JEE Chemistry Equilibrium MCQs Set 21

Chemistry Objective Questions and Answers: Equilibrium

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

Download Equilibrium MCQs with Answers

Access the complete set of multiple-choice questions for Equilibrium below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: The molar solubility of Hg(CN)\(_2\) (\(K_{sp} = 1.35 \times 10^{-23}\)) in pure water, assuming no hydrolysis, is:
(a) \(2.3 \times 10^{-8}\) M
(b) \(1.5 \times 10^{-12}\) M
(c) \(3.7 \times 10^{-12}\) M
(d) \(1.5 \times 10^{-8}\) M
Show Answer & Explanation

Answer: (d) \(1.5 \times 10^{-8}\) M

Explanation:
1. Hg(CN)\(_2\) gives Hg\(^{2+}\) and 2CN\(^-\), so \(K_{sp} = s(2s)^2 = 4s^3\).
2. \(s^3 = 1.35 \times 10^{-23}/4 = 3.375 \times 10^{-24}\), so \(s = 1.5 \times 10^{-8}\) M.

Teacher's Note:
1. For an AB\(_2\) salt, \(K_{sp} = 4s^3\).

Question 2: The volume of pure water needed to dissolve 1.134 g of Hg(CN)\(_2\) (molar mass 252 g/mol; solubility \(1.5 \times 10^{-8}\) mol/L) is:
(a) \(3 \times 10^{11}\) mL
(b) \(3 \times 10^{8}\) mL
(c) \(3 \times 10^{5}\) mL
(d) \(3 \times 10^{6}\) mL
Show Answer & Explanation

Answer: (b) \(3 \times 10^{8}\) mL

Explanation:
1. Moles = 1.134/252 = 0.0045 mol.
2. Volume = \(0.0045/(1.5 \times 10^{-8}) = 3 \times 10^{5}\) L = \(3 \times 10^{8}\) mL.

Teacher's Note:
1. Such a salt is almost insoluble: hundreds of cubic metres of water are needed.

Question 3: The mass of Au\(^{3+}\) (atomic mass 197) per litre in a saturated solution of AuI\(_3\) (\(K_{sp} = 2.7 \times 10^{-47}\)), assuming no hydrolysis, is:
(a) \(1.97 \times 10^{-12}\) mg
(b) \(1.97 \times 10^{-4}\) mg
(c) \(5.9 \times 10^{-7}\) mg
(d) \(1.97 \times 10^{-7}\) mg
Show Answer & Explanation

Answer: (d) \(1.97 \times 10^{-7}\) mg

Explanation:
1. For AuI\(_3\), \(K_{sp} = s(3s)^3 = 27s^4\).
2. \(s^4 = 10^{-48}\), so \(s = 10^{-12}\) M.
3. Mass = \(10^{-12} \times 197 \times 1000 = 1.97 \times 10^{-7}\) mg/L.

Teacher's Note:
1. AB\(_3\) salts have \(K_{sp} = 27s^4\).

Question 4: Equimolar solutions of FeSO\(_4\) and Na\(_2\)S are mixed in equal volumes (\(K_{sp}\) of FeS = \(6.25 \times 10^{-18}\)). The maximum initial concentration of each so that no precipitate forms is:
(a) \(2.5 \times 10^{-9}\) M
(b) \(1 \times 10^{-8}\) M
(c) \(5 \times 10^{-9}\) M
(d) \(2.5 \times 10^{-18}\) M
Show Answer & Explanation

Answer: (c) \(5 \times 10^{-9}\) M

Explanation:
1. After mixing, each ion is at C/2 and no precipitation requires \((C/2)^2 \le 6.25 \times 10^{-18}\).
2. \(C/2 = 2.5 \times 10^{-9}\) M, so \(C = 5 \times 10^{-9}\) M.

Teacher's Note:
1. Mixing equal volumes halves each concentration.

Question 5: The solubility of Ag\(_3\)PO\(_4\) (\(K_{sp} = 2.7 \times 10^{-23}\), molar mass 419 g/mol) in pure water, assuming no hydrolysis, is:
(a) \(4.19 \times 10^{-2}\) mg/mL
(b) \(4.19 \times 10^{-7}\) mg/mL
(c) \(4.19 \times 10^{-4}\) mg/mL
(d) \(1.0 \times 10^{-6}\) mg/mL
Show Answer & Explanation

Answer: (c) \(4.19 \times 10^{-4}\) mg/mL

Explanation:
1. \(K_{sp} = (3s)^3 s = 27s^4\), so \(s = 10^{-6}\) M.
2. Mass = \(10^{-6} \times 419 = 4.19 \times 10^{-4}\) g/L = \(4.19 \times 10^{-4}\) mg/mL.

