). A rise in temperature lifts K_p in accordance with the van’t Hoff equation: for ΔH⊖ > 0, ln K increases with T.

Conclusion: (a) K_p = p_COp_H2³/p_CH4p_H2O; (b) pressure: shifts backward, K_p same; temperature: forward, K_p rises; catalyst: no change.

Concept. Le Chatelier’s principle for concentration changes: if the concentration of a species is raised, the equilibrium shifts in the direction that consumes it; if it is lowered, the equilibrium shifts in the direction that produces it. K_c is unchanged by these changes at constant temperature.

1. (a) Addition of H₂: increases a reactant, so the equilibrium shifts forward: more CH₃OH is formed (and CO is used up).
2. (b) Addition of CH₃OH: increases the product, so the equilibrium shifts backward: CH₃OH decomposes into H₂ and CO.
3. (c) Removal of CO: a reactant is withdrawn, so the equilibrium shifts backward (to produce more CO), and the amount of CH₃OH decreases.
4. (d) Removal of CH₃OH: the product is withdrawn, so the equilibrium shifts forward to make up for it — more CH₃OH is formed.

Extra depth. In the industrial synthesis of methanol, CH₃OH is continuously removed (as liquid, by cooling) and unreacted H₂ and CO are recycled — actions (a) and (d) put to practical use to push the equilibrium forward. The same idea is the basis for removing water in esterification and removing NH₃ in the Haber process.

Conclusion: (a) forward; (b) backward; (c) backward; (d) forward.

Concept. K_c is a function of temperature alone. Concentration changes and pressure changes alter the position of equilibrium (the composition) but not the value of K_c. A temperature change alters K_c in the direction given by the sign of Δ_rH (van’t Hoff).

1. (a) K_c = [PCl₃][Cl₂]/[PCl₅].
2. (b) Reverse reaction (PCl₃ + Cl₂ ⇌ PCl₅): K′_c = 1/K_c = 1/(8.3 × 10⁻³) = 120.5 (about 1.2 × 10²).
3. (c)(i) More PCl₅: the equilibrium shifts forward to use the added PCl₅, but K_c is unchanged.
4. (ii) Increased pressure: the equilibrium shifts backward (fewer gas moles on the left), but K_c is unchanged.
5. (iii) Raised temperature: the reaction is endothermic (Δ_rH = +124 kJ mol⁻¹), so the equilibrium shifts forward and K_c increases.

Extra depth. The reverse reaction is exothermic (−124 kJ mol⁻¹) and its K′_c falls with temperature. The numerical value 120.5 shows that at 473 K the formation of PCl₅ from PCl₃ and Cl₂ is strongly favoured when the concentrations are of the order of 1 M, even though the decomposition itself is small. Note that K_c for the reverse reaction is written for the equation as given in part (b), so the stoichiometry is the same, not doubled.

Conclusion: (a) K_c = [PCl₃][Cl₂]/[PCl₅]; (b) 1/(8.3 × 10⁻³) = 120.5; (c) unchanged, unchanged, increases.

Concept. With an equimolar start, p(CO) = p(H₂O) throughout, and p(CO₂) = p(H₂) = x. Because Δn_g = 0, the expression K_p = p(CO₂)p(H₂)/(p(CO)p(H₂O)) = x²/(4.0 − x)² is a perfect square.

1. ICE (bar): p(CO) = p(H₂O) = 4.0 − x; p(CO₂) = p(H₂) = x.
2. Expression: x²/(4.0 − x)² = 10.1, so x/(4.0 − x) = √10.1 = 3.178.
3. Solve: x = 3.178(4.0 − x) = 12.71 − 3.178x ⇒ 4.178x = 12.71 ⇒ x = 3.04 bar.
4. Result: p(H₂) = p(CO₂) = 3.04 bar; p(CO) = p(H₂O) = 4.0 − 3.04 = 0.96 bar.

Extra depth. Check: (3.04 × 3.04)/(0.96 × 0.96) = 9.24/0.92 = 10.0 ≈ 10.1. About 76 % of the CO is converted in this stage. Industrial plants use two shift reactors — a high-temperature one (fast) and a low-temperature one (K is larger because the reaction is exothermic) — to maximise the H₂ yield before it is used in the Haber process.

Conclusion: p(H₂) = 3.04 bar at equilibrium.

Concept. The magnitude of K_c indicates the extent of reaction (section 6.5 of the chapter): if K_c > 10³, products predominate (reaction nearly complete); if K_c < 10⁻³, reactants predominate (reaction hardly occurs); if K_c lies between 10⁻³ and 10³, both reactants and products are present in appreciable amounts.

1. (a) K_c = 5 × 10⁻³⁹ (<< 10⁻³): the equilibrium lies far to the left; the mixture is essentially all Cl₂, with a negligible concentration of Cl atoms.
2. (b) K_c = 3.7 × 10⁸ (>> 10³): the reaction goes nearly to completion; mostly NOCl, with negligible Cl₂ and NO left.
3. (c) K_c = 1.8 (between 10⁻³ and 10³): both reactants and products are present in appreciable concentrations — this is the reaction asked for.

