JEE Chemistry Ionic Equilibrium MCQs Set 28

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Multiple Choice Questions

Question 1: The solubility product expression for Hg\(_2\)SO\(_4\) is:
(a) \([Hg_2^{2+}][SO_4^{2-}]\)
(b) \([Hg^{2+}]^2[SO_4^{2-}]\)
(c) \([Hg_2^{2+}]^2[SO_4^{2-}]\)
(d) \([Hg^{+}]^2[SO_4^{2-}]\)
Show Answer & Explanation

Answer: (a) \([Hg_2^{2+}][SO_4^{2-}]\)

Explanation:
1. Mercury(I) exists as the dimer Hg\(_2^{2+}\), so the salt dissociates into Hg\(_2^{2+}\) and SO\(_4^{2-}\).

Teacher's Note:
1. Remember Hg\(_2^{2+}\).

Question 2: The solubility of PbI\(_2\) (\(K_{sp} = 8 \times 10^{-9}\)) in 0.1 M Pb(NO\(_3\))\(_2\) is:
(a) \(8 \times 10^{-7}\) M
(b) \(1.414 \times 10^{-4}\) M
(c) \(1.26 \times 10^{-3}\) M
(d) \(2 \times 10^{-3}\) M
Show Answer & Explanation

Answer: (b) \(1.414 \times 10^{-4}\) M

Explanation:
1. \(0.1 \times (2s)^2 = 8 \times 10^{-9}\), so \(s^2 = 2 \times 10^{-8}\).
2. \(s = 1.414 \times 10^{-4}\) M.

Teacher's Note:
1. Pb\(^{2+}\) is the common ion.

Question 3: The order of solubility of Ag\(_2\)CO\(_3\) (\(K_{sp} = 4.3 \times 10^{-13}\)) in (a) 0.01 M AgNO\(_3\), (b) 0.04 M K\(_2\)CO\(_3\), (c) pure water, (d) a buffer of pH 4 is:
(a) c > d > b > a
(b) d > b > c > a
(c) d > c > b > a
(d) a > b > c > d
Show Answer & Explanation

Answer: (c) d > c > b > a

Explanation:
1. (a) \(s = 4.3 \times 10^{-13}/10^{-4} = 4.3 \times 10^{-9}\).
2. (b) \([Ag^+] = 1.04 \times 10^{-6}\), \(s = 5.2 \times 10^{-7}\).
3. (c) \(s = 4.8 \times 10^{-5}\).
4. (d) the acid buffer protonates CO\(_3^{2-}\), so the solubility is the highest.

Teacher's Note:
1. Common ions reduce and acid increases the solubility of a carbonate.

Question 4: M(OH)\(_x\) (giving M\(^{x+}\) and OH\(^-\)) has \(K_{sp} = 4 \times 10^{-12}\) and solubility \(10^{-4}\) M. The value of x is:
(a) 1
(b) 2
(c) 4
(d) 3
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. \(K_{sp} = s(xs)^x\).
2. For x = 2: \(10^{-4} \times (2 \times 10^{-4})^2 = 4 \times 10^{-12}\), which matches.

Teacher's Note:
1. Try the possible values of x.

Question 5: Which of MnS (\(K_{sp} = 2.5 \times 10^{-13}\)), ZnS (\(1.6 \times 10^{-24}\)), Bi\(_2\)S\(_3\) (\(1.6 \times 10^{-72}\)) and Ag\(_2\)S (\(10^{-51}\)) is the most soluble in water (no reactions of ions)?
(a) Ag\(_2\)S
(b) Bi\(_2\)S\(_3\)
(c) ZnS
(d) MnS
Show Answer & Explanation

Answer: (d) MnS

Explanation:
1. Solubilities: MnS \(5 \times 10^{-7}\); ZnS \(1.3 \times 10^{-12}\); Bi\(_2\)S\(_3\) about \(10^{-15}\); Ag\(_2\)S about \(6 \times 10^{-18}\).
2. MnS is clearly the largest.

Teacher's Note:
1. Compute each solubility with the right formula type.