Teacher's Note:
1. Remember to convert g/L to mg/mL (they are numerically equal).

Question 6: The solubility of Ag\(_3\)PO\(_4\) (\(K_{sp} = 2.7 \times 10^{-23}\), molar mass 419 g/mol) in 0.03 M AgNO\(_3\) is:
(a) \(4.19 \times 10^{-16}\) mg/mL
(b) \(4.19 \times 10^{-13}\) mg/mL
(c) \(1.0 \times 10^{-6}\) mg/mL
(d) \(4.19 \times 10^{-4}\) mg/mL
Show Answer & Explanation

Answer: (a) \(4.19 \times 10^{-16}\) mg/mL

Explanation:
1. [Ag\(^+\)] is about 0.03 M, so \(s' = K_{sp}/(0.03)^3 = 2.7 \times 10^{-23}/(2.7 \times 10^{-5}) = 10^{-18}\) M.
2. Mass = \(10^{-18} \times 419\) g/L = \(4.19 \times 10^{-16}\) mg/mL.

Teacher's Note:
1. The common ion Ag\(^+\) lowers solubility enormously.

Question 7: In a saturated solution containing both AgCl (\(K_{sp} = 10^{-10}\)) and AgI (\(K_{sp} = 8.5 \times 10^{-17}\)), the concentrations of Cl\(^-\) and Ag\(^+\) are respectively:
(a) \(10^{-5}\) M and \(10^{-5}\) M
(b) \(8.5 \times 10^{-12}\) M and \(10^{-5}\) M
(c) \(10^{-5}\) M and \(8.5 \times 10^{-12}\) M
(d) \(10^{-10}\) M and \(10^{-5}\) M
Show Answer & Explanation

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

Explanation:
1. AgCl is far more soluble, so [Ag\(^+\)] is set by it: \([Ag^+] \approx [Cl^-] = \sqrt{10^{-10}} = 10^{-5}\) M.
2. \([I^-] = 8.5 \times 10^{-17}/10^{-5} = 8.5 \times 10^{-12}\) M.

Teacher's Note:
1. I\(^-\) is the one at \(8.5 \times 10^{-12}\) M, not Ag\(^+\).

Question 8: Solid Pb(NO\(_3\))\(_2\) is added gradually to a solution that is 0.1 M in NaI and 0.1 M in Na\(_2\)SO\(_4\) (\(K_{sp}\) of PbI\(_2\) = \(9 \times 10^{-9}\), of PbSO\(_4\) = \(1.8 \times 10^{-8}\)). What precipitates first and what is [SO\(_4^{2-}\)] when the second salt starts to precipitate?
(a) PbSO\(_4\); 0.02 M
(b) PbI\(_2\); 0.02 M
(c) PbI\(_2\); 0.1 M
(d) PbSO\(_4\); 0.2 M
Show Answer & Explanation

Answer: (a) PbSO\(_4\); 0.02 M

Explanation:
1. [Pb\(^{2+}\)] needed: PbSO\(_4\) = \(1.8 \times 10^{-8}/0.1 = 1.8 \times 10^{-7}\) M; PbI\(_2\) = \(9 \times 10^{-9}/0.01 = 9 \times 10^{-7}\) M.
2. The smaller value is for PbSO\(_4\), so it precipitates first.
3. When PbI\(_2\) starts, \([SO_4^{2-}] = 1.8 \times 10^{-8}/(9 \times 10^{-7}) = 0.02\) M.

Teacher's Note:
1. Compare the required [Pb\(^{2+}\)], not the \(K_{sp}\) values directly.

Question 9: 0.1 mol AgCl (\(K_{sp} = 1.75 \times 10^{-10}\)) is in 1 L water, and NaBr is added until 75% of the AgCl is converted to AgBr (\(K_{sp} = 5.25 \times 10^{-13}\)). The final [Br\(^-\)] is:
(a) \(2.25 \times 10^{-2}\) M
(b) \(2.25 \times 10^{-4}\) M
(c) \(5.25 \times 10^{-4}\) M
(d) \(7.5 \times 10^{-2}\) M
Show Answer & Explanation

Answer: (b) \(2.25 \times 10^{-4}\) M

Explanation:
1. 75% conversion releases 0.075 M Cl\(^-\) (the rest of the AgCl, 0.025 M, stays solid).
2. \([Ag^+] = 1.75 \times 10^{-10}/0.075 = 2.33 \times 10^{-9}\) M.
3. \([Br^-] = 5.25 \times 10^{-13}/(2.33 \times 10^{-9}) = 2.25 \times 10^{-4}\) M.

