JEE Chemistry Equilibrium MCQs Set 16

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Chapter-wise Objective Questions: Equilibrium

Navigate directly to the 50 objective questions for 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: The solubility product of sodium chloride is 36 \( mol^2\,L^{-2} \). The solubility of sodium chloride is
(a) \( 1/36 \)
(b) \( 1/6 \)
(c) 6
(d) 3600
Show Answer & Explanation

Answer: (c) 6

Explanation:
1. For NaCl, \( K_{sp}=s^2 \), so \( s=\sqrt{36}=6 \) mol/L.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 2: If the concentration of lead iodide in its saturated solution at \( 25^\circ C \) is \( 2\times10^{-3} \) mol/L, its solubility product is
(a) \( 4\times10^{-6} \)
(b) \( 8\times10^{-12} \)
(c) \( 6\times10^{-9} \)
(d) \( 32\times10^{-9} \)
Show Answer & Explanation

Answer: (d) \( 32\times10^{-9} \)

Explanation:
1. \( \ce{PbI2} \) gives \( \ce{Pb^2+} \) = s and \( \ce{I-} \) = 2s.
2. \( K_{sp}=s(2s)^2=4s^3=4\times(2\times10^{-3})^3=3.2\times10^{-8}=32\times10^{-9} \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 3: A precipitate of \( \ce{CaF2} \) (\( K_{sp}=1.7\times10^{-10} \)) is obtained when equal volumes of the following are mixed
(a) \( 10^{-4}\ M\ Ca^{2+}+10^{-4}\ M\ F^- \)
(b) \( 10^{-2}\ M\ Ca^{2+}+10^{-3}\ M\ F^- \)
(c) both
(d) none of these
Show Answer & Explanation

Answer: (b) \( 10^{-2}\ M\ Ca^{2+}+10^{-3}\ M\ F^- \)

Explanation:
1. Mixing equal volumes halves each concentration.
2. (a): \( Q=5\times10^{-5}\times(5\times10^{-5})^2=1.25\times10^{-13}3. (b): \( Q=5\times10^{-3}\times(5\times10^{-4})^2=1.25\times10^{-9}>K_{sp} \), precipitate forms.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 4: What is the minimum concentration of \( \ce{SO4^2-} \) required to precipitate \( \ce{BaSO4} \) in a solution containing \( 1.0\times10^{-4} \) mol of \( \ce{Ba^2+} \) per litre? (\( K_{sp} \) of \( \ce{BaSO4} \) = \( 4\times10^{-10} \))
(a) \( 4\times10^{-10}\ M \)
(b) \( 2\times10^{-7}\ M \)
(c) \( 4\times10^{-6}\ M \)
(d) \( 2\times10^{-3}\ M \)
Show Answer & Explanation

Answer: (c) \( 4\times10^{-6}\ M \)

Explanation:
1. Precipitation starts when \( [Ba^{2+}][SO_4^{2-}]=K_{sp} \).
2. \( [SO_4^{2-}]=4\times10^{-10}/10^{-4}=4\times10^{-6}\ M \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 5: The solubility product of salt \( \ce{AB2} \) is \( 4\times10^{-12} \). Its solubility is
(a) \( 1\times10^{-3} \) mol/L
(b) \( 1\times10^{-5} \) mol/L
(c) \( 1\times10^{-4} \) mol/L
(d) \( 1\times10^{-2} \) mol/L
Show Answer & Explanation

Answer: (c) \( 1\times10^{-4} \) mol/L

Explanation:
1. \( K_{sp}=4s^3 \), so \( s^3=10^{-12} \) and \( s=10^{-4} \) mol/L.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 6: The solubility product of a salt AB is \( 1\times10^{-8} \). In a solution in which the concentration of A is \( 10^{-3} \) M, the salt will precipitate when the concentration of B becomes more than
(a) \( 10^{-4}\ M \)
(b) \( 10^{-7}\ M \)
(c) \( 10^{-6}\ M \)
(d) \( 10^{-5}\ M \)
Show Answer & Explanation

Answer: (d) \( 10^{-5}\ M \)

Explanation:
1. Precipitation needs \( [A][B]>K_{sp} \).
2. \( [B]>10^{-8}/10^{-3}=10^{-5}\ M \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 7: The solubility product of \( \ce{BaSO4} \) is \( 1.3\times10^{-9} \). The solubility of this salt in pure water is
(a) \( 1.69\times10^{-9} \) mol/L
(b) \( 1.69\times10^{-18} \) mol/L
(c) \( 3.6\times10^{-18} \) mol/L
(d) \( 3.6\times10^{-5} \) mol/L
Show Answer & Explanation

Answer: (d) \( 3.6\times10^{-5} \) mol/L

Explanation:
1. \( s=\sqrt{K_{sp}}=\sqrt{1.3\times10^{-9}}=3.6\times10^{-5} \) mol/L.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 8: The solubility of an \( \ce{MX2} \) type electrolyte is \( 0.5\times10^{-4} \) mol/L. The value of \( K_{sp} \) of the electrolyte is
(a) \( 5\times10^{-13} \)
(b) \( 25\times10^{-10} \)
(c) \( 1.25\times10^{-13} \)
(d) \( 5\times10^{12} \)
Show Answer & Explanation