Extra depth. Only reaction (c) has both reactants and products at significant concentration. A very large or small K does not say anything about the rate: a reaction with K = 10⁸ may be very slow without a catalyst. The scale here is only a guide: a K of 10³ or 10⁻³ means 99.9 % or 0.1 % conversion for a simple A ⇌ B system.

Conclusion: Reaction (c), with K_c = 1.8, has appreciable concentrations of both reactants and products.

Concept. For 3O₂ ⇌ 2O₃, K_c = [O₃]²/[O₂]³. Given K_c and [O₂], solve for [O₃]. No ICE table is needed because the equilibrium concentration of O₂ is given directly.

1. Rearrange: [O₃]² = K_c × [O₂]³.
2. Substitute: [O₃]² = 2.0 × 10⁻⁵⁰ × (1.6 × 10⁻²)³ = 2.0 × 10⁻⁵⁰ × 4.096 × 10⁻⁶ = 8.19 × 10⁻⁵⁶.
3. Square root: [O₃] = √(8.19 × 10⁻⁵⁶) = 2.86 × 10⁻²⁸ mol L⁻¹.

Extra depth. The concentration is almost unimaginably small — far fewer than one molecule per litre (about 10⁻⁴ molecules per litre, since 2.9 × 10⁻²⁸ × 6.02 × 10²³ ≈ 1.7 × 10⁻⁴). This shows that at 25°C, O₃ is thermodynamically unstable relative to O₂; atmospheric ozone is formed by photochemical processes (UV light), not by thermal equilibrium.

Conclusion: [O₃] = √(2.0 × 10⁻⁵⁰ × (1.6 × 10⁻²)³) = 2.86 × 10⁻²⁸ M.

Concept. The flask volume is 1 L, so the number of moles equals the molar concentration. We write the K_c expression and solve for the unknown [CH₄] — the only variable.

1. Expression: K_c = [CH₄][H₂O]/([CO][H₂]³) = 3.90.
2. Rearrange: [CH₄] = K_c[CO][H₂]³/[H₂O].
3. Substitute: [CH₄] = 3.90 × 0.30 × (0.10)³/0.02 = 3.90 × 0.30 × 1.0 × 10⁻³/0.02 = 1.17 × 10⁻³/0.02 = 0.0585 mol L⁻¹.

Extra depth. Check by substituting back: (0.0585 × 0.02)/(0.30 × 0.001) = 0.00117/0.00030 = 3.90. This is the methanation reaction (the reverse of steam reforming); at 1300 K K_c is small because the forward reaction is exothermic, as the data show. Since the flask is 1 L, mol and mol L⁻¹ are numerically equal — if the volume were not 1 L we would have to divide first.

Conclusion: [CH₄] = 3.90 × 0.30 × (0.10)³/0.02 = 0.0585 mol L⁻¹.

Concept. In the Brønsted–Lowry theory an acid is a proton (H⁺) donor and a base is a proton acceptor. When an acid donates a proton, the species left behind can take the proton back — it is a base. The acid and the base that differ by one proton form a conjugate acid–base pair: HA (acid) ⇌ H⁺ + A⁻ (its conjugate base). A strong acid has a weak conjugate base, and vice versa.

1. Definition: a conjugate pair = two species differing by one proton: acid → conjugate base (loses H⁺); base → conjugate acid (gains H⁺). Example: NH₄⁺/NH₃.
2. HNO₂ is an acid; its conjugate base is NO₂⁻.
3. CN⁻ is a base; its conjugate acid is HCN.
4. HClO₄ is an acid; its conjugate base is ClO₄⁻.
5. F⁻ is a base; its conjugate acid is HF.
6. OH⁻ is a base; its conjugate acid is H₂O (as an acid it can also lose a proton to give its conjugate base O²⁻).
7. CO₃²⁻ is a base; its conjugate acid is HCO₃⁻.
8. S²⁻ is a base; its conjugate acid is HS⁻.

Extra depth. Textbook note: in the printed question the species before ‘and S²⁻’ appears as ‘CO’ with its subscript lost in printing; it is the carbonate ion CO₃²⁻, as in the NCERT answer. Strong acids such as HClO₄ have extremely weak conjugate bases (ClO₄⁻), while weak acids like HCN have relatively strong conjugate bases (CN⁻, giving a basic solution of NaCN). The relation K_a × K_b = K_w connects the strengths of a conjugate pair (see Exercise 6.43).

Conclusion: HNO₂/NO₂⁻; HCN/CN⁻; HClO₄/ClO₄⁻; HF/F⁻; H₂O/OH⁻; HCO₃⁻/CO₃²⁻; HS⁻/S²⁻.

Concept. A Lewis acid is a species that can accept an electron pair; a Lewis base is one that can donate an electron pair. So Lewis acids are electron-deficient species: positive ions with vacant orbitals, or molecules with an incomplete octet (like BF₃).

