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AP Chemistry Help: Introduction To Solubility Equilibria

Review real example questions for Introduction To Solubility Equilibria in AP Chemistry.

Question 1 / 10

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A student adds excess solid barium sulfate, BaSO4(s)\text{BaSO}_4(s), to pure water at 25∘C25^\circ\text{C}. After equilibrium is established, the solubility product constant is Ksp(BaSO4)=1.0×10−10K_{sp}(\text{BaSO}_4)=1.0\times10^{-10}. What is the equilibrium concentration of Ba2+(aq)\text{Ba}^{2+}(aq)? (Assume volume change is negligible.)

All questions

Question 1

A student adds excess solid barium sulfate, BaSO4(s)\text{BaSO}_4(s), to pure water at 25∘C25^\circ\text{C}. After equilibrium is established, the solubility product constant is Ksp(BaSO4)=1.0×10−10K_{sp}(\text{BaSO}_4)=1.0\times10^{-10}. What is the equilibrium concentration of Ba2+(aq)\text{Ba}^{2+}(aq)? (Assume volume change is negligible.)

  1. 2.0×10−5 M2.0\times10^{-5}\,\text{M}
  2. 1.0×10−20 M1.0\times10^{-20}\,\text{M}
  3. 1.0×10−10 M1.0\times10^{-10}\,\text{M}
  4. 5.0×10−11 M5.0\times10^{-11}\,\text{M}
  5. 1.0×10−5 M1.0\times10^{-5}\,\text{M} (correct answer)

Explanation: This question tests your understanding of solubility equilibria (quantitative). The dissolution equation is BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq). To construct the ICE table, we set initial concentrations of Ba²⁺ and SO₄²⁻ to 0 M, the change as +x for both ions due to 1:1 stoichiometry, and equilibrium concentrations as x for both. The Ksp expression is [Ba²⁺][SO₄²⁻] = x · x = x² = 1.0×10^{-10}, so we solve for the single variable x by taking the square root to get x = 1.0×10^{-5} M. We use only one variable because the equal stoichiometric coefficients link the ion concentrations directly. A tempting distractor is 1.0×10^{-20} M (choice E), which results from squaring the Ksp instead of taking the square root. For simple salts, Ksp problems reduce to one variable on AP MCQs, so always set up the expression based on stoichiometry and solve accordingly.

Question 2

Excess solid copper(I) chloride, CuCl(s)\text{CuCl}(s), is placed in pure water at 25∘C25^\circ\text{C} until equilibrium is established. The KspK_{sp} of CuCl\text{CuCl} is 1.0×10−61.0\times10^{-6}.

Dissolution: CuCl(s)⇌Cu+(aq)+Cl−(aq)\text{CuCl}(s)\rightleftharpoons \text{Cu}^+(aq)+\text{Cl}^-(aq)

Starting with zero ion concentrations, what is the equilibrium concentration of Cl−(aq)\text{Cl}^-(aq)?

ICE table (in M):

Cu+\text{Cu}^+Cl−\text{Cl}^-
I00
C+ss+ss
Essss
  1. 1.0×10−6 M1.0\times10^{-6}\,\text{M}
  2. 1.0×10−3 M1.0\times10^{-3}\,\text{M} (correct answer)
  3. 5.0×10−4 M5.0\times10^{-4}\,\text{M}
  4. 2.0×10−3 M2.0\times10^{-3}\,\text{M}
  5. 1.0×103 M1.0\times10^{3}\,\text{M}

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation CuCl(s) ⇌ Cu⁺(aq) + Cl⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Cu⁺ and Cl⁻. The Ksp expression is Ksp = [Cu⁺][Cl⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻⁶, so s = √(1.0×10⁻⁶) = 1.0×10⁻³ M, which equals [Cl⁻]. Choice A (1.0×10⁻⁶) incorrectly uses Ksp as the concentration. For any 1:1 salt, both ion concentrations equal the molar solubility at equilibrium.

Question 3

Excess solid barium sulfate, BaSO4(s)\text{BaSO}_4(s), is stirred in pure water at 25∘C25^\circ\text{C} until equilibrium is reached. The KspK_{sp} of BaSO4\text{BaSO}_4 is 1.0×10−101.0\times10^{-10}.