Answer: (a) \( 5\times10^{-13} \)

Explanation:
1. \( K_{sp}=4s^3=4(0.5\times10^{-4})^3=4\times1.25\times10^{-13}=5\times10^{-13} \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 9: According to the reaction \( \ce{PbCl2 <=> Pb^2+ + 2Cl-} \), the solubility product expression is
(a) \( [Pb^{2+}][Cl^-]^2 \)
(b) \( [Pb^{2+}][Cl^-] \)
(c) \( [Pb^{2+}]^2[Cl^-] \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( [Pb^{2+}][Cl^-]^2 \)

Explanation:
1. \( K_{sp}=[Pb^{2+}][Cl^-]^2 \) because the concentration of each ion is raised to its coefficient.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 10: \( K_{sp} \) values of \( \ce{Al(OH)3} \) and \( \ce{Zn(OH)2} \) are \( 8.5\times10^{-23} \) and \( 1.8\times10^{-14} \) respectively. If \( \ce{NH4OH} \) is added to a solution of \( \ce{Al^3+} \) and \( \ce{Zn^2+} \) (equal concentrations), which will precipitate earlier?
(a) \( \ce{Al(OH)3} \)
(b) \( \ce{Zn(OH)2} \)
(c) both together
(d) none
Show Answer & Explanation

Answer: (a) \( \ce{Al(OH)3} \)

Explanation:
1. The salt with the much smaller Ksp reaches its ionic product at a far lower hydroxide concentration.
2. So \( \ce{Al(OH)3} \) precipitates first.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 11: Pure NaCl is precipitated when HCl gas is passed through a saturated solution of NaCl because
(a) impurities dissolve in HCl
(b) the value of \( [Na^+][Cl^-] \) becomes smaller than \( K_{sp} \) of NaCl
(c) the value of \( [Na^+][Cl^-] \) becomes greater than \( K_{sp} \) of NaCl
(d) HCl dissolves in the water
Show Answer & Explanation

Answer: (c) the value of \( [Na^+][Cl^-] \) becomes greater than \( K_{sp} \) of NaCl

Explanation:
1. HCl adds \( \ce{Cl-} \) ions (common ion effect).
2. The ionic product exceeds Ksp, so pure NaCl crystallises.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 12: The solubility of a salt \( \ce{M2X3} \) is y mol \( dm^{-3} \). Its solubility product is
(a) \( 6y^4 \)
(b) \( 64y^4 \)
(c) \( 36y^5 \)
(d) \( 108y^5 \)
Show Answer & Explanation

Answer: (d) \( 108y^5 \)

Explanation:
1. Dissolution gives \( 2y \) of \( \ce{M^3+} \) and \( 3y \) of \( \ce{X^2-} \).
2. \( K_{sp}=(2y)^2(3y)^3=4\cdot27\,y^5=108y^5 \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 13: Which of the following is most soluble?
(a) \( \ce{CuS} \) (\( K_{sp}=8\times10^{-37} \))
(b) \( \ce{MnS} \) (\( K_{sp}=7\times10^{-16} \))
(c) \( \ce{Bi2S3} \) (\( K_{sp}=1\times10^{-70} \))
(d) \( \ce{Ag2S} \) (\( K_{sp}=6\times10^{-51} \))
Show Answer & Explanation

Answer: (b) \( \ce{MnS} \) (\( K_{sp}=7\times10^{-16} \))

Explanation:
1. Compare solubilities: CuS and MnS (same type) have \( s=\sqrt{K_{sp}} \), giving \( 9\times10^{-19} \) and \( 2.6\times10^{-8} \).
2. The other salts have extremely small Ksp, so MnS is the most soluble.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 14: The solubility product of \( \ce{PbCl2} \) at \( 20^\circ C \) is \( 1.5\times10^{-4} \). The solubility is
(a) \( 3.75\times10^{-4} \)
(b) \( 3.34\times10^{-2} \)
(c) \( 3.34\times10^{2} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( 3.34\times10^{-2} \)

Explanation:
1. \( s=(K_{sp}/4)^{1/3}=(3.75\times10^{-5})^{1/3}=3.34\times10^{-2} \) mol/L.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 15: The pH of a buffer made from 25 mL of 1 M \( \ce{CH3COONa} \) and 25 mL of 1 M \( \ce{CH3COOH} \) will be appreciably affected by 5 mL of
(a) 1 M \( \ce{CH3COOH} \)
(b) 5 M \( \ce{CH3COOH} \)
(c) 5 M HCl
(d) 1 M \( \ce{NH4OH} \)
Show Answer & Explanation

Answer: (c) 5 M HCl

Explanation:
1. A buffer resists small additions of acid or base.
2. 5 mL of 5 M HCl supplies 25 mmol of strong acid, equal to the whole acetate in the buffer, so it destroys the buffer capacity.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 16: Which of the following sulphides has the lowest solubility product?
(a) \( \ce{FeS} \)
(b) \( \ce{MnS} \)
(c) \( \ce{PbS} \)
(d) \( \ce{ZnS} \)
Show Answer & Explanation