1. H₂O: oxygen has two lone pairs which it can donate — it is a Lewis base (not a Lewis acid).
2. BF₃: boron has only six electrons in its valence shell (incomplete octet) and an empty 2p orbital, so it accepts a pair, e.g. BF₃ + :NH₃ → F₃B←NH₃ — a Lewis acid.
3. H⁺: it has an empty 1s orbital and accepts a lone pair (as in H⁺ + :NH₃ → NH₄⁺) — a Lewis acid.
4. NH₄⁺: nitrogen already has a complete octet with no lone pair and no vacant orbital, so it cannot accept a pair — it is not a Lewis acid by itself (it is only a Brønsted acid, because it can release H⁺).

Extra depth. The Lewis concept is broader than Brønsted–Lowry because it covers reactions without protons, such as BF₃ + F⁻ → BF₄⁻. Every Brønsted base is also a Lewis base, but not every Brønsted acid is a Lewis acid: the Brønsted acid NH₄⁺ donates H⁺ (which is the real Lewis acid).

Conclusion: Lewis acids: BF₃ and H⁺.

Concept. The conjugate base of a Brønsted acid is obtained by removing one proton (H⁺): HA → A⁻ + H⁺. The charge on the species decreases by one unit.

1. HF → H⁺ + F⁻.
2. H₂SO₄ → H⁺ + HSO₄⁻.
3. HCO₃⁻ → H⁺ + CO₃²⁻.

Extra depth. Notice that HSO₄⁻ and HCO₃⁻ are amphiprotic: they can also act as bases (accepting a proton) — see Exercise 6.39. H₂SO₄ is a strong acid, so its conjugate base HSO₄⁻ is a very weak base; HF is a weak acid, so F⁻ is a stronger base and its solutions in water are slightly basic.

Conclusion: F⁻, HSO₄⁻ and CO₃²⁻.

Concept. The conjugate acid of a Brønsted base is obtained by adding one proton: B + H⁺ → BH⁺. The charge increases by one unit.

1. NH₂⁻ + H⁺ → NH₃.
2. NH₃ + H⁺ → NH₄⁺.
3. HCOO⁻ + H⁺ → HCOOH.

Extra depth. The amide ion NH₂⁻ is an extremely strong base (its conjugate acid NH₃ is a very weak acid), whereas formate HCOO⁻ is a weak base (conjugate acid HCOOH, K_a = 1.8 × 10⁻⁴). NH₃ is amphiprotic: it is both the conjugate base of NH₄⁺ and the conjugate acid of NH₂⁻.

Conclusion: NH₃, NH₄⁺ and HCOOH.

Concept. A species that can both donate and accept a proton is amphiprotic (amphoteric in the Brønsted sense). When it acts as a base it gains H⁺ (giving its conjugate acid); when it acts as an acid it loses H⁺ (giving its conjugate base).

1. H₂O: conjugate acid H₃O⁺ (H₂O + H⁺); conjugate base OH⁻ (H₂O − H⁺).
2. HCO₃⁻: conjugate acid H₂CO₃; conjugate base CO₃²⁻.
3. HSO₄⁻: conjugate acid H₂SO₄; conjugate base SO₄²⁻.
4. NH₃: conjugate acid NH₄⁺; conjugate base NH₂⁻.

Extra depth. In water these species mostly behave as follows: H₂O is a very weak acid and base (K_w = 10⁻¹⁴); HCO₃⁻ is slightly basic in water (the basis of the blood buffer H₂CO₃/HCO₃⁻); HSO₄⁻ acts mainly as an acid because H₂SO₄ is a strong acid; and NH₃ acts mainly as a base, as the acid role needs a very strong base to remove its proton.

Conclusion: H₂O: H₃O⁺/OH⁻; HCO₃⁻: H₂CO₃/CO₃²⁻; HSO₄⁻: H₂SO₄/SO₄²⁻; NH₃: NH₄⁺/NH₂⁻.

Concept. A Lewis base donates an electron pair (it has a lone pair); a Lewis acid accepts an electron pair (it has a vacant orbital). A Lewis acid–base reaction results in a coordinate (dative) covalent bond in which both electrons come from the base.

1. (a) OH⁻ — Lewis base: it has lone pairs on oxygen and donates one to an acceptor, e.g. H⁺ + :OH⁻ → H–OH, or OH⁻ + BF₃ → [F₃B←OH]⁻.
2. (b) F⁻ — Lewis base: it has four lone pairs and can donate one, e.g. BF₃ + F⁻ → BF₄⁻.
3. (c) H⁺ — Lewis acid: its 1s orbital is empty; it accepts a pair from, for example, NH₃ (H⁺ + :NH₃ → NH₄⁺) or from OH⁻.
4. (d) BCl₃ — Lewis acid: boron has only six valence electrons and a vacant 2p orbital; it accepts a pair, e.g. BCl₃ + :NH₃ → Cl₃B←NH₃ (or with Cl⁻ to form BCl₄⁻).