Question 6: When different types of salts have nearly the same \(K_{sp}\) (less than one), the most soluble is the one that:
(a) produces the maximum number of ions per formula unit
(b) produces ions of maximum charge
(c) produces ions of minimum charge
(d) produces the minimum number of ions
Show Answer & Explanation

Answer: (a) produces the maximum number of ions per formula unit

Explanation:
1. For \(K_{sp} < 1\), solubility is \(K^{1/n}\) divided by a constant that grows with n, so salts with more ions per formula unit have larger solubility (for example AB\(_3\) over AB).

Teacher's Note:
1. Compare \(\sqrt{K}\), \((K/4)^{1/3}\) and \((K/27)^{1/4}\).

Question 7: In a saturated solution of Ag\(_2\)SO\(_4\), [Ag\(^+\)] is \(3 \times 10^{-2}\) M. Its \(K_{sp}\) is:
(a) \(4.5 \times 10^{-4}\)
(b) \(1.35 \times 10^{-5}\)
(c) \(2.7 \times 10^{-5}\)
(d) \(1.08 \times 10^{-4}\)
Show Answer & Explanation

Answer: (b) \(1.35 \times 10^{-5}\)

Explanation:
1. \([SO_4^{2-}] = 1.5 \times 10^{-2}\) M.
2. \(K_{sp} = (3 \times 10^{-2})^2 \times 1.5 \times 10^{-2} = 1.35 \times 10^{-5}\).

Teacher's Note:
1. [SO\(_4^{2-}\)] is half of [Ag\(^+\)].

Question 8: The minimum volume of water needed to dissolve 1 g of BaSO\(_4\) (\(K_{sp} = 10^{-10}\), molar mass 233) is about:
(a) 430 L
(b) 105 L
(c) 43 L
(d) 4300 L
Show Answer & Explanation

Answer: (a) 430 L

Explanation:
1. \(s = 10^{-5}\) M.
2. 1 g = \(4.29 \times 10^{-3}\) mol, so volume = \(4.29 \times 10^{-3}/10^{-5} = 429\) L.

Teacher's Note:
1. Volume = moles / solubility.

Question 9: Precipitation of AgCl (\(K_{sp} = 1.8 \times 10^{-10}\)) occurs only when equal volumes of which solutions are mixed?
(a) \(2 \times 10^{-5}\) M of each
(b) \(10^{-7}\) M of each
(c) \(10^{-5}\) M of each
(d) \(10^{-4}\) M Ag\(^+\) and \(10^{-4}\) M Cl\(^-\)
Show Answer & Explanation

Answer: (d) \(10^{-4}\) M Ag\(^+\) and \(10^{-4}\) M Cl\(^-\)

Explanation:
1. After mixing, concentrations halve.
2. For \(10^{-4}\) M: \((5 \times 10^{-5})^2 = 2.5 \times 10^{-9} > K_{sp}\).
3. For \(2 \times 10^{-5}\) M: \((10^{-5})^2 = 10^{-10} < K_{sp}\), so no precipitation.

Teacher's Note:
1. Halve the concentrations on mixing.

Question 10: The solubilities of AgCl in pure water, 0.01 M CaCl\(_2\), 0.01 M NaCl and 0.05 M AgNO\(_3\) are s\(_1\), s\(_2\), s\(_3\) and s\(_4\). The correct order is:
(a) s\(_4\) > s\(_2\) > s\(_3\) > s\(_1\)
(b) s\(_1\) > s\(_4\) > s\(_3\) > s\(_2\)
(c) s\(_1\) > s\(_2\) = s\(_3\) > s\(_4\)
(d) s\(_1\) > s\(_3\) > s\(_2\) > s\(_4\)
Show Answer & Explanation

Answer: (d) s\(_1\) > s\(_3\) > s\(_2\) > s\(_4\)

Explanation:
1. Take \(K_{sp} = 1.8 \times 10^{-10}\): s\(_1\) = \(1.3 \times 10^{-5}\); in 0.01 M NaCl s\(_3\) = \(1.8 \times 10^{-8}\); in 0.01 M CaCl\(_2\) (0.02 M Cl\(^-\)) s\(_2\) = \(9 \times 10^{-9}\); in 0.05 M AgNO\(_3\) s\(_4\) = \(3.6 \times 10^{-9}\).