Answer: (c) \( \ce{PbS} \)

Explanation:
1. Group II sulphides such as PbS have extremely small Ksp (about \( 10^{-28} \)), much lower than the group IV sulphides FeS, MnS and ZnS.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 17: The compound insoluble in acetic acid is
(a) calcium oxide
(b) calcium carbonate
(c) calcium oxalate
(d) calcium hydroxide
Show Answer & Explanation

Answer: (c) calcium oxalate

Explanation:
1. CaO, \( \ce{CaCO3} \) and \( \ce{Ca(OH)2} \) react with acetic acid.
2. Calcium oxalate is the salt of a stronger acid (oxalic) and does not dissolve in acetic acid.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 18: The solubility product of \( \ce{Ag2SO4} \) is obtained from its solubility of \( 2.5\times10^{-2} \) M. Its value is
(a) \( 62.5\times10^{-6} \)
(b) \( 6.25\times10^{-4} \)
(c) \( 15.625\times10^{-6} \)
(d) \( 3.125\times10^{-6} \)
Show Answer & Explanation

Answer: (a) \( 62.5\times10^{-6} \)

Explanation:
1. \( K_{sp}=4s^3=4(2.5\times10^{-2})^3=4\times1.5625\times10^{-5}=6.25\times10^{-5}=62.5\times10^{-6} \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 19: The solubility product of AgCl is \( 1\times10^{-6} \) at 298 K. Its solubility in mol/L would be
(a) \( 1\times10^{-6} \)
(b) \( 1\times10^{-3} \)
(c) \( 1\times10^{-12} \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( 1\times10^{-3} \)

Explanation:
1. \( s=\sqrt{10^{-6}}=10^{-3} \) mol/L.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 20: A litre of solution is saturated with AgCl. To this solution, if \( 1.0\times10^{-4} \) mole of solid NaCl is added, the concentration of \( \ce{Ag+} \) will be
(a) more
(b) less
(c) equal
(d) zero
Show Answer & Explanation

Answer: (b) less

Explanation:
1. Added \( \ce{Cl-} \) shifts \( \ce{AgCl(s) <=> Ag+ + Cl-} \) to the left (common ion effect), so \( \ce{Ag+} \) falls.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 21: The solubility product of a sparingly soluble salt \( \ce{AX2} \) is \( 3.2\times10^{-11} \). Its solubility in moles per litre is
(a) \( 2\times10^{-4} \)
(b) \( 4\times10^{-4} \)
(c) \( 5.6\times10^{-6} \)
(d) \( 3.1\times10^{-4} \)
Show Answer & Explanation

Answer: (a) \( 2\times10^{-4} \)

Explanation:
1. \( s=(K_{sp}/4)^{1/3}=(8\times10^{-12})^{1/3}=2\times10^{-4} \) mol/L.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 22: The solubility of \( \ce{Sb2S3} \) in water is \( 1.0\times10^{-5} \) mol/L at 298 K. Its solubility product will be
(a) \( 108\times10^{-25} \)
(b) \( 1.0\times10^{-25} \)
(c) \( 144\times10^{-25} \)
(d) \( 126\times10^{-24} \)
Show Answer & Explanation

Answer: (a) \( 108\times10^{-25} \)

Explanation:
1. \( K_{sp}=(2s)^2(3s)^3=108s^5=108\times10^{-25} \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 23: Which of the following aqueous solutions has a pH less than 7?
(a) \( \ce{KNO3} \)
(b) \( \ce{NaOH} \)
(c) \( \ce{FeCl3} \)
(d) \( \ce{NaCN} \)
Show Answer & Explanation

Answer: (c) \( \ce{FeCl3} \)

Explanation:
1. \( \ce{Fe^3+} \) is hydrolysed, producing \( \ce{H+} \), so \( \ce{FeCl3} \) is acidic.
2. KNO3 is neutral; NaOH and NaCN are basic.

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 24: Which salt will give a basic solution on hydrolysis?
(a) \( \ce{KCN} \)
(b) \( \ce{KCl} \)
(c) \( \ce{NH4Cl} \)
(d) \( \ce{CH3COONH4} \)
Show Answer & Explanation

Answer: (a) \( \ce{KCN} \)

Explanation:
1. \( \ce{CN-} \) is the conjugate base of the weak acid HCN and takes protons from water, giving \( \ce{OH-} \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

Question 25: The hydrolysis constant for a salt of a weak acid and a weak base is
(a) \( K_h=K_w/K_a \)
(b) \( K_h=K_w/K_b \)
(c) \( K_h=K_w/(K_aK_b) \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( K_h=K_w/(K_aK_b) \)

Explanation:
1. Both ions hydrolyse, so \( K_h=\frac{K_w}{K_aK_b} \).

Teacher's Note:
1. Write the relevant expression (Ksp, Kh, pH or Henderson equation) and substitute the data carefully.
2. Check the final value against the nature of the solution (acidic, neutral or basic).

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Practice MCQs for JEE Chemistry Equilibrium

JEE Chemistry Equilibrium Objective Test Questions

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FAQs

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