Extra depth. The reaction between a Lewis acid and a Lewis base is neutralisation in the widest sense, and the product is called an adduct or coordination compound. Note that OH⁻ and F⁻ are also Brønsted bases, but BCl₃ shows the full power of the Lewis idea: it has no proton at all yet behaves as an acid. This is why anhydrous AlCl₃ and BF₃ are such widely used catalysts in organic chemistry (Friedel–Crafts reactions).

Conclusion: Lewis bases: OH⁻, F⁻; Lewis acids: H⁺, BCl₃.

Concept. pH is defined as the negative common logarithm of the hydrogen ion concentration: pH = −log[H⁺] (Sørensen, 1909); [H⁺] here is in mol L⁻¹ (strictly, activity).

1. Substitute: pH = −log(3.8 × 10⁻³).
2. Split the logarithm: log(3.8 × 10⁻³) = log 3.8 + log 10⁻³ = 0.58 − 3 = −2.42.
3. pH = 2.42.

Extra depth. A pH of 2.4 is quite acidic — similar to lemon juice (2.2) and cola drinks; the acidity comes from phosphoric acid and carbonic acid. Since pH is logarithmic, a drop of one pH unit means a tenfold increase in [H⁺]. The value of pH has no unit, and the number of decimal places in pH equals the number of significant figures in [H⁺] (two here).

Conclusion: pH = −log(3.8 × 10⁻³) = 2.42.

Concept. The relationship is reversed: [H⁺] = 10^−pH = antilog(−pH). To avoid a negative-log antilog on a calculator, split pH into an integer part and a decimal part: write −3.76 = −4 + 0.24.

1. [H⁺] = 10^−3.76.
2. Rewrite: 10^−3.76 = 10^0.24 × 10⁻⁴.
3. Antilog: 10^0.24 = 1.74, so [H⁺] = 1.74 × 10⁻⁴ mol L⁻¹.

Extra depth. Vinegar is about 5 % acetic acid but its pH is only 3.76 (in a diluted sample) since acetic acid is a weak acid — only a small fraction ionises. Check: with K_a = 1.74 × 10⁻⁵, [H⁺] = 1.74 × 10⁻⁴ M would require a concentration of about 1.7 × 10⁻³ M, i.e. the sample is a dilute one.

Conclusion: [H⁺] = antilog(−3.76) = 1.74 × 10⁻⁴ M.

Concept. For a conjugate acid–base pair in water, K_a × K_b = K_w = 1.0 × 10⁻¹⁴ at 298 K. Therefore K_b of the conjugate base = K_w/K_a (eq. 6.45 of the chapter).

1. F⁻ (conjugate base of HF): K_b = 10⁻¹⁴/(6.8 × 10⁻⁴) = 1.47 × 10⁻¹¹.
2. HCOO⁻: K_b = 10⁻¹⁴/(1.8 × 10⁻⁴) = 5.56 × 10⁻¹¹.
3. CN⁻: K_b = 10⁻¹⁴/(4.8 × 10⁻⁹) = 2.08 × 10⁻⁶.

Extra depth. The weaker the acid (smaller K_a), the stronger its conjugate base (larger K_b): CN⁻ is the strongest of the three bases, F⁻ the weakest. The same relation gives pK_a + pK_b = 14. Strengths: HF > HCOOH > HCN as acids, but F⁻ < HCOO⁻ < CN⁻ as bases — an inverse order.

Conclusion: K_b(F⁻) = 1.47 × 10⁻¹¹; K_b(HCOO⁻) = 5.56 × 10⁻¹¹; K_b(CN⁻) = 2.08 × 10⁻⁶.

Concept. Phenol (C₆H₅OH) is a very weak acid: C₆H₅OH ⇌ C₆H₅O⁻ + H⁺, K_a = [C₆H₅O⁻][H⁺]/[C₆H₅OH]. For a weak acid of concentration c, [A⁻] = [H⁺] = √(K_ac), and α = √(K_a/c). The second part is the common-ion effect: added sodium phenolate (fully ionised) suppresses the ionisation of phenol.

1. (i) Without common ion: [C₆H₅O⁻] = [H⁺] = √(1.0 × 10⁻¹⁰ × 0.05) = √(5 × 10⁻¹²) = 2.24 × 10⁻⁶ M.
2. Degree of ionisation in pure phenol = 2.24 × 10⁻⁶/0.05 = 4.5 × 10⁻⁵.
3. (ii) With 0.01 M sodium phenolate: [C₆H₅O⁻] ≈ 0.01 M (the salt supplies it; the phenol’s own contribution is negligible) and [C₆H₅OH] ≈ 0.05 M. Then [H⁺] = K_a[C₆H₅OH]/[C₆H₅O⁻] = 10⁻¹⁰ × 0.05/0.01 = 5 × 10⁻¹⁰ M.
4. Degree of ionisation of phenol = [H⁺]/0.05 (each H⁺ comes from one phenol molecule) = 5 × 10⁻¹⁰/0.05 = 1.0 × 10⁻⁸.

