JEE Chemistry Ionic Equilibrium MCQs Set 31

Multiple Choice Questions (MCQs) for JEE Chemistry: Ionic Equilibrium

Review structured MCQ sets for JEE Chemistry Ionic Equilibrium. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Multiple Choice Questions

Question 1: The pH of a solution having 2 × 10−3 moles of OH− ions in 2 litre of solution is:
(a) 3
(b) 3 + log 2
(c) 3 − log 2
(d) 11
Show Answer & Explanation

Answer: (d) 11

Explanation:
1. [OH−] = 2 × 10−3/2 = 10−3 M, so pOH = 3.
2. pH = 14 − 3 = 11.
3. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 2: If α is the degree of dissociation of a weak dibasic organic acid and y is the hydrogen ion concentration (assuming each molecule gives 2H+), the initial concentration of the acid is:
(a) (α/y)/2 M
(b) y/α M
(c) y/(2α) M
(d) None of these
Show Answer & Explanation

Answer: (c) y/(2α) M

Explanation:
1. [H+] = 2Cα = y.
2. C = y/(2α) M.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 3: The Ka for 10−2 M HCN acid, having pOH = 10, is:
(a) Ka = 10−4
(b) Ka = 10−2
(c) Ka = 10−5
(d) None of these
Show Answer & Explanation

Answer: (d) None of these

Explanation:
1. pOH = 10 gives pH = 4 and [H+] = 10−4 M.
2. Ka = [H+]2/C = 10−8/10−2 = 10−6, which is not among (a) to (c).
3. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 4: The extent of ionisation of weak electrolytes increases:
(a) With an increase in the concentration of solute
(b) On decreasing the temperature of the solution
(c) On addition of excess of water to the solution
(d) On stirring the solution vigorously
Show Answer & Explanation

Answer: (c) On addition of excess of water to the solution

Explanation:
1. Ostwald's dilution law: α = √(K/C), so α increases as the solution is diluted.
2. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 5: The correct statement(s) for the weak acid HCN at 25°C: (i) α = Ka/[H+] (for small α); (ii) α = Ka[OH−]/Kw; (iii) Kb = Cα2. Choose the correct option:
(a) (i) only
(b) (ii) only
(c) (i) and (ii) both
(d) (iii) only
Show Answer & Explanation

Answer: (c) (i) and (ii) both

Explanation:
1. Ka = [H+][CN−]/[HCN] ≈ [H+]α, so α = Ka/[H+].
2. Since [H+] = Kw/[OH−], this equals Ka[OH−]/Kw.
3. Cα2 gives Ka of the acid, not Kb.
4. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 6: The ionic product of water is equal to:
(a) Dissociation constant of water × [H2O]
(b) Dissociation constant of water × [H+]
(c) Product of [H2O] and [H+]
(d) Product of [OH−]2 and [H+]
Show Answer & Explanation

Answer: (a) Dissociation constant of water × [H2O]

Explanation:
1. K = [H+][OH−]/[H2O], so Kw = K[H2O] = [H+][OH−].
2. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 7: At 90°C, pure water has [H3O+] = 10−6.7 mol L−1. The value of Kw at 90°C is:
(a) 10−6
(b) 10−12
(c) 10−6.7
(d) 10−13.4
Show Answer & Explanation

Answer: (d) 10−13.4

Explanation:
1. Pure water has [H+] = [OH−].
2. Kw = 10−6.7 × 10−6.7 = 10−13.4.
3. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 8: At 373 K the pH of pure H2O can be:
(a) < 7
(b) > 7
(c) = 7
(d) = 0
Show Answer & Explanation

Answer: (a) < 7

Explanation:
1. Kw increases with temperature (about 5.5 × 10−13 at 373 K), so [H+] rises and pH of neutral water falls below 7 (about 6.1).
2. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 9: The basic strength of NH4OH in the presence of NH4Cl:
(a) Increases
(b) Remains unchanged
(c) Decreases
(d) Sometimes increases or sometimes decreases
Show Answer & Explanation

Answer: (c) Decreases

Explanation:
1. NH4Cl supplies NH4+ (common ion), which suppresses the ionisation of NH4OH, so [OH−] and basic strength decrease.
2. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 10: H2S is a weak acid ionised into 2H+ and S2−. If HCl is added so that the pH becomes less, then:
(a) Decrease in S2− ion concentration
(b) The concentration of S2− is not affected
(c) Increase in S2− ion concentration
(d) It is not possible to add HCl in the solution
Show Answer & Explanation

Answer: (a) Decrease in S2− ion concentration

Explanation:
1. Added H+ (common ion) shifts the equilibrium H2S ⇌ 2H+ + S2− to the left, so [S2−] decreases.
2. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 11: Which is a basic salt?
(a) PbS
(b) PbCO3
(c) PbSO4
(d) 2PbCO3·Pb(OH)2
Show Answer & Explanation

Answer: (d) 2PbCO3·Pb(OH)2

Explanation:
1. A basic salt contains hydroxide groups in addition to the anion; 2PbCO3·Pb(OH)2 (white lead) is basic lead carbonate.
2. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 12: Which of the following is an acid salt?
(a) Na2S
(b) Na2SO3
(c) NaHSO3
(d) Na2SO4
Show Answer & Explanation