Dissolution: BaSO4(s)⇌Ba2+(aq)+SO42−(aq)\text{BaSO}_4(s)\rightleftharpoons \text{Ba}^{2+}(aq)+\text{SO}_4^{2-}(aq)

Starting from zero ion concentrations, what is the equilibrium concentration of Ba2+(aq)\text{Ba}^{2+}(aq)?

ICE table (in M):

Ba2+\text{Ba}^{2+}SO42−\text{SO}_4^{2-}
I00
C+ss+ss
Essss
  1. 1.0×10−10 M1.0\times10^{-10}\,\text{M}
  2. 1.0×10−5 M1.0\times10^{-5}\,\text{M} (correct answer)
  3. 5.0×10−6 M5.0\times10^{-6}\,\text{M}
  4. 2.0×10−5 M2.0\times10^{-5}\,\text{M}
  5. 1.0×105 M1.0\times10^{5}\,\text{M}

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Ba²⁺ and SO₄²⁻. The Ksp expression is Ksp = [Ba²⁺][SO₄²⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻¹⁰, so s = √(1.0×10⁻¹⁰) = 1.0×10⁻⁵ M. Choice A (1.0×10⁻¹⁰) incorrectly uses Ksp as the concentration without solving. For 1:1 salts, always take the square root of Ksp to find the molar solubility.

Question 4

Excess solid barium sulfate, BaSO4(s)\text{BaSO}_4(s), is mixed with pure water at 25∘C25^\circ\text{C} until equilibrium is established. For BaSO4(s)⇌Ba2+(aq)+SO42−(aq)\text{BaSO}_4(s)\rightleftharpoons \text{Ba}^{2+}(aq)+\text{SO}_4^{2-}(aq), Ksp=1.0×10−10K_{sp}=1.0\times10^{-10}. What is the equilibrium concentration of Ba2+\text{Ba}^{2+} in the saturated solution?

  1. 1.0×10−5 M1.0\times10^{-5}\,\text{M} (correct answer)
  2. 1.0×10−10 M1.0\times10^{-10}\,\text{M}
  3. 5.0×10−6 M5.0\times10^{-6}\,\text{M}
  4. 1.0×10−20 M1.0\times10^{-20}\,\text{M}
  5. 2.0×10−5 M2.0\times10^{-5}\,\text{M}

Explanation: This problem tests solubility equilibria (quantitative). For BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq), both ions form in a 1:1 ratio, so [Ba²⁺] = [SO₄²⁻] = x at equilibrium. Substituting into Ksp = [Ba²⁺][SO₄²⁻] = x² = 1.0×10⁻¹⁰. Solving gives x = √(1.0×10⁻¹⁰) = 1.0×10⁻⁵ M. A common mistake is reporting the Ksp value (1.0×10⁻¹⁰ M) as the concentration. For simple 1:1 salts, molar solubility equals the square root of Ksp.

Question 5

Excess solid barium sulfate, BaSO4(s)\text{BaSO}_4(s), is added to pure water at 25∘C25^\circ\text{C} and stirred to equilibrium. The solubility-product constant is Ksp(BaSO4)=1.0×10−10K_{sp}(\text{BaSO}_4)=1.0\times 10^{-10}. What is the equilibrium concentration of Ba2+(aq)\text{Ba}^{2+}(aq)?\n\nUse the following ICE table setup:\n\n| | BaSO4(s)⇌Ba2++SO42−\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-} | [Ba2+][\text{Ba}^{2+}] (M) | [SO42−][\text{SO}_4^{2-}] (M) |\n|------|-------------------------------------------------------------------------|------------------------|--------------------------|\n| I | | 0 | 0 |\n| C | | +s+s | +s+s |\n| E | | ss | ss |

  1. 1.0×10−5 M1.0\times 10^{-5}\,\text{M} (correct answer)
  2. 1.0×10−10 M1.0\times 10^{-10}\,\text{M}
  3. 3.2×10−5 M3.2\times 10^{-5}\,\text{M}
  4. 5.0×10−6 M5.0\times 10^{-6}\,\text{M}
  5. 2.0×10−5 M2.0\times 10^{-5}\,\text{M}

