CBSE Class 11 Chemistry Chemical Equilibrium MCQs Set 10

Download CBSE MCQs for Class 11 Chemistry: Chapter 06 Equilibrium

Access targeted multiple-choice questions for Chapter 06 Equilibrium designed to align with the latest CBSE academic syllabus for Class 11 Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: Chapter 06 Equilibrium

Navigate directly to the 50 objective questions for Chapter 06 Equilibrium using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: For preparing a buffer of pH 6 by mixing sodium acetate and acetic acid, the ratio of the concentration of salt to acid should be: (Ka of acetic acid = 10⁻⁵)
(a) 1 : 10
(b) 100 : 1
(c) 10 : 1
(d) 1 : 100
Show Answer & Explanation

Answer: (c) 10 : 1

Explanation:
1. pH = pKa + log([salt] ÷ [acid]), with pKa = 5.
2. 6 = 5 + log([salt] ÷ [acid]), so log([salt] ÷ [acid]) = 1 and [salt] ÷ [acid] = 10.

Teacher's Note:
1. Write pKa from Ka first.
2. A pH above pKa means more salt than acid.

Question 2: In a mixture of a weak acid and its salt, the ratio of the concentration of acid to salt is increased tenfold. The pH of the solution:
(a) increases tenfold
(b) increases by one
(c) decreases tenfold
(d) decreases by one
Show Answer & Explanation

Answer: (d) decreases by one

Explanation:
1. pH = pKa + log([salt] ÷ [acid]).
2. If [acid] ÷ [salt] becomes 10 times larger, [salt] ÷ [acid] becomes one-tenth, so log falls by 1 and the pH falls by one unit.

Teacher's Note:
1. The pH changes by the logarithm of the ratio change, not by the ratio itself.
2. log (1/10) = −1.

Question 3: The dissociation constant of HCN is 5 × 10⁻¹⁰. The pH of a solution made by mixing 1.5 mol of HCN and 0.15 mol of KCN in water and making up the total volume to 0.5 dm³ is:
(a) 7.302
(b) 9.302
(c) 8.302
(d) 10.302
Show Answer & Explanation

Answer: (c) 8.302

Explanation:
1. pKa = −log(5 × 10⁻¹⁰) = 9.301.
2. [salt] ÷ [acid] = 0.15 ÷ 1.5 = 0.1, so pH = 9.301 + log 0.1 = 9.301 − 1 = 8.301 ≈ 8.30.

Teacher's Note:
1. The volume cancels in the ratio of salt to acid.
2. KCN is the salt of the weak acid HCN and a strong base.

Question 4: 0.1 mol of CH3NH2 (Kb = 5 × 10⁻⁴) is mixed with 0.08 mol of HCl and the solution is diluted to one litre. The H⁺ ion concentration in the solution will be:
(a) 1.6 × 10⁻¹¹ M
(b) 8 × 10⁻² M
(c) 5 × 10⁻⁵ M
(d) 8 × 10⁻¹¹ M
Show Answer & Explanation

Answer: (d) 8 × 10⁻¹¹ M

Explanation:
1. 0.08 mol HCl converts 0.08 mol of base into its salt CH3NH3⁺. The base left is 0.1 − 0.08 = 0.02 mol, so this is a basic buffer.
2. [OH⁻] = Kb × [base] ÷ [salt] = 5 × 10⁻⁴ × 0.02 ÷ 0.08 = 1.25 × 10⁻⁴ M.
3. [H⁺] = 10⁻¹⁴ ÷ 1.25 × 10⁻⁴ = 8 × 10⁻¹¹ M.

Teacher's Note:
1. HCl is the limiting reagent. Find the amount of each component left.
2. Convert [OH⁻] to [H⁺] with Kw.

Question 5: How many moles of sodium formate must be added to 1 L of 0.1 M HCOOH to prepare a buffer of pH 3.4? (Ka of HCOOH = 2.0 × 10⁻⁴; log 2 = 0.30)
(a) 0.01
(b) 0.05
(c) 0.10
(d) 0.20
Show Answer & Explanation

Answer: (b) 0.05

Explanation:
1. pKa = 4 − 0.30 = 3.70.
2. 3.4 = 3.70 + log([salt] ÷ 0.1), so log([salt] ÷ 0.1) = −0.30 and [salt] ÷ 0.1 = 0.5.
3. [salt] = 0.05 M, so 0.05 mol is needed in 1 L.