Extra depth. Notice the common-ion effect in numbers: the extent of ionisation falls by a factor of about 4500 (4.5 × 10⁻⁵ to 1.0 × 10⁻⁸). The pH of the first solution is −log(2.24 × 10⁻⁶) = 5.65 (the water contribution 10⁻⁷ is small but not totally negligible); in the second, the solution is a buffer with pH = pK_a + log(0.01/0.05) = 10 − 0.70 = 9.30, an application of the Henderson–Hasselbalch equation.

Conclusion: [C₆H₅O⁻] = 2.24 × 10⁻⁶ M; with 0.01 M sodium phenolate α = 1.0 × 10⁻⁸.

Concept. H₂S is a diprotic acid that ionises in two steps: H₂S ⇌ H⁺ + HS⁻ (K_a1 = 9.1 × 10⁻⁸) and HS⁻ ⇌ H⁺ + S²⁻ (K_a2 = 1.2 × 10⁻¹³). Because K_a2 << K_a1, nearly all the H⁺ comes from the first step.

1. (a) 0.1 M H₂S alone: [H⁺] = [HS⁻] = x; x²/0.1 = 9.1 × 10⁻⁸ ⇒ x = √(9.1 × 10⁻⁹) = 9.54 × 10⁻⁵ M.
2. (b) With 0.1 M HCl: [H⁺] ≈ 0.1 M from HCl; [HS⁻] = K_a1[H₂S]/[H⁺] = 9.1 × 10⁻⁸ × 0.1/0.1 = 9.1 × 10⁻⁸ M, i.e. about 1000 times lower (common-ion effect).
3. [S²⁻] in case (a): K_a2 = [H⁺][S²⁻]/[HS⁻]; since [H⁺] = [HS⁻], [S²⁻] = K_a2 = 1.2 × 10⁻¹³ M.
4. [S²⁻] in case (b): [S²⁻] = K_a2[HS⁻]/[H⁺] = 1.2 × 10⁻¹³ × 9.1 × 10⁻⁸/0.1 = 1.09 × 10⁻¹⁹ M.

Extra depth. This is the basis of qualitative analysis: in acidic solution (0.1 M HCl) the sulphide concentration is so low (10⁻¹⁹ M) that only the very insoluble sulphides (Group II: CuS, CdS, etc.) precipitate, while those of Zn, Mn, Fe (Group IV) need a basic medium to be precipitated — see Exercise 6.73. Overall K for H₂S ⇌ 2H⁺ + S²⁻ is K_a1K_a2 = 1.09 × 10⁻²⁰, so [S²⁻] = 1.09 × 10⁻²⁰[H₂S]/[H⁺]² — a quick check of the 10⁻¹⁹ value.

Conclusion: [HS⁻] = 9.54 × 10⁻⁵ M (9.1 × 10⁻⁸ M with HCl); [S²⁻] = 1.2 × 10⁻¹³ M (1.09 × 10⁻¹⁹ M with HCl).

Concept. For a weak acid HA of concentration c and small α: K_a = cα²/(1 − α) ≈ cα², so α = √(K_a/c); [A⁻] = [H⁺] = cα. The approximation 1 − α ≈ 1 is valid when α < 0.05.

1. Degree of dissociation: α = √(1.74 × 10⁻⁵/0.05) = √(3.48 × 10⁻⁴) = 0.0187 (1.87 %).
2. Acetate ion concentration: [CH₃COO⁻] = cα = 0.05 × 0.0187 = 9.33 × 10⁻⁴ M (= [H⁺]).
3. pH = −log(9.33 × 10⁻⁴) = 4 − 0.97 = 3.03.

Extra depth. Verification: K_a = (9.33 × 10⁻⁴)²/0.0491 = 1.77 × 10⁻⁵ ≈ 1.74 × 10⁻⁵. Only 1.9 % of the acid is ionised, which is why acetic acid is classified as weak. As the solution is diluted, α increases (Ostwald’s dilution law): at 0.005 M, α would be 0.059.

Conclusion: α = √(K_a/c) = 0.0187; [CH₃COO⁻] = 9.33 × 10⁻⁴ M; pH = 3.03.

Concept. For a monoprotic acid HA, each H⁺ is accompanied by one anion A⁻, so [A⁻] = [H⁺] = 10^−pH (water’s contribution neglected). The ionisation constant follows from K_a = [H⁺][A⁻]/[HA] with [HA] = c − [H⁺] ≈ c.

1. [H⁺] = 10^−4.15 = 7.08 × 10⁻⁵ M.
2. Concentration of the anion: [A⁻] = [H⁺] = 7.08 × 10⁻⁵ M.
3. [HA] at equilibrium = 0.01 − 7.08 × 10⁻⁵ ≈ 0.01 M (the correction is only 0.7 %).
4. K_a = (7.08 × 10⁻⁵)²/0.01 = 5.01 × 10⁻⁹/0.01 = 5.0 × 10⁻⁷.
5. pK_a = −log(5.0 × 10⁻⁷) = 7 − 0.70 = 6.30.