Explanation: This problem tests your understanding of solubility equilibria (quantitative). When BaSO4\text{BaSO}_4 dissolves, it produces equal moles of Ba2+\text{Ba}^{2+} and SO42−\text{SO}_4^{2-}, so if ss = molar solubility, then [Ba2+]=[SO42−]=s[\text{Ba}^{2+}] = [\text{SO}_4^{2-}] = s at equilibrium. The Ksp expression is Ksp=[Ba2+][SO42−]=s×s=s2K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] = s \times s = s^2. Substituting: 1.0×10−10=s21.0 \times 10^{-10} = s^2, so s=1.0×10−10=1.0×10−5 Ms = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} \, \text{M}. A common error is confusing this 1:1 salt with more complex stoichiometries or using Ksp directly as the concentration. For simple 1:1 salts, the calculation reduces to taking the square root of Ksp.

Question 6

Excess solid zinc sulfide, ZnS(s)\text{ZnS}(s), is added to pure water at 25∘C25^\circ\text{C} and allowed to reach equilibrium. The KspK_{sp} of ZnS\text{ZnS} is 1.0×10−241.0\times10^{-24}.

Dissolution: ZnS(s)⇌Zn2+(aq)+S2−(aq)\text{ZnS}(s)\rightleftharpoons \text{Zn}^{2+}(aq)+\text{S}^{2-}(aq)

Assuming initial ion concentrations are zero, what is the molar solubility of ZnS\text{ZnS}?

ICE table (in M):

Zn2+\text{Zn}^{2+}S2−\text{S}^{2-}
I00
C+ss+ss
Essss
  1. 1.0×10−12 M1.0\times10^{-12}\,\text{M} (correct answer)
  2. 1.0×10−24 M1.0\times10^{-24}\,\text{M}
  3. 5.0×10−13 M5.0\times10^{-13}\,\text{M}
  4. 1.0×10−8 M1.0\times10^{-8}\,\text{M}
  5. 2.0×10−12 M2.0\times10^{-12}\,\text{M}

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation ZnS(s) ⇌ Zn²⁺(aq) + S²⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Zn²⁺ and S²⁻. The Ksp expression is Ksp = [Zn²⁺][S²⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻²⁴, so s = √(1.0×10⁻²⁴) = 1.0×10⁻¹² M. Choice B (1.0×10⁻²⁴) incorrectly uses the Ksp value as the molar solubility without taking the square root. For extremely insoluble salts like ZnS, the molar solubility is still found by taking the square root of Ksp.

Question 7

A student adds excess solid silver chloride, AgCl(s)\text{AgCl}(s), to 1.00 L1.00\,\text{L} of pure water at 25∘C25^\circ\text{C} and stirs until equilibrium is established. The KspK_{sp} of AgCl\text{AgCl} at 25∘C25^\circ\text{C} is 1.6×10−101.6\times10^{-10}.

Dissolution: AgCl(s)⇌Ag+(aq)+Cl−(aq)\text{AgCl}(s)\rightleftharpoons \text{Ag}^+(aq)+\text{Cl}^-(aq)

Assuming the initial ion concentrations are zero, what is the equilibrium concentration of Ag+(aq)\text{Ag}^+(aq)?

ICE table (in M):

Ag+\text{Ag}^+Cl−\text{Cl}^-
I00
C+ss+ss
Essss
  1. 4.0×10−5 M4.0\times10^{-5}\,\text{M}
  2. 1.3×10−5 M1.3\times10^{-5}\,\text{M} (correct answer)
  3. 8.0×10−11 M8.0\times10^{-11}\,\text{M}
  4. 1.6×10−10 M1.6\times10^{-10}\,\text{M}
  5. 2.5×105 M2.5\times10^{5}\,\text{M}

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) shows a 1:1 stoichiometry, so if s moles of AgCl dissolve, we get s moles of Ag⁺ and s moles of Cl⁻. The Ksp expression is Ksp = [Ag⁺][Cl⁻] = (s)(s) = s². Solving: s² = 1.6×10⁻¹⁰, so s = √(1.6×10⁻¹⁰) = 1.26×10⁻⁵ M. Choice A (8.0×10⁻¹¹) incorrectly divides Ksp by 2 instead of taking the square root. For simple 1:1 salts like AgCl, the molar solubility equals the square root of Ksp.