Teacher's Note:
1. pH below pKa means less salt than acid.
2. The antilog of −0.30 is 0.5.

Question 6: What is the pH of a buffer that is 0.250 M in benzoic acid and 0.150 M in sodium benzoate? (Ka of benzoic acid = 6.5 × 10⁻⁵)
(a) 3.40
(b) 4.19
(c) 3.97
(d) 4.41
Show Answer & Explanation

Answer: (c) 3.97

Explanation:
1. pKa = −log(6.5 × 10⁻⁵) = 4.187.
2. pH = 4.187 + log(0.150 ÷ 0.250) = 4.187 + log 0.6 = 4.187 − 0.222 = 3.97.

Teacher's Note:
1. More acid than salt gives a pH below pKa.
2. log 0.6 = −0.22.

Question 7: A buffer is made by mixing 0.10 mol of sodium formate and 0.05 mol of formic acid in 1.0 L of solution. Its pH is: (Ka of HCOOH = 1.8 × 10⁻⁴)
(a) 1.8 × 10⁻⁴
(b) 3.44
(c) 5.31
(d) 4.05
Show Answer & Explanation

Answer: (d) 4.05

Explanation:
1. pKa = −log(1.8 × 10⁻⁴) = 3.745.
2. pH = 3.745 + log(0.10 ÷ 0.05) = 3.745 + log 2 = 3.745 + 0.301 = 4.05.

Teacher's Note:
1. A pH is not a value like 1.8 × 10⁻⁴; that is a Ka.
2. Salt is twice the acid, so pH = pKa + 0.30.

Question 8: To 1.0 L of a solution containing 0.10 mol each of NH3 and NH4Cl, 0.05 mol of solid NaOH is added. The change in pH is: (assume no change in volume; log 3 = 0.48)
(a) +0.48
(b) −0.30
(c) +0.30
(d) −0.48
Show Answer & Explanation

Answer: (a) +0.48

Explanation:
1. NaOH converts NH4⁺ into NH3: NH4⁺ + OH⁻ → NH3 + H2O. After the addition: NH3 = 0.15 mol and NH4⁺ = 0.05 mol.
2. Initially [base] = [salt], so pOH = pKb. Finally pOH = pKb + log(0.05 ÷ 0.15) = pKb − 0.48.
3. pOH falls by 0.48, so the pH rises by 0.48.

Teacher's Note:
1. Added base converts the acid component of the buffer into the base component.
2. pH + pOH = 14, so a fall in pOH means a rise in pH.

Question 9: The pH of blood is maintained at about 7.4 mainly by the buffer pair:
(a) NaH2PO4 + Na2HPO4
(b) H2CO3 + HCO3⁻
(c) CH3COOH + CH3COONa
(d) H3PO4 + NaH2PO4
Show Answer & Explanation

Answer: (b) H2CO3 + HCO3⁻

Explanation:
1. The carbonic acid–bicarbonate pair removes added H⁺ (as H2CO3) or OH⁻ (as HCO3⁻ is converted to CO3²⁻ and water).
2. It works with the lungs and kidneys to keep blood pH between 7.35 and 7.45.

Teacher's Note:
1. Blood pH changes of even 0.4 units are dangerous.
2. If CO2 builds up, the blood pH falls.

Question 10: An aqueous solution of sodium cyanide, NaCN, is:
(a) acidic
(b) amphoteric
(c) neutral
(d) basic
Show Answer & Explanation

Answer: (d) basic

Explanation:
1. NaCN is the salt of a strong base (NaOH) and a weak acid (HCN).
2. CN⁻ takes a proton from water: CN⁻ + H2O ⇌ HCN + OH⁻. The excess OH⁻ makes the solution basic.

Teacher's Note:
1. The anion of a weak acid hydrolyses to give OH⁻.
2. Like Na2CO3 and CH3COONa, NaCN gives a pH above 7.