Extra depth. The acid is weak (K_a ≈ 10⁻⁷, comparable to carbonic acid’s first ionisation, pK_a1 = 6.35). Degree of ionisation is 0.7 %, so the approximation [HA] ≈ c is fully justified. If we used the more exact value 0.01 − 7.08 × 10⁻⁵ = 0.009929, K_a would be 5.05 × 10⁻⁷ — the same to two significant figures.

Conclusion: [A⁻] = 7.08 × 10⁻⁵ M; K_a = 5.0 × 10⁻⁷; pK_a = 6.30.

Concept. HCl and HBr are strong acids; NaOH and KOH are strong bases — all completely dissociated. So [H⁺] (for the acids) or [OH⁻] (for the bases) equals the stated concentration, because each formula unit gives one H⁺ or one OH⁻. For bases use pOH = −log[OH⁻] and pH = 14 − pOH (at 298 K).

1. (a) 0.003 M HCl: [H⁺] = 3 × 10⁻³; pH = −log(3 × 10⁻³) = 3 − 0.477 = 2.52.
2. (b) 0.005 M NaOH: [OH⁻] = 5 × 10⁻³; pOH = 3 − 0.699 = 2.30; pH = 14 − 2.30 = 11.70.
3. (c) 0.002 M HBr: [H⁺] = 2 × 10⁻³; pH = 3 − 0.301 = 2.70.
4. (d) 0.002 M KOH: [OH⁻] = 2 × 10⁻³; pOH = 2.70; pH = 14 − 2.70 = 11.30.

Extra depth. The symmetry of (c) and (d) — same concentration, pH values 2.70 and 11.30, adding up to 14 — follows from K_w = [H⁺][OH⁻] = 10⁻¹⁴. These simple formulas hold when the concentration is much larger than 10⁻⁷ M so water’s own ionisation can be ignored; for a 10⁻⁸ M HCl solution the pH would be about 6.98, not 8.

Conclusion: (a) 2.52; (b) 11.70; (c) 2.70; (d) 11.30.

Concept. In each case convert the mass (or volume) to molarity, find [OH⁻] (or [H⁺]) allowing for the number of hydroxide ions per formula unit (TlOH and NaOH: 1; Ca(OH)₂: 2), then compute pOH and pH = 14 − pOH. TlOH is a strong base (thallous hydroxide dissociates completely).

1. (a) TlOH: molar mass = 204.4 + 17 = 221.4 g mol⁻¹; moles = 2/221.4 = 9.03 × 10⁻³; M = 9.03 × 10⁻³/2 = 4.52 × 10⁻³; [OH⁻] = 4.52 × 10⁻³; pOH = 2.35; pH = 11.65.
2. (b) Ca(OH)₂: M = 74 g mol⁻¹; moles = 0.3/74 = 4.05 × 10⁻³; M = 4.05 × 10⁻³/0.5 = 8.11 × 10⁻³; [OH⁻] = 2 × 8.11 × 10⁻³ = 1.62 × 10⁻²; pOH = 1.79; pH = 12.21.
3. (c) NaOH: M = 40 g mol⁻¹; moles = 0.3/40 = 7.5 × 10⁻³; M = 7.5 × 10⁻³/0.2 = 3.75 × 10⁻²; pOH = 1.43; pH = 12.57.
4. (d) HCl dilution: M₁V₁ = M₂V₂: 13.6 × 1 mL = M₂ × 1000 mL ⇒ M₂ = 1.36 × 10⁻² M; [H⁺] = 1.36 × 10⁻²; pH = 1.87.

Extra depth. Ca(OH)₂ is the trap: it yields two OH⁻ per formula unit, so [OH⁻] is double the molarity. Although Ca(OH)₂ is only sparingly soluble (about 1.7 g per litre), 0.3 g in 500 mL is below the solubility limit (0.6 g L⁻¹), so complete dissolution can be assumed. Thallium compounds are highly toxic — the exercise is purely theoretical.

Conclusion: (a) 11.65; (b) 12.21; (c) 12.57; (d) 1.87.

Concept. For a weak acid of concentration c and degree of ionisation α: [H⁺] = cα, and K_a = cα²/(1 − α). Here α = 0.132 is not small (> 5 %), so the exact form should be used.

1. [H⁺] = cα = 0.1 × 0.132 = 1.32 × 10⁻² M.
2. pH = −log(1.32 × 10⁻²) = 2 − 0.12 = 1.88.
3. K_a = cα²/(1 − α) = 0.1 × (0.132)²/(1 − 0.132) = 1.742 × 10⁻³/0.868 = 2.01 × 10⁻³.
4. pK_a = −log(2.01 × 10⁻³) = 2.70.