Teacher's Note:
1. Higher common-ion concentration gives lower solubility.

Question 11: The solubility of BaF\(_2\) in a solution of Ba(NO\(_3\))\(_2\) is represented by which concentration term (no reaction of ions)?
(a) \([Ba^{2+}]\)
(b) \([F^-]/2\)
(c) \(2[F^-]\)
(d) \([F^-]\)
Show Answer & Explanation

Answer: (b) \([F^-]/2\)

Explanation:
1. All the F\(^-\) comes from BaF\(_2\), with two ions per formula unit.
2. So solubility = \([F^-]/2\); [Ba\(^{2+}\)] includes the Ba from the nitrate.

Teacher's Note:
1. Only the salt of interest contributes the other ion.

Question 12: The \(K_{sp}\) expressions in terms of solubility s are: Hg\(_2\)I\(_2\) = ?, Cr(OH)\(_3\) = ?
(a) \(27s^4\) and \(4s^3\)
(b) \(4s^3\) and \(27s^4\)
(c) \(4s^3\) and \(16s^4\)
(d) \(108s^5\) and \(4s^3\)
Show Answer & Explanation

Answer: (b) \(4s^3\) and \(27s^4\)

Explanation:
1. Hg\(_2\)I\(_2\) gives Hg\(_2^{2+}\) and 2I\(^-\): \(s(2s)^2 = 4s^3\).
2. Cr(OH)\(_3\) gives Cr\(^{3+}\) and 3OH\(^-\): \(s(3s)^3 = 27s^4\).

Teacher's Note:
1. Count ions and apply the exponents.

Question 13: The maximum pH of Ca(OH)\(_2\) solution (\(K_{sp} = 5.324 \times 10^{-6}\), \(\log 11 = 1.04\)) is:
(a) 14
(b) 10.68
(c) 12.04
(d) 12.34
Show Answer & Explanation

Answer: (d) 12.34

Explanation:
1. \(4s^3 = 5.324 \times 10^{-6}\) gives \(s = 0.011\) M.
2. \([OH^-] = 0.022\) M, so pOH = 1.66 and pH = 12.34.

Teacher's Note:
1. Ca(OH)\(_2\) gives two OH\(^-\).

Question 14: The solubility of Ag\(_2\)CO\(_3\) in water is \(1.26 \times 10^{-4}\) mol/L. Its solubility in 0.02 M Na\(_2\)CO\(_3\) is (take \(\sqrt[3]{2} = 1.26\)):
(a) \(10^{-5}\) M
(b) \(5 \times 10^{-5}\) M
(c) \(5 \times 10^{-6}\) M
(d) \(2 \times 10^{-5}\) M
Show Answer & Explanation

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

Explanation:
1. \(K_{sp} = 4s^3 = 4 \times 2 \times 10^{-12} = 8 \times 10^{-12}\).
2. In 0.02 M CO\(_3^{2-}\): \([Ag^+]^2 = 8 \times 10^{-12}/0.02 = 4 \times 10^{-10}\), so \([Ag^+] = 2 \times 10^{-5}\) M.
3. Solubility = \([Ag^+]/2 = 10^{-5}\) M.

Teacher's Note:
1. Solubility is half of [Ag\(^+\)].

Question 15: The solubility product of Mg(OH)\(_2\) with solubility s mol/L is:
(a) \(3s^4\)
(b) \(4s^2\)
(c) \(4s^3\)
(d) \(s^3\)
Show Answer & Explanation

Answer: (c) \(4s^3\)

Explanation:
1. \(K_{sp} = s(2s)^2 = 4s^3\).

Teacher's Note:
1. Standard AB\(_2\) result.

Question 16: The solubility in water of a sparingly soluble salt AB\(_2\) is \(1.0 \times 10^{-5}\) mol/L. Its solubility product is:
(a) \(4 \times 10^{-10}\)
(b) \(4 \times 10^{-15}\)
(c) \(1 \times 10^{-10}\)
(d) \(1 \times 10^{-15}\)
Show Answer & Explanation

Answer: (b) \(4 \times 10^{-15}\)

Explanation:
1. \(K_{sp} = 4s^3 = 4 \times 10^{-15}\).