Teacher's Note:
1. Both solids are in equilibrium with the same [Ag\(^+\)].

Question 10: The minimum pH at which Mg(OH)\(_2\) (\(K_{sp} = 1.8 \times 10^{-11}\)) starts precipitating from 0.02 M MgCl\(_2\) is:
(a) 9.48
(b) 4.52
(c) 9.04
(d) 8.48
Show Answer & Explanation

Answer: (a) 9.48

Explanation:
1. \([OH^-] = \sqrt{1.8 \times 10^{-11}/0.02} = 3 \times 10^{-5}\) M.
2. pOH = 4.52, so pH = 9.48.

Teacher's Note:
1. Precipitation starts when the ionic product equals \(K_{sp}\).

Question 11: The solubility of each of two sparingly soluble salts with a common ion, when both are present in the same saturated solution, compared with its solubility in a separate saturated solution, is:
(a) greater in the mixture
(b) in between the two separate solubilities
(c) exactly the same
(d) less in the mixture
Show Answer & Explanation

Answer: (d) less in the mixture

Explanation:
1. The common ion from one salt suppresses the dissolution of the other.
2. So each salt dissolves to a smaller extent than it does alone.

Teacher's Note:
1. This is simply the common-ion effect acting on both salts.

Question 12: A solution is saturated with AgCl (\(K_{sp} = 1.8 \times 10^{-10}\)) and AgCNS (\(K_{sp} = 1.6 \times 10^{-11}\)). The ratio \([Cl^-]/[CNS^-]\) and \([Ag^+]\) are:
(a) 1.25 and \(4 \times 10^{-6}\) M
(b) 1.25 and \(4 \times 10^{-5}\) M
(c) 11.25 and \(1.4 \times 10^{-5}\) M
(d) 1.125 and \(4 \times 10^{-6}\) M
Show Answer & Explanation

Answer: (c) 11.25 and \(1.4 \times 10^{-5}\) M

Explanation:
1. Ratio = \(K_{sp}(AgCl)/K_{sp}(AgCNS) = 11.25\).
2. Let \([CNS^-] = c\): \([Ag^+] = 12.25c\) and \(12.25c \times 11.25c = 1.8 \times 10^{-10}\).
3. \(c = 1.14 \times 10^{-6}\) M, so \([Ag^+] = 1.4 \times 10^{-5}\) M.

Teacher's Note:
1. Same [Ag\(^+\)] for both solids means the anion ratio equals the \(K_{sp}\) ratio.

Question 13: In a solution saturated with AgCl and AgBr, their separate molar solubilities are x (AgCl) and y (AgBr), with x > y. Which is true?
(a) \(x > [Cl^-] > [Br^-]\)
(b) \([Ag^+] < [Br^-] + [Cl^-]\)
(c) \([Ag^+] = x + y\)
(d) \([Br^-] > y\)
Show Answer & Explanation

Answer: (a) \(x > [Cl^-] > [Br^-]\)

Explanation:
1. Charge balance gives \([Ag^+] = [Cl^-] + [Br^-]\), ruling out the second option.
2. Each anion is lower than its own solubility because of the common Ag\(^+\), so \([Cl^-] < x\) and \([Br^-] < y\).
3. Since AgCl is far more soluble, \([Cl^-] > [Br^-]\).

Teacher's Note:
1. Common ion effect lowers both anion concentrations.