Extra depth. If the approximation K_a ≈ cα² = 1.74 × 10⁻³ were used, pK_a would be 2.76 — about 15 % error in K_a, because 1 − α = 0.868 is not close to 1. Bromoacetic acid is a much stronger acid than acetic acid (pK_a 4.76) because the electron-withdrawing bromine atom stabilises the anion (inductive effect), exactly as the chloroacetic acid in Exercise 6.64.

Conclusion: pH = 1.88; K_a = 2.0 × 10⁻³; pK_a = 2.70.

Concept. Codeine is a weak organic base, B + H₂O ⇌ BH⁺ + OH⁻, with K_b = [BH⁺][OH⁻]/[B]. From the pH we get pOH = 14 − pH, hence [OH⁻]; and [BH⁺] = [OH⁻] because each OH⁻ comes with one BH⁺.

1. pOH = 14 − 9.95 = 4.05; [OH⁻] = 10^−4.05 = 8.91 × 10⁻⁵ M.
2. [BH⁺] = [OH⁻] = 8.91 × 10⁻⁵ M; [B] = 0.005 − 0.0000891 ≈ 0.005 M (only 1.8 % ionised).
3. K_b = (8.91 × 10⁻⁵)²/0.005 = 7.94 × 10⁻⁹/0.005 = 1.6 × 10⁻⁶.
4. pK_b = −log(1.6 × 10⁻⁶) = 6 − 0.20 = 5.8.

Extra depth. Codeine is a much weaker base than ammonia (K_b = 1.77 × 10⁻⁵) but stronger than aniline (4.27 × 10⁻¹⁰). The degree of ionisation is 8.91 × 10⁻⁵/0.005 = 1.8 %, so ignoring the small amount ionised in [B] is justified. Its conjugate acid has K_a = K_w/K_b = 6.3 × 10⁻⁹ (pK_a = 8.2).

Conclusion: K_b = 1.6 × 10⁻⁶; pK_b = 5.8.

Concept. Aniline, C₆H₅NH₂, is a very weak base: K_b = 4.27 × 10⁻¹⁰ (Table 6.7). For a weak base of concentration c: [OH⁻] = √(K_bc) and α = √(K_b/c). The conjugate acid (anilinium ion) has K_a = K_w/K_b.

1. [OH⁻] = √(4.27 × 10⁻¹⁰ × 0.001) = √(4.27 × 10⁻¹³) = 6.53 × 10⁻⁷ M.
2. pOH = −log(6.53 × 10⁻⁷) = 6.19; pH = 14 − 6.19 = 7.81.
3. Degree of ionisation: α = 6.53 × 10⁻⁷/0.001 = 6.53 × 10⁻⁴ (0.065 %).
4. K_a of the conjugate acid (C₆H₅NH₃⁺) = K_w/K_b = 10⁻¹⁴/(4.27 × 10⁻¹⁰) = 2.34 × 10⁻⁵.

Extra depth. The result is very close to neutral because aniline is so weak. Strictly, the OH⁻ from water (10⁻⁷ M) is no longer negligible beside 6.5 × 10⁻⁷ M; including it, [OH⁻]² ≈ K_bc + K_w gives pH = 7.82 — practically the same. The lone pair on aniline’s nitrogen is delocalised into the benzene ring (resonance), which makes it a far weaker base than aliphatic amines and ammonia.

Conclusion: pH = 7.81; α = 6.53 × 10⁻⁴; K_a(C₆H₅NH₃⁺) = 2.34 × 10⁻⁵.

Concept. First convert pK_a to K_a: K_a = 10^−4.74 = 1.82 × 10⁻⁵. For the acid alone α = √(K_a/c). When a strong acid (HCl) is present, [H⁺] ≈ [HCl] and the acid’s own ionisation is suppressed; then K_a = [H⁺]cα/c(1 − α) ≈ [H⁺]α, so α = K_a/[H⁺].

1. Acetic acid alone: α = √(1.82 × 10⁻⁵/0.05) = √(3.64 × 10⁻⁴) = 0.0191.
2. (a) With 0.01 M HCl: [H⁺] ≈ 0.01 M, so α = K_a/[H⁺] = 1.82 × 10⁻⁵/0.01 = 1.82 × 10⁻³.
3. (b) With 0.1 M HCl: α = 1.82 × 10⁻⁵/0.1 = 1.82 × 10⁻⁴.

Extra depth. This is the common-ion effect: the extra H⁺ from HCl shifts CH₃COOH ⇌ H⁺ + CH₃COO⁻ to the left, so the acid ionises less. The degree of ionisation decreases in inverse proportion to the H⁺ concentration of the strong acid. Note that approximating [H⁺] by [HCl] is valid because the H⁺ supplied by acetic acid itself (about 10⁻⁶ M) is negligible.

Conclusion: α = 0.0191 (alone); 1.82 × 10⁻³ with 0.01 M HCl; 1.82 × 10⁻⁴ with 0.1 M HCl.

