RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Chemistry · 10 questions · 24 marks
Most reactions you meet in the lab never finish — they stall at a point where forward and backward changes run at the same rate and nothing more seems to happen. This chapter teaches you to put a number on that balance point through Kc and Kp, to predict how it shifts when you squeeze, heat or add something using Le Chatelier's principle, and then to apply the same ideas to acids, bases, buffers and sparingly soluble salts.
For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the relation between Kp and Kc is:
Answer
Kp = Kc(RT)⁻¹. The relation is Kp = Kc(RT)^Δn, where Δn counts only gaseous species. The products give 2 mol of gas and the reactants give 2 + 1 = 3 mol of gas. So Δn = 2 − 3 = −1 and Kp = Kc(RT)⁻¹, that is Kp = Kc ÷ RT. Whenever a gaseous reaction contracts, Kp comes out smaller than Kc at ordinary temperatures; Kp equals Kc only when Δn = 0.
The endothermic equilibrium 2X(g) ⇌ Y(g) + Z(g) is disturbed. Which change will shift it to the right?
Answer
Raising the temperature of the mixture shifts it to the right. For an endothermic reaction heat behaves like a reactant, so supplying more heat drives the system forward to absorb it, and the value of Kc itself increases with temperature. Compression does nothing here because both sides have 2 mol of gas, so Δn = 0 and pressure has no effect. A catalyst speeds the forward and backward reactions by the same factor and cannot move the position of equilibrium. Adding an inert gas at constant volume leaves every partial pressure and every concentration unchanged, so the system stays exactly where it was.
The conjugate base of the dihydrogen phosphate ion H₂PO₄⁻ is:
Answer
HPO₄²⁻ is the conjugate base of H₂PO₄⁻. A conjugate base is what remains after the species donates exactly one proton, so its formula has one fewer hydrogen and one more unit of negative charge: H₂PO₄⁻ → H⁺ + HPO₄²⁻. H₃PO₄ is the conjugate acid of H₂PO₄⁻, not its base, since it has gained a proton. PO₄³⁻ is two protons removed, so it is the conjugate base of HPO₄²⁻ instead. Because H₂PO₄⁻ can both donate and accept a proton it is amphiprotic, which is why buffers of this ion are so common in biology.
The solubility of silver chromate in pure water is s mol/L. Its solubility in 0.10 M silver nitrate solution will be:
Answer
Less than s, because of the common ion effect. Silver chromate dissolves as Ag₂CrO₄(s) ⇌ 2Ag⁺(aq) + CrO₄²⁻(aq) with Ksp = [Ag⁺]²[CrO₄²⁻]. Silver nitrate is fully soluble and floods the solution with Ag⁺, an ion already on the right-hand side. Since Ksp must keep its fixed value at that temperature, a much larger [Ag⁺] forces [CrO₄²⁻] to fall, and the amount of salt that can dissolve drops. Le Chatelier's principle says the same thing: the extra Ag⁺ pushes the dissolution equilibrium backwards. The solubility becomes smaller but never zero, since some chromate always remains in solution.
Assertion (A): Adding a catalyst to a reaction mixture at equilibrium does not change the equilibrium constant. Reason (R): A catalyst lowers the activation energy of the forward and the backward reaction by exactly the same amount.
Answer
Both A and R are true and R is the correct explanation of A. A catalyst works by offering an alternative path over a lower energy barrier. That new path is available to molecules travelling in either direction, so the activation energy falls by the same amount for the forward and the backward step. The equilibrium constant depends on the difference between the two activation energies, which is fixed by the energy gap between reactants and products. Since both barriers drop equally, that difference is untouched and K stays the same. The only effect is that equilibrium is reached sooner — useful in industry, where a catalyst lets a plant reach the same yield at a workable rate.
Calculate the pH and the degree of dissociation of a 0.050 M solution of a weak monobasic acid whose Ka is 2.0 × 10⁻⁵ at 298 K.
Answer
For a weak acid where dissociation is small, [H⁺] = √(Ka × C). [H⁺] = √(2.0 × 10⁻⁵ × 0.050) = √(1.0 × 10⁻⁶) = 1.0 × 10⁻³ mol/L. pH = −log(1.0 × 10⁻³) = 3.0. Degree of dissociation α = [H⁺] ÷ C = (1.0 × 10⁻³) ÷ 0.050 = 0.020, that is 2.0%. Since α is only 2%, the approximation of ignoring the dissociated fraction in the denominator was justified. Note that diluting this solution would raise α but lower [H⁺], so the pH would rise.