Question 11: The aqueous solution of FeCl3 is acidic because of:
(a) acidic impurities
(b) hydrolysis
(c) ionisation
(d) dissociation
Show Answer & Explanation

Answer: (b) hydrolysis

Explanation:
1. FeCl3 is the salt of a weak base, Fe(OH)3, and a strong acid, HCl.
2. Fe³⁺ reacts with water (hydrolysis) and releases H⁺, so the solution becomes acidic.

Teacher's Note:
1. Cations of weak bases hydrolyse to give an acidic solution.
2. NH4Cl and Al2(SO4)3 behave in the same way.

Question 12: The pH values of 0.1 M solutions of the following salts and acid increase in the order:
(a) NaCl < NH4Cl < NaCN < HCl
(b) NaCN < NH4Cl < NaCl < HCl
(c) HCl < NH4Cl < NaCl < NaCN
(d) HCl < NaCl < NaCN < NH4Cl
Show Answer & Explanation

Answer: (c) HCl < NH4Cl < NaCl < NaCN

Explanation:
1. HCl is a strong acid, with pH about 1.
2. NH4Cl is acidic (about 5), NaCl is neutral (7) and NaCN is basic (above 7).
3. So the order of increasing pH is HCl < NH4Cl < NaCl < NaCN.

Teacher's Note:
1. Sort by the nature of the solution: strongly acidic, weakly acidic, neutral, basic.
2. Salts of a strong acid and a strong base are neutral.

Question 13: If the solubility of Mg(OH)2 is x mol L⁻¹, its solubility product Ksp is:
(a) 4x³
(b) 108x⁵
(c) 9x
(d) 27x⁴
Show Answer & Explanation

Answer: (a) 4x³

Explanation:
1. Mg(OH)2 ⇌ Mg²⁺ + 2OH⁻, so [Mg²⁺] = x and [OH⁻] = 2x.
2. Ksp = x × (2x)² = 4x³.

Teacher's Note:
1. Put the coefficient into each concentration and then raise to the power.
2. AB2 salts have Ksp = 4S³.

Question 14: The solubility of a sparingly soluble salt AB2 in water is 1.0 × 10⁻⁵ mol L⁻¹. Its solubility product is:
(a) 4 × 10⁻¹⁰
(b) 4 × 10⁻¹⁵
(c) 1 × 10⁻¹⁵
(d) 1 × 10⁻¹⁰
Show Answer & Explanation

Answer: (b) 4 × 10⁻¹⁵

Explanation:
1. AB2 ⇌ A²⁺ + 2B⁻, so [A²⁺] = S and [B⁻] = 2S.
2. Ksp = 4S³ = 4 × (1.0 × 10⁻⁵)³ = 4 × 10⁻¹⁵.

Teacher's Note:
1. (10⁻⁵)³ = 10⁻¹⁵.
2. Do not forget the factor 4.

Question 15: A precipitate of AgCl (Ksp = 10⁻¹⁰) will form when equal volumes of which of the following solutions are mixed?
(a) 10⁻⁴ M AgNO3 and 10⁻⁷ M HCl
(b) 10⁻⁵ M AgNO3 and 10⁻⁶ M HCl
(c) 10⁻⁵ M AgNO3 and 10⁻⁴ M HCl
(d) 10⁻⁶ M AgNO3 and 10⁻⁵ M HCl
Show Answer & Explanation

Answer: (c) 10⁻⁵ M AgNO3 and 10⁻⁴ M HCl

Explanation:
1. Mixing equal volumes halves each concentration.
2. For 10⁻⁵ M and 10⁻⁴ M: [Ag⁺][Cl⁻] = (5 × 10⁻⁶)(5 × 10⁻⁵) = 2.5 × 10⁻¹⁰ > 10⁻¹⁰, so AgCl precipitates.
3. Each of the other three mixtures gives an ionic product of only 2.5 × 10⁻¹², which is below Ksp, so no precipitate forms.

Teacher's Note:
1. Precipitation needs Q > Ksp.
2. Always halve the concentrations on mixing equal volumes.