Question 14: Equal volumes of 0.02 M AgNO\(_3\) and 0.16 M KCN are mixed (Ag\(^+\) + 2CN\(^-\) \(\rightleftharpoons\) [Ag(CN)\(_2\)]\(^-\), \(K_c = 10^{18}\)). Which is correct?
(a) [NO\(_3^-\)] = 0.01 M, [K\(^+\)] = 0.08 M, [CN\(^-\)] = 0.06 M and [Ag(CN)\(_2^-\)] = 0.01 M
(b) only [Ag(CN)\(_2^-\)] = 0.02 M is correct
(c) [CN\(^-\)] = 0.07 M and [Ag(CN)\(_2^-\)] = 0.01 M
(d) only [CN\(^-\)] = 0.06 M is correct
Show Answer & Explanation

Answer: (a) [NO\(_3^-\)] = 0.01 M, [K\(^+\)] = 0.08 M, [CN\(^-\)] = 0.06 M and [Ag(CN)\(_2^-\)] = 0.01 M

Explanation:
1. After mixing: Ag\(^+\) 0.01 M, CN\(^-\) 0.08 M, NO\(_3^-\) 0.01 M, K\(^+\) 0.08 M.
2. Ag\(^+\) is consumed, using 0.02 M CN\(^-\) and forming 0.01 M complex.
3. CN\(^-\) left = 0.06 M.

Teacher's Note:
1. Dilution on mixing halves every concentration first.

Question 15: MCl(s) is added gradually to a solution having equal concentrations of X\(^-\), Y\(^-\) and Z\(^-\), with \(K_{sp}\)(MX) > \(K_{sp}\)(MY) > \(K_{sp}\)(MZ). Which precipitates first?
(a) MX
(b) MZ
(c) cannot be said
(d) MY
Show Answer & Explanation

Answer: (b) MZ

Explanation:
1. The required [M\(^+\)] is \(K_{sp}/C\), smallest for the smallest \(K_{sp}\).
2. MZ needs the least M\(^+\), so it precipitates first.

Teacher's Note:
1. Equal anion concentrations make the \(K_{sp}\) order decide.

Question 16: In the previous situation the initial concentrations are \([X^-] : [Y^-] : [Z^-] = 100 : 10 : 1\) and \(K_{sp}\)(MX) : \(K_{sp}\)(MY) : \(K_{sp}\)(MZ) = 10 : 0.1 : 1. The order of precipitation is:
(a) MX, then MY, then MZ
(b) MY, then MX, then MZ
(c) MY, then MZ, then MX
(d) MZ, then MX, then MY
Show Answer & Explanation

Answer: (b) MY, then MX, then MZ

Explanation:
1. Required [M\(^+\)] is proportional to \(K_{sp}/C\): MX = 10/100 = 0.1; MY = 0.1/10 = 0.01; MZ = 1/1 = 1.
2. Order of increasing value: MY, MX, MZ.

Teacher's Note:
1. Always compare \(K_{sp}/C\), not \(K_{sp}\) alone.

Question 17: For \([Ag(CN)_2]^- \rightleftharpoons Ag^+ + 2CN^-\), \(K_c = 4 \times 10^{-19}\). A solution originally 0.1 M in KCN and 0.03 M in AgNO\(_3\) has [Ag\(^+\)] equal to:
(a) \(1.5 \times 10^{-17}\) M
(b) \(7.5 \times 10^{-18}\) M
(c) \(3.0 \times 10^{-18}\) M
(d) \(4.69 \times 10^{-19}\) M
Show Answer & Explanation

Answer: (b) \(7.5 \times 10^{-18}\) M

Explanation:
1. Complexation uses 0.06 M CN\(^-\) and forms 0.03 M complex, leaving 0.04 M CN\(^-\).
2. \([Ag^+] = K_c[complex]/[CN^-]^2 = 4 \times 10^{-19} \times 0.03/0.0016 = 7.5 \times 10^{-18}\) M.

Teacher's Note:
1. Use the dissociation form of the constant here.

Question 18: Na\(_2\)CO\(_3\) solution is added drop by drop to 1 L containing \(10^{-4}\) mol Ba\(^{2+}\) and \(10^{-5}\) mol Ag\(^+\) (\(K_{sp}\) BaCO\(_3\) = \(8.1 \times 10^{-9}\), Ag\(_2\)CO\(_3\) = \(6.9 \times 10^{-12}\)). Which statement is NOT true?
(a) Ag\(_2\)CO\(_3\) does not precipitate until [CO\(_3^{2-}\)] reaches \(6.9 \times 10^{-2}\) M
(b) BaCO\(_3\) does not precipitate until [CO\(_3^{2-}\)] reaches \(8.1 \times 10^{-5}\) M
(c) BaCO\(_3\) precipitates first
(d) Ag\(_2\)CO\(_3\) precipitates when [CO\(_3^{2-}\)] reaches \(6.9 \times 10^{-5}\) M
Show Answer & Explanation

Answer: (d) Ag\(_2\)CO\(_3\) precipitates when [CO\(_3^{2-}\)] reaches \(6.9 \times 10^{-5}\) M

Explanation:
1. BaCO\(_3\) needs \(8.1 \times 10^{-9}/10^{-4} = 8.1 \times 10^{-5}\) M.
2. Ag\(_2\)CO\(_3\) needs \(6.9 \times 10^{-12}/(10^{-5})^2 = 6.9 \times 10^{-2}\) M, not \(6.9 \times 10^{-5}\).
3. So the first statement is false, the others are true.