Answer: (c) NaHSO3

Explanation:
1. An acid salt still has a replaceable hydrogen; NaHSO3 contains HSO3−.
2. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 13: Which salt will NOT undergo hydrolysis?
(a) KCl
(b) Na2SO4
(c) NaCl
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Each is a salt of a strong acid and a strong base; neither ion reacts with water.
2. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 14: If pKb for CN− at 25°C is 4.7, the pH of a 0.5 M aqueous NaCN solution is:
(a) 12
(b) 10
(c) 11.5
(d) 11
Show Answer & Explanation

Answer: (c) 11.5

Explanation:
1. pOH = ½(pKb − log C) = ½(4.7 + 0.3) = 2.5.
2. pH = 14 − 2.5 = 11.5.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 15: For anionic hydrolysis, pH is given by:
(a) ½pKw − ½pKb − ½ log c
(b) ½pKw + ½pKa − ½pKb
(c) ½pKw + ½pKa + ½ log c
(d) None of these
Show Answer & Explanation

Answer: (c) ½pKw + ½pKa + ½ log c

Explanation:
1. For the salt of a weak acid and strong base, pH = 7 + ½pKa + ½ log c = ½pKw + ½pKa + ½ log c.
2. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 16: Ka for cyanoacetic acid is 3.5 × 10−3. The degree of hydrolysis of 0.05 M sodium cyanoacetate solution will be:
(a) 4.559 × 10−6
(b) 5.559 × 10−6
(c) 6.559 × 10−6
(d) 7.559 × 10−6
Show Answer & Explanation

Answer: (d) 7.559 × 10−6

Explanation:
1. h = √(Kw/(KaC)) = √(10−14/(3.5 × 10−3 × 0.05)).
2. = √(5.71 × 10−11) = 7.559 × 10−6.
3. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 17: The degree of hydrolysis of N/100 solution of KCN (Ka of HCN = 1.4 × 10−9) is:
(a) 2.7 × 10−3
(b) 2.7 × 10−2
(c) 2.7 × 10−4
(d) 2.7 × 10−5
Show Answer & Explanation

Answer: (b) 2.7 × 10−2

Explanation:
1. h = √(Kw/(KaC)) = √(10−14/(1.4 × 10−9 × 0.01)).
2. = √(7.14 × 10−4) = 2.7 × 10−2.
3. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 18: If the solubilities of the salts M2X, QY2 and PZ2 are equal, then the relation between their Ksp is:
(a) Ksp(M2X) > Ksp(QY2) > Ksp(PZ2)
(b) Ksp(M2X) = Ksp(QY2) < Ksp(PZ2)
(c) Ksp(M2X) > Ksp(QY2) = Ksp(PZ2)
(d) Ksp(M2X) = Ksp(QY2) = Ksp(PZ2)
Show Answer & Explanation

Answer: (d) Ksp(M2X) = Ksp(QY2) = Ksp(PZ2)

Explanation:
1. All three are of the AB2/A2B type, so Ksp = 4s3 in every case.
2. With equal s the Ksp are equal.
3. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 19: How many grams of CaC2O4 will dissolve in distilled water to make one litre of saturated solution? Ksp of CaC2O4 is 2.5 × 10−9 mol2 L−2 and its molecular weight is 128.
(a) 0.0064 g
(b) 0.0128 g
(c) 0.0032 g
(d) 0.0640 g
Show Answer & Explanation

Answer: (a) 0.0064 g

Explanation:
1. s = √Ksp = √(2.5 × 10−9) = 5 × 10−5 mol L−1.
2. Mass = 5 × 10−5 × 128 = 6.4 × 10−3 g.
3. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 20: If the solubility of AgCl (formula mass 143) in water at 25°C is 1.43 × 10−4 g per 100 mL of solution, the value of Ksp will be:
(a) 1 × 10−5
(b) 2 × 10−5
(c) 1 × 10−10
(d) 2 × 10−10
Show Answer & Explanation

Answer: (c) 1 × 10−10

Explanation:
1. s = 1.43 × 10−3 g L−1/143 = 10−5 mol L−1.
2. Ksp = s2 = 10−10.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 21: If the sparingly soluble salts M2X, QY2 and PZ3 have the same (very small) solubility S, their Ksp values are related as:
(a) Ksp(M2X) = Ksp(QY2) < Ksp(PZ3)
(b) Ksp(M2X) > Ksp(QY2) = Ksp(PZ3)
(c) Ksp(M2X) = Ksp(QY2) > Ksp(PZ3)
(d) Ksp(M2X) > Ksp(QY2) > Ksp(PZ3)
Show Answer & Explanation

Answer: (c) Ksp(M2X) = Ksp(QY2) > Ksp(PZ3)