Concept. Dimethylamine, (CH₃)₂NH, is a weak base: (CH₃)₂NH + H₂O ⇌ (CH₃)₂NH₂⁺ + OH⁻. For base alone α = √(K_b/c) (approximation). In the presence of NaOH, which is a source of the common ion OH⁻, ionisation is suppressed.

1. Alone: α = √(K_b/c) = √(5.4 × 10⁻⁴/0.02) = √(0.027) = 0.164 (16.4 %).
2. With 0.1 M NaOH: [OH⁻] ≈ 0.1 M. K_b = [BH⁺][OH⁻]/[B] with [BH⁺] = cα = 0.02α and [B] ≈ 0.02: 5.4 × 10⁻⁴ = (0.02α)(0.1)/0.02 = 0.1α.
3. α = 5.4 × 10⁻⁴/0.1 = 5.4 × 10⁻³, i.e. 0.54 % is ionised.

Extra depth. Because α = 0.164 is not small compared with 1, the quadratic is more accurate: 0.02α² + 5.4 × 10⁻⁴α − 5.4 × 10⁻⁴ = 0 gives α = 0.151. The NCERT-style approximation (0.164) is acceptable but overestimates by about 8 %. In the NaOH solution the approximation is perfect (α << 1). The 30-fold fall (16.4 % to 0.54 %) shows how strongly a common ion suppresses ionisation.

Conclusion: α = 0.164 (16.4 %); with 0.1 M NaOH 0.54 % of dimethylamine is ionised.

Concept. [H⁺] = 10^−pH. Write pH as an integer minus a fraction: for example, pH 6.83 = 7 − 0.17, so [H⁺] = 10^0.17 × 10⁻⁷ = 1.48 × 10⁻⁷.

1. (a) Muscle fluid, pH 6.83: [H⁺] = 10^−6.83 = 1.48 × 10⁻⁷ M.
2. (b) Stomach fluid, pH 1.2: [H⁺] = 10^−1.2 = 6.31 × 10⁻² M.
3. (c) Blood, pH 7.38: [H⁺] = 10^−7.38 = 4.17 × 10⁻⁸ M.
4. (d) Saliva, pH 6.4: [H⁺] = 10^−6.4 = 3.98 × 10⁻⁷ M.

Extra depth. Stomach fluid is more than a million times as acidic as blood: [H⁺] ratio = 6.31 × 10⁻²/4.17 × 10⁻⁸ = 1.5 × 10⁶. Blood pH is held narrowly between 7.35 and 7.45 by the H₂CO₃/HCO₃⁻ buffer; a fall below 7.0 or rise above 7.8 is life-threatening.

Conclusion: (a) 1.48 × 10⁻⁷; (b) 6.31 × 10⁻²; (c) 4.17 × 10⁻⁸; (d) 3.98 × 10⁻⁷ mol L⁻¹.

Concept. Again [H⁺] = 10^−pH. The only skill needed is the antilog of the decimal part.

1. Milk, pH 6.8: [H⁺] = 10^−6.8 = 10^0.2 × 10⁻⁷ = 1.58 × 10⁻⁷ M.
2. Black coffee, pH 5.0: [H⁺] = 1.0 × 10⁻⁵ M.
3. Tomato juice, pH 4.2: 10^−4.2 = 10^0.8 × 10⁻⁵ = 6.31 × 10⁻⁵ M.
4. Lemon juice, pH 2.2: 10^−2.2 = 10^0.8 × 10⁻³ = 6.31 × 10⁻³ M.
5. Egg white, pH 7.8: 10^−7.8 = 10^0.2 × 10⁻⁸ = 1.58 × 10⁻⁸ M.

Extra depth. Lemon juice is about 100 times more acidic than tomato juice (2 pH units apart) and about 4 × 10⁵ times as acidic as egg white (5.6 units apart). Egg white (7.8) is the only one above pH 7, so it alone is basic. This shows how a small change in pH reflects a large change in [H⁺] because of the logarithmic scale.

Conclusion: [H⁺]: milk 1.58 × 10⁻⁷; coffee 1.0 × 10⁻⁵; tomato juice 6.31 × 10⁻⁵; lemon juice 6.31 × 10⁻³; egg white 1.58 × 10⁻⁸ M.

Concept. KOH is a strong base, completely dissociated: KOH → K⁺ + OH⁻. So [K⁺] = [OH⁻] = molarity of KOH, and [H⁺] follows from K_w = [H⁺][OH⁻] = 10⁻¹⁴ at 298 K.

1. Moles: molar mass of KOH = 39 + 16 + 1 = 56.1 g mol⁻¹; n = 0.561/56.1 = 0.0100 mol.
2. Molarity: 0.0100 mol/0.200 L = 0.05 M.
3. Ions: [K⁺] = [OH⁻] = 0.05 M; [H⁺] = 10⁻¹⁴/0.05 = 2.0 × 10⁻¹³ M.
4. pH = −log(2.0 × 10⁻¹³) = 13 − 0.30 = 12.70 (check: pOH = 1.30; 14 − 1.30 = 12.70).