Teacher's Note:
1. Square the Ag\(^+\) concentration for Ag\(_2\)CO\(_3\).

Question 19: Pure water is saturated with CaCO\(_3\) and CaC\(_2\)O\(_4\), giving [Ca\(^{2+}\)] = \(8 \times 10^{-5}\) M. If \(K_{sp}\)(CaCO\(_3\)) : \(K_{sp}\)(CaC\(_2\)O\(_4\)) = 3, the \(K_{sp}\) of CaCO\(_3\) is:
(a) \(6.4 \times 10^{-9}\)
(b) \(4.8 \times 10^{-9}\)
(c) \(3.2 \times 10^{-9}\)
(d) \(2.4 \times 10^{-9}\)
Show Answer & Explanation

Answer: (b) \(4.8 \times 10^{-9}\)

Explanation:
1. \([Ca^{2+}] = [CO_3^{2-}] + [C_2O_4^{2-}] = (K_1 + K_2)/[Ca^{2+}]\).
2. \([Ca^{2+}]^2 = K_1 + K_1/3 = 4K_1/3\).
3. \(K_1 = 0.75 \times 6.4 \times 10^{-9} = 4.8 \times 10^{-9}\).

Teacher's Note:
1. Charge balance links the three concentrations.

Question 20: Equal volumes of 0.1 M AgNO\(_3\) and 0.8 M KCN are mixed (\(K_d\) of [Ag(CN)\(_2\)]\(^-\) = \(10^{-18}\)). The percentage of the original Ag\(^+\) present as free Ag\(^+\) is:
(a) \(5.6 \times 10^{-19}\)%
(b) \(1.11 \times 10^{-17}\)%
(c) \(2.2 \times 10^{-15}\)%
(d) \(1.11 \times 10^{-15}\)%
Show Answer & Explanation

Answer: (d) \(1.11 \times 10^{-15}\)%

Explanation:
1. After mixing: Ag\(^+\) 0.05 M, CN\(^-\) 0.4 M; complex 0.05 M and CN\(^-\) 0.3 M left.
2. \([Ag^+] = 10^{-18} \times 0.05/0.09 = 5.56 \times 10^{-19}\) M.
3. Percentage = \(5.56 \times 10^{-19}/0.05 \times 100 = 1.11 \times 10^{-15}\)%.

Teacher's Note:
1. Do not forget to convert the fraction into a percentage.

Question 21: 0.2 mol of MgCl\(_2\) is added to 1 L of a solution already containing 0.2 mol NaOH (\(K_{sp}\) of Mg(OH)\(_2\) = \(1.6 \times 10^{-12}\); volume change ignored). Then 0.04 mol of NaOH is added to the solution. The new pH is:
(a) 8.65
(b) 8.5
(c) 8.6
(d) 8.56
Show Answer & Explanation

Answer: (a) 8.65

Explanation:
1. 0.04 mol OH\(^-\) precipitates 0.02 mol more Mg(OH)\(_2\), so \([Mg^{2+}] = 0.08\) M.
2. \([OH^-] = \sqrt{1.6 \times 10^{-12}/0.08} = 4.47 \times 10^{-6}\) M.
3. pOH = 5.35, pH = 8.65.

Teacher's Note:
1. Added base lowers [Mg\(^{2+}\)] and raises [OH\(^-\)] only slightly.

Question 22: 0.2 mol of MgCl\(_2\) is added to 1 L of a solution already containing 0.2 mol NaOH (\(K_{sp}\) of Mg(OH)\(_2\) = \(1.6 \times 10^{-12}\); volume change ignored). Then 0.04 mol of HCl gas is added to the original solution (the one before NaOH was added). The pH becomes:
(a) 8.56
(b) 8.65
(c) 8.5
(d) 8.6
Show Answer & Explanation

Answer: (a) 8.56

Explanation:
1. 0.04 mol H\(^+\) dissolves 0.02 mol Mg(OH)\(_2\), so \([Mg^{2+}] = 0.12\) M.
2. \([OH^-] = \sqrt{1.6 \times 10^{-12}/0.12} = 3.65 \times 10^{-6}\) M.
3. pOH = 5.44, pH = 8.56.