Question 16: The solubility of AgCl will be minimum in:
(a) 0.01 M CaCl2
(b) pure water
(c) 0.001 M AgNO3
(d) 0.01 M NaCl
Show Answer & Explanation

Answer: (a) 0.01 M CaCl2

Explanation:
1. Solubility is lowest where the concentration of a common ion is highest.
2. 0.01 M CaCl2 gives [Cl⁻] = 0.02 M, which is more than the Cl⁻ in 0.01 M NaCl (0.01 M) or the Ag⁺ in 0.001 M AgNO3.

Teacher's Note:
1. CaCl2 gives two Cl⁻ per formula unit.
2. Compare the common-ion concentrations.

Question 17: The solubility product of PbCl2 at 298 K is 1.0 × 10⁻⁶. What is its solubility in mol L⁻¹?
(a) 1.0 × 10⁻³
(b) 6.3 × 10⁻³
(c) 3.0 × 10⁻⁵
(d) 4.6 × 10⁻¹⁴
Show Answer & Explanation

Answer: (b) 6.3 × 10⁻³

Explanation:
1. PbCl2 is an AB2-type salt: Ksp = 4S³.
2. S³ = 1.0 × 10⁻⁶ ÷ 4 = 2.5 × 10⁻⁷, so S = 6.3 × 10⁻³ mol L⁻¹.

Teacher's Note:
1. Take the cube root: (2.5 × 10⁻⁷)^(1/3) = 6.3 × 10⁻³.
2. Check by cubing the answer.

Question 18: Ksp for Ca(OH)2 is 5.5 × 10⁻⁶. What is the maximum pH that can be reached in a sewage tank treated with slaked lime, Ca(OH)2?
(a) 12.35
(b) 10.35
(c) 11.35
(d) 9.35
Show Answer & Explanation

Answer: (a) 12.35

Explanation:
1. 4S³ = 5.5 × 10⁻⁶, so S³ = 1.375 × 10⁻⁶ and S = 1.11 × 10⁻² M.
2. [OH⁻] = 2S = 2.22 × 10⁻² M, so pOH = 1.65 and pH = 14 − 1.65 = 12.35.

Teacher's Note:
1. A saturated solution gives the maximum [OH⁻].
2. The OH⁻ concentration is twice the solubility.

Question 19: The solubility products of XY, X2Y and XY3 are 4 × 10⁻²⁰, 3.2 × 10⁻¹¹ and 2.7 × 10⁻³¹ respectively. The order of increasing solubility is:
(a) XY3 < XY < X2Y
(b) X2Y < XY < XY3
(c) XY < X2Y < XY3
(d) XY < XY3 < X2Y
Show Answer & Explanation

Answer: (d) XY < XY3 < X2Y

Explanation:
1. XY: S² = 4 × 10⁻²⁰, so S = 2 × 10⁻¹⁰. X2Y: 4S³ = 3.2 × 10⁻¹¹, so S = 2 × 10⁻⁴. XY3: 27S⁴ = 2.7 × 10⁻³¹, so S = 1 × 10⁻⁸.
2. In increasing order: XY (2 × 10⁻¹⁰) < XY3 (10⁻⁸) < X2Y (2 × 10⁻⁴).

Teacher's Note:
1. Salts of different types cannot be compared by Ksp alone; compute S.
2. Use the formula that matches the number of ions.

Question 20: Ksp of Mg(OH)2 is 4.0 × 10⁻⁶. At what minimum pH will Mg²⁺ ions start precipitating from a 0.01 M solution of MgCl2? (log 2 = 0.30)
(a) 12 + log 2
(b) 2 − log 2
(c) 2 + log 2
(d) 12 − log 2
Show Answer & Explanation

Answer: (a) 12 + log 2

Explanation:
1. Precipitation starts when [Mg²⁺][OH⁻]² = Ksp.
2. [OH⁻]² = 4.0 × 10⁻⁶ ÷ 0.01 = 4.0 × 10⁻⁴, so [OH⁻] = 2 × 10⁻² M.
3. pOH = 2 − log 2, so pH = 14 − (2 − log 2) = 12 + log 2.

Teacher's Note:
1. Use the ionic product just reaching Ksp.
2. A higher pH is needed for precipitation, so 12 + log 2 is reasonable.

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