2.0 mol of PCl₅ is heated in a closed 4.0 L vessel at 500 K until equilibrium is reached, at which point 40% of it has dissociated according to PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Calculate Kc and Kp. (R = 0.0821 L atm K⁻¹ mol⁻¹)
Answer
Moles dissociated = 40% of 2.0 = 0.80 mol. At equilibrium: PCl₅ = 2.0 − 0.80 = 1.20 mol, PCl₃ = 0.80 mol, Cl₂ = 0.80 mol. Dividing by the 4.0 L volume gives the equilibrium concentrations: [PCl₅] = 0.30 M, [PCl₃] = 0.20 M, [Cl₂] = 0.20 M. Kc = ([PCl₃][Cl₂]) ÷ [PCl₅] = (0.20 × 0.20) ÷ 0.30 = 0.040 ÷ 0.30 = 0.133 mol/L. For Kp, Δn = 2 − 1 = +1, so Kp = Kc(RT)¹ = 0.133 × 0.0821 × 500 = 5.46 atm. Because Δn is positive, Kp comes out numerically larger than Kc here.
0.80 mol of N₂O₄ is introduced into an evacuated 2.0 L flask at 400 K, where it establishes the equilibrium N₂O₄(g) ⇌ 2NO₂(g). At equilibrium the N₂O₄ is found to be 50% dissociated. (i) Set up the ICE table and find all equilibrium concentrations. (ii) Calculate Kc. (iii) Calculate Kp at 400 K. (iv) If the volume of the flask were suddenly halved, predict and justify the direction of the shift. (v) The forward reaction is endothermic; explain the effect of cooling the flask on the colour of the mixture, given that NO₂ is brown and N₂O₄ is colourless. (R = 0.0821 L atm K⁻¹ mol⁻¹)
Answer
(i) Moles dissociated = 50% of 0.80 = 0.40 mol. Initial: N₂O₄ = 0.80 mol, NO₂ = 0. Change: N₂O₄ = −0.40 mol, NO₂ = +2 × 0.40 = +0.80 mol (the stoichiometry gives two NO₂ for every N₂O₄ lost). Equilibrium: N₂O₄ = 0.40 mol, NO₂ = 0.80 mol. Dividing by the 2.0 L volume: [N₂O₄] = 0.20 M and [NO₂] = 0.40 M. (ii) Kc = [NO₂]² ÷ [N₂O₄] = (0.40)² ÷ 0.20 = 0.16 ÷ 0.20 = 0.80 mol/L. (iii) Δn = 2 − 1 = +1, so Kp = Kc(RT)^1 = 0.80 × 0.0821 × 400 = 26.3 atm. (iv) Halving the volume doubles every concentration at that instant, so check Q. New instantaneous values are [N₂O₄] = 0.40 M and [NO₂] = 0.80 M, giving Q = (0.80)² ÷ 0.40 = 1.60, which is greater than Kc = 0.80. Since Q > Kc the system shifts backwards, towards N₂O₄. This agrees with Le Chatelier's principle: compression raises the pressure, and the system relieves it by moving to the side with fewer gas moles, which is the single mole of N₂O₄ rather than the two moles of NO₂. (v) The forward dissociation absorbs heat, so cooling removes heat and the system responds by shifting in the exothermic direction — backwards, forming more N₂O₄. Since NO₂ is the coloured species, its concentration falls and the brown colour fades noticeably; the value of Kc also decreases on cooling. Warming the flask reverses the observation and the gas darkens, which is the classic demonstration of this equilibrium using two sealed tubes in hot and ice-cold water.
(i) Define a buffer solution and explain, with equations, how a mixture of a weak acid HA and its sodium salt NaA resists a change in pH. (ii) Derive the Henderson–Hasselbalch relation for such a buffer. (iii) A buffer is made by dissolving 0.20 mol of HA (Ka = 1.0 × 10⁻⁵) and 0.20 mol of NaA in water to make 1.00 L of solution. Calculate its pH. (iv) Calculate the new pH after 0.020 mol of solid NaOH is dissolved in it, assuming no volume change. (v) Compare this with the pH change that 0.020 mol of NaOH would cause in 1.00 L of pure water at pH 7.