Teacher's Note:
1. AB\(_2\): \(4s^3\).

Question 17: The molar solubility of a sparingly soluble salt MX\(_4\) is s. The relation between s and \(K_{sp}\) is:
(a) \(s = (128K_{sp})^{1/4}\)
(b) \(s = (K_{sp}/128)^{1/4}\)
(c) \(s = (K_{sp}/256)^{1/5}\)
(d) \(s = (256K_{sp})^{1/5}\)
Show Answer & Explanation

Answer: (c) \(s = (K_{sp}/256)^{1/5}\)

Explanation:
1. \(K_{sp} = s(4s)^4 = 256s^5\), so \(s = (K_{sp}/256)^{1/5}\).

Teacher's Note:
1. Count 1 + 4 = 5 ions.

Question 18: The solubility product of MX\(_2\) is \(4 \times 10^{-12}\). The concentration of M\(^{2+}\) in the saturated solution is:
(a) \(2.0 \times 10^{-6}\) M
(b) \(1.0 \times 10^{-4}\) M
(c) \(1.6 \times 10^{-4}\) M
(d) \(4.0 \times 10^{-10}\) M
Show Answer & Explanation

Answer: (b) \(1.0 \times 10^{-4}\) M

Explanation:
1. \(4s^3 = 4 \times 10^{-12}\) gives \(s = 10^{-4}\) M, and [M\(^{2+}\)] = s.

Teacher's Note:
1. Cation concentration equals the solubility.

Question 19: The \(K_{sp}\) of AgIO\(_3\) (molar mass 283) is \(1.0 \times 10^{-8}\). The mass of AgIO\(_3\) in 100 mL of its saturated solution is:
(a) \(2.83 \times 10^{-3}\) g
(b) \(1.0 \times 10^{-4}\) g
(c) \(28.3 \times 10^{-2}\) g
(d) \(1.0 \times 10^{-7}\) g
Show Answer & Explanation

Answer: (a) \(2.83 \times 10^{-3}\) g

Explanation:
1. \(s = 10^{-4}\) mol/L.
2. Mass in 100 mL = \(10^{-4} \times 0.1 \times 283 = 2.83 \times 10^{-3}\) g.

Teacher's Note:
1. Do not forget the 100 mL.

Question 20: Solid Ba(NO\(_3\))\(_2\) is gradually dissolved in \(1.0 \times 10^{-4}\) M Na\(_2\)CO\(_3\) (\(K_{sp}\) of BaCO\(_3\) = \(5.1 \times 10^{-9}\)). Precipitation begins at [Ba\(^{2+}\)] =
(a) \(5.1 \times 10^{-5}\) M
(b) \(4.1 \times 10^{-5}\) M
(c) \(8.1 \times 10^{-7}\) M
(d) \(8.1 \times 10^{-8}\) M
Show Answer & Explanation

Answer: (a) \(5.1 \times 10^{-5}\) M

Explanation:
1. \([Ba^{2+}] = 5.1 \times 10^{-9}/10^{-4} = 5.1 \times 10^{-5}\) M.

Teacher's Note:
1. Ionic product reaching \(K_{sp}\).

Question 21: The \(K_{sp}\) of AgBr is \(5.0 \times 10^{-13}\). The mass of KBr (molar mass 120) to be added to 1 L of 0.05 M AgNO\(_3\) to start precipitation of AgBr is:
(a) \(1.2 \times 10^{-9}\) g
(b) \(1.2 \times 10^{-10}\) g
(c) \(5.0 \times 10^{-8}\) g
(d) \(6.2 \times 10^{-5}\) g
Show Answer & Explanation

Answer: (a) \(1.2 \times 10^{-9}\) g

Explanation:
1. \([Br^-] = 5 \times 10^{-13}/0.05 = 10^{-11}\) M.
2. Mass = \(10^{-11} \times 120 = 1.2 \times 10^{-9}\) g.

Teacher's Note:
1. Needed [Br\(^-\)] times molar mass.