Explanation:
1. Ksp of M2X = 4S3, QY2 = 4S3, PZ3 = 27S4.
2. For very small S (S < 4/27), 27S4 < 4S3, so the first two are equal and larger than the third.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 22: One litre of a saturated solution of CaCO3 is evaporated to dryness, when 7.0 g of residue is left. The solubility product of CaCO3 is:
(a) 4.9 × 10−3
(b) 4.9 × 10−5
(c) 4.9 × 10−9
(d) 4.9 × 10−7
Show Answer & Explanation

Answer: (a) 4.9 × 10−3

Explanation:
1. Molar mass of CaCO3 = 100, so s = 7.0/100 = 0.07 mol L−1.
2. Ksp = s2 = 4.9 × 10−3 (with the data as given).
3. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 23: At 30°C, in which one-litre solution will the solubility of Ag2CO3 (Ksp = 8 × 10−12) be maximum?
(a) 0.05 M Na2CO3
(b) Pure water
(c) 0.05 M AgNO3
(d) 0.05 M NH3
Show Answer & Explanation

Answer: (d) 0.05 M NH3

Explanation:
1. Na2CO3 and AgNO3 supply common ions and reduce the solubility.
2. NH3 forms the complex [Ag(NH3)2]+, removing Ag+ and increasing the solubility.
3. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 24: The solubility of AgBr will be minimum in:
(a) Pure water
(b) 0.1 M CaBr2
(c) 0.1 M NaBr
(d) 0.1 M AgNO3
Show Answer & Explanation

Answer: (b) 0.1 M CaBr2

Explanation:
1. The common-ion effect is greatest for the highest concentration of the common ion.
2. 0.1 M CaBr2 gives 0.2 M Br−, more than 0.1 M NaBr or 0.1 M Ag+.
3. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 25: If the maximum concentration of PbCl2 in water is 0.01 M at 298 K, its maximum concentration in 0.1 M NaCl will be:
(a) 4 × 10−3 M
(b) 0.4 × 10−4 M
(c) 4 × 10−2 M
(d) 4 × 10−4 M
Show Answer & Explanation

Answer: (d) 4 × 10−4 M

Explanation:
1. Ksp = 4s3 = 4 × (0.01)3 = 4 × 10−6.
2. In 0.1 M NaCl: s′ = Ksp/[Cl−]2 = 4 × 10−6/10−2 = 4 × 10−4 M.
3. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 26: If the solubility products of AgBrO3 and Ag2SO4 are 5.5 × 10−5 and 2 × 10−5 respectively, the relationship between their solubilities is:
(a) s(AgBrO3) > s(Ag2SO4)
(b) s(AgBrO3) = s(Ag2SO4)
(c) s(AgBrO3) < s(Ag2SO4)
(d) Can't predict
Show Answer & Explanation

Answer: (c) s(AgBrO3) < s(Ag2SO4)

Explanation:
1. AgBrO3: s = √(5.5 × 10−5) = 7.4 × 10−3 M.
2. Ag2SO4: s = (2 × 10−5/4)1/3 = 1.7 × 10−2 M.
3. So s(AgBrO3) < s(Ag2SO4).
4. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 27: The solubility product of Mg(OH)2 is 1 × 10−11. At what pH will precipitation of Mg(OH)2 begin from a 0.1 M Mg2+ solution?
(a) 9
(b) 5
(c) 3
(d) 7
Show Answer & Explanation

Answer: (a) 9

Explanation:
1. [OH−]2 = Ksp/[Mg2+] = 10−11/0.1 = 10−10, so [OH−] = 10−5.
2. pOH = 5 and pH = 9.
3. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 28: What is the molar concentration of chloride ion in the solution obtained by mixing 300 mL of 3.0 M NaCl and 200 mL of 4.0 M BaCl2?
(a) 5.0 M
(b) 1.8 M
(c) 1.6 M
(d) None of these
Show Answer & Explanation

Answer: (a) 5.0 M

Explanation:
1. Cl− from NaCl = 300 × 3 = 900 mmol; from BaCl2 = 200 × 4 × 2 = 1600 mmol.
2. Total 2500 mmol in 500 mL = 5.0 M.
3. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 29: The minimum pH is shown by an aqueous solution of:
(a) 0.1 M BaCl2
(b) 0.1 M Ba(NO3)2
(c) 0.1 M BeCl2
(d) 0.1 M Ba(OH)2
Show Answer & Explanation

Answer: (c) 0.1 M BeCl2

Explanation:
1. BaCl2 and Ba(NO3)2 are neutral, Ba(OH)2 is strongly basic.
2. BeCl2 hydrolyses (small, highly polarising Be2+) to give an acidic solution, so it has the lowest pH.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 30: Given (a) 0.005 M H2SO4, (b) 0.1 M Na2SO4, (c) 10−2 M NaOH, (d) 0.01 M HCl. Choose the correct code of solutions having the same pH:
(a) a, c, d
(b) b, d
(c) a, d
(d) a, c
Show Answer & Explanation

Answer: (c) a, d

Explanation:
1. (a) [H+] = 2 × 0.005 = 0.01, pH = 2.
2. (d) [H+] = 0.01, pH = 2.
3. (b) is neutral (pH 7) and (c) has pH 12.
4. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

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Multiple Choice Questions (MCQs) for JEE Chemistry Ionic Equilibrium

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