Answer
(i) A buffer is a solution that resists a change in pH when a small amount of strong acid or strong base is added to it, or when it is moderately diluted. An acidic buffer contains a large reserve of the un-ionised weak acid and a large reserve of its conjugate base supplied by the fully dissociated salt. If H⁺ is added, the conjugate base mops it up: A⁻ + H⁺ → HA, so the free H⁺ is converted to un-ionised acid and the pH barely moves. If OH⁻ is added, the weak acid neutralises it: HA + OH⁻ → A⁻ + H₂O, so the strong base is replaced by the far weaker base A⁻. (ii) For the ionisation HA ⇌ H⁺ + A⁻, Ka = ([H⁺][A⁻]) ÷ [HA]. Rearranging, [H⁺] = Ka × ([HA] ÷ [A⁻]). Taking negative logarithms of both sides: −log[H⁺] = −log Ka − log([HA] ÷ [A⁻]). So pH = pKa + log([A⁻] ÷ [HA]), which is written pH = pKa + log([salt] ÷ [acid]). The salt is assumed completely dissociated and the weak acid barely dissociated, so their formal amounts may be used directly. (iii) pKa = −log(1.0 × 10⁻⁵) = 5.00. [salt] = [acid] = 0.20 M, so log(0.20 ÷ 0.20) = log 1 = 0. pH = 5.00 + 0 = 5.00. A buffer is at its most effective at exactly this point, where pH = pKa. (iv) The added NaOH consumes an equal number of moles of HA and creates the same number of moles of A⁻. HA remaining = 0.20 − 0.020 = 0.18 mol; A⁻ present = 0.20 + 0.020 = 0.22 mol. The volume is 1.00 L, so these are also the molarities. pH = 5.00 + log(0.22 ÷ 0.18) = 5.00 + log(1.222) = 5.00 + 0.087 = 5.09. The pH has risen by only 0.09 units. (v) In pure water, 0.020 mol of NaOH in 1.00 L gives [OH⁻] = 0.020 M directly, since NaOH is a strong base. pOH = −log(0.020) = 1.70, so pH = 14.00 − 1.70 = 12.30. The pH has jumped from 7.00 to 12.30, a change of 5.3 units, compared with 0.09 units in the buffer — the buffer is roughly sixty times more resistant here. This is exactly why blood is buffered by the carbonic acid–hydrogencarbonate system: a swing of even half a pH unit would be dangerous.
A water-testing technician studies a sparingly soluble metal hydroxide M(OH)₂ whose Ksp is 4.0 × 10⁻¹² at 298 K. She first measures its solubility in pure water, then repeats the measurement in a solution already containing 0.010 M NaOH, and finally checks whether a precipitate forms when equal volumes of 2.0 × 10⁻⁴ M M(NO₃)₂ and 2.0 × 10⁻⁴ M NaOH are mixed. Answer: (a) Calculate the molar solubility of M(OH)₂ in pure water and the pH of the saturated solution. (b) Calculate its solubility in 0.010 M NaOH. (c) Decide, with a calculation, whether a precipitate forms in the mixing experiment. (d) Suggest one practical way to remove M²⁺ ions from waste water, and state one limitation of using Ksp values in real water samples.
Answer
(a) Let the solubility be s mol/L. Dissolution gives M(OH)₂(s) ⇌ M²⁺(aq) + 2OH⁻(aq), so [M²⁺] = s and [OH⁻] = 2s. Ksp = [M²⁺][OH⁻]² = s × (2s)² = 4s³. 4s³ = 4.0 × 10⁻¹², so s³ = 1.0 × 10⁻¹² and s = 1.0 × 10⁻⁴ mol/L. [OH⁻] = 2s = 2.0 × 10⁻⁴ M, so pOH = −log(2.0 × 10⁻⁴) = 4 − 0.301 = 3.70 and pH = 14.00 − 3.70 = 10.30. The saturated solution is distinctly alkaline. (b) Now the hydroxide ion comes almost entirely from the NaOH, so [OH⁻] = 0.010 M and the tiny contribution from the dissolving salt can be ignored. Ksp = [M²⁺](0.010)², so [M²⁺] = (4.0 × 10⁻¹²) ÷ (1.0 × 10⁻⁴) = 4.0 × 10⁻⁸ mol/L. Since every dissolved formula unit gives one M²⁺, the solubility is 4.0 × 10⁻⁸ mol/L — about 2500 times smaller than in pure water. This is the common ion effect at work. (c) Mixing equal volumes halves each concentration, so immediately after mixing [M²⁺] = 1.0 × 10⁻⁴ M and [OH⁻] = 1.0 × 10⁻⁴ M. Ionic product Q = [M²⁺][OH⁻]² = (1.0 × 10⁻⁴) × (1.0 × 10⁻⁴)² = 1.0 × 10⁻¹². Comparing, Q = 1.0 × 10⁻¹² is less than Ksp = 4.0 × 10⁻¹², so the solution is unsaturated and no precipitate forms. Students often forget the dilution on mixing, which would have wrongly predicted a precipitate. (d) Raising the pH of the waste water by adding a cheap alkali such as slaked lime forces [OH⁻] up, which by the calculation in part (b) drives [M²⁺] down to a very low level, and the hydroxide can then be filtered off as sludge. The chief limitation is that real water is not a simple two-ion system: it contains complexing agents such as ammonia, chloride or organic ligands that tie up M²⁺ as soluble complexes, and its high total ionic strength alters effective ion concentrations. The measured solubility can therefore be considerably higher than the ideal Ksp calculation predicts, so treated water must always be tested rather than assumed clean.
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