Teacher's Note:
1. Acid dissolves some of the solid and raises [Mg\(^{2+}\)].

Question 23: For a sparingly soluble salt A\(_p\)B\(_q\) of solubility S, the solubility product is:
(a) \(S^{pq} (pq)^{p+q}\)
(b) \(S^{p+q} p^p q^q\)
(c) \(S^{pq} p^p q^q\)
(d) \(S^{p+q} p^q q^p\)
Show Answer & Explanation

Answer: (b) \(S^{p+q} p^p q^q\)

Explanation:
1. A\(_p\)B\(_q\) gives \(p\) cations at concentration \(pS\) and \(q\) anions at \(qS\).
2. \(K_{sp} = (pS)^p (qS)^q = S^{p+q} p^p q^q\).

Teacher's Note:
1. Each ion's concentration is raised to its own stoichiometric coefficient.

Question 24: A solution is \(10^{-3}\) M each in Mn\(^{2+}\), Fe\(^{2+}\), Zn\(^{2+}\) and Hg\(^{2+}\) and is treated with \(10^{-16}\) M sulphide ion (\(K_{sp}\): MnS \(10^{-15}\), FeS \(10^{-23}\), ZnS \(10^{-20}\), HgS \(10^{-54}\)). Which precipitates first?
(a) MnS
(b) ZnS
(c) FeS
(d) HgS
Show Answer & Explanation

Answer: (d) HgS

Explanation:
1. The sulphide needed is \(K_{sp}/10^{-3}\): MnS \(10^{-12}\), FeS \(10^{-20}\), ZnS \(10^{-17}\), HgS \(10^{-51}\) M.
2. HgS needs by far the least sulphide, so it precipitates first.

Teacher's Note:
1. Smallest required [S\(^{2-}\)] precipitates first.

Question 25: The solubility of Ag\(_2\)CrO\(_4\) (\(K_{sp} = 1.1 \times 10^{-12}\)) in 0.1 M AgNO\(_3\) is:
(a) \(1.1 \times 10^{-9}\) mol/L
(b) \(1.1 \times 10^{-10}\) mol/L
(c) \(1.1 \times 10^{-11}\) mol/L
(d) \(1.1 \times 10^{-12}\) mol/L
Show Answer & Explanation

Answer: (b) \(1.1 \times 10^{-10}\) mol/L

Explanation:
1. \(K_{sp} = (0.1)^2 s\), taking [Ag\(^+\)] = 0.1 M.
2. \(s = 1.1 \times 10^{-12}/10^{-2} = 1.1 \times 10^{-10}\) mol/L.

Teacher's Note:
1. The Ag\(^+\) from the dissolving salt is negligible.

Question 26: The solubility of the salt AB (\(K_{sp} = 2 \times 10^{-10}\)) of a weak acid HB (\(K_a = 10^{-8}\)) at pH 3 is \(Y \times 10^{-3}\) mol/L. Y is nearest to:
(a) 2.00
(b) 0.45
(c) 1.41
(d) 4.47
Show Answer & Explanation

Answer: (d) 4.47

Explanation:
1. B\(^-\) is protonated: \([HB]/[B^-] = [H^+]/K_a = 10^{-3}/10^{-8} = 10^5\).
2. Total B = \(S = [B^-](1 + 10^5)\) and \([A^+] = S\).
3. \(K_{sp} = S^2/(1 + 10^5)\), so \(S = \sqrt{2 \times 10^{-10} \times 100001} = 4.47 \times 10^{-3}\) mol/L.

Teacher's Note:
1. Acidic pH greatly raises the solubility of salts of weak acids.

Question 27: Which of the following is not a Lewis acid?
(a) Zn\(^{2+}\)
(b) SO\(_3\)
(c) SiCl\(_4\)
(d) CO
Show Answer & Explanation

Answer: (d) CO

Explanation:
1. SiCl\(_4\) can expand its octet using d orbitals, SO\(_3\) has an electron-deficient sulphur and Zn\(^{2+}\) has vacant orbitals.
2. CO has no vacant orbital, and the lone pair on carbon makes it a Lewis base.

Teacher's Note:
1. A Lewis acid must be able to accept an electron pair.

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