Question 22: The \(K_{sp}\) of Mg(OH)\(_2\) is \(1.0 \times 10^{-11}\). At what pH does Mg\(^{2+}\) begin to precipitate from 0.001 M Mg\(^{2+}\)?
(a) 11
(b) 10
(c) 8
(d) 9
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. \([OH^-] = \sqrt{10^{-11}/10^{-3}} = 10^{-4}\) M.
2. pOH = 4, pH = 10.

Teacher's Note:
1. Ionic product equals \(K_{sp}\).

Question 23: An aqueous solution contains an unknown concentration of Ba\(^{2+}\). When 50 mL of 1 M Na\(_2\)SO\(_4\) is added, BaSO\(_4\) just begins to precipitate. The final volume is 500 mL and \(K_{sp}\) = \(1 \times 10^{-10}\). The original [Ba\(^{2+}\)] is:
(a) \(1.1 \times 10^{-9}\) M
(b) \(1.0 \times 10^{-10}\) M
(c) \(2 \times 10^{-9}\) M
(d) \(5 \times 10^{-9}\) M
Show Answer & Explanation

Answer: (a) \(1.1 \times 10^{-9}\) M

Explanation:
1. Final \([SO_4^{2-}] = 50/500 = 0.1\) M, so final \([Ba^{2+}] = 10^{-9}\) M.
2. Original volume was 450 mL: \(10^{-9} \times 500/450 = 1.1 \times 10^{-9}\) M.

Teacher's Note:
1. Allow for dilution of the barium solution.

Question 24: The minimum volume of water required to dissolve 0.1 g of PbCl\(_2\) (\(K_{sp} = 3.2 \times 10^{-8}\); Pb = 207, Cl = 35.5) to get a saturated solution is:
(a) 0.36 L
(b) 0.18 L
(c) 17.98 L
(d) 1.798 L
Show Answer & Explanation

Answer: (b) 0.18 L

Explanation:
1. \(s = (3.2 \times 10^{-8}/4)^{1/3} = 2 \times 10^{-3}\) M.
2. 0.1 g = \(3.6 \times 10^{-4}\) mol (molar mass 278), so V = 0.18 L.

Teacher's Note:
1. Volume = moles / solubility.

Question 25: A mixture of 100 mmol Ca(OH)\(_2\) and 2 g Na\(_2\)SO\(_4\) is dissolved in water to make 100 mL (\(K_{sp}\) of Ca(OH)\(_2\) = \(5.5 \times 10^{-6}\); molar masses: Ca(OH)\(_2\) 74, Na\(_2\)SO\(_4\) 143, CaSO\(_4\) 136). The mass of CaSO\(_4\) formed and [OH\(^-\)] in the solution are:
(a) 1.9 g and 0.14 M
(b) 13.6 g and 0.14 M
(c) 13.6 g and 0.28 M
(d) 1.9 g and 0.28 M
Show Answer & Explanation

Answer: (d) 1.9 g and 0.28 M

Explanation:
1. Na\(_2\)SO\(_4\) = 2/143 = 14 mmol, which converts to 14 mmol CaSO\(_4\) (1.9 g) since Ca(OH)\(_2\) is in excess.
2. 14 mmol Na\(_2\)SO\(_4\) releases 28 mmol OH\(^-\) as NaOH, giving 0.28 M in 100 mL.

Teacher's Note:
1. Solid Ca(OH)\(_2\) is exchanged for CaSO\(_4\).

Question 26: If \(K_{sp}\) of Ag\(_2\)CO\(_3\) is \(8 \times 10^{-12}\), its molar solubility in 0.1 M AgNO\(_3\) is:
(a) \(8 \times 10^{-12}\) M
(b) \(8 \times 10^{-11}\) M
(c) \(8 \times 10^{-13}\) M
(d) \(8 \times 10^{-10}\) M
Show Answer & Explanation

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

Explanation:
1. \(s = K_{sp}/[Ag^+]^2 = 8 \times 10^{-12}/0.01 = 8 \times 10^{-10}\) M.

Teacher's Note:
1. Ag\(^+\) common ion at 0.1 M.

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