Question 8

A student adds excess solid silver chloride, AgCl(s)\text{AgCl}(s), to 1.00 L1.00\,\text{L} of pure water at 25∘C25^\circ\text{C} and stirs until equilibrium is established. The KspK_{sp} of AgCl\text{AgCl} at 25∘C25^\circ\text{C} is 1.6×10−101.6\times10^{-10}.

Dissolution: AgCl(s)⇌Ag+(aq)+Cl−(aq)\text{AgCl}(s)\rightleftharpoons \text{Ag}^+(aq)+\text{Cl}^-(aq)

Assuming the initial ion concentrations are zero, what is the equilibrium concentration of Ag+(aq)\text{Ag}^+(aq)?

ICE table (in M):

Ag+\text{Ag}^+Cl−\text{Cl}^-
I00
C+ss+ss
Essss
  1. 2.5×105 M2.5\times10^{5}\,\text{M}
  2. 4.0×10−5 M4.0\times10^{-5}\,\text{M}
  3. 8.0×10−11 M8.0\times10^{-11}\,\text{M}
  4. 1.3×10−5 M1.3\times10^{-5}\,\text{M} (correct answer)
  5. 1.6×10−10 M1.6\times10^{-10}\,\text{M}

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation AgCl(s)⇌Ag+(aq)+Cl−(aq)\text{AgCl}(s) \rightleftharpoons \text{Ag}^{+}(aq) + \text{Cl}^{-}(aq) shows a 1:1 stoichiometry, so if s moles of AgCl dissolve, we get s moles of Ag⁺ and s moles of Cl⁻. The Ksp expression is Ksp=[Ag+][Cl−]=(s)(s)=s2K_{sp} = [\text{Ag}^{+}] [\text{Cl}^{-}] = (s)(s) = s^2. Solving: s2=1.6×10−10s^2 = 1.6\times10^{-10}, so s=1.6×10−10=1.26×10−5 Ms = \sqrt{1.6\times10^{-10}} = 1.26\times10^{-5} \, \text{M}. Choice A (8.0×10−118.0\times10^{-11}) incorrectly divides Ksp by 2 instead of taking the square root. For simple 1:1 salts like AgCl, the molar solubility equals the square root of Ksp.

Question 9

Excess solid iron(II) hydroxide, Fe(OH)2(s)\text{Fe(OH)}_2(s), is added to pure water at 25∘C25^\circ\text{C} until equilibrium is established. The KspK_{sp} of Fe(OH)2\text{Fe(OH)}_2 is 4.0×10−144.0\times10^{-14}.

Dissolution: Fe(OH)2(s)⇌Fe2+(aq)+2OH−(aq)\text{Fe(OH)}_2(s)\rightleftharpoons \text{Fe}^{2+}(aq)+2\text{OH}^-(aq)

If the initial ion concentrations are zero, what is the equilibrium concentration of OH−(aq)\text{OH}^-(aq)?

ICE table (in M):

Fe2+\text{Fe}^{2+}OH−\text{OH}^-
I00
C+ss+2s2s
Ess2s2s
  1. 2.0×10−5 M2.0\times10^{-5}\,\text{M} (correct answer)
  2. 1.0×10−10 M1.0\times10^{-10}\,\text{M}
  3. 1.0×10−5 M1.0\times10^{-5}\,\text{M}
  4. 5.0×10−5 M5.0\times10^{-5}\,\text{M}
  5. 4.0×10−14 M4.0\times10^{-14}\,\text{M}

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation Fe(OH)2(s)⇌Fe2+(aq)+2OH−(aq)\text{Fe(OH)}_2(s) \rightleftharpoons \text{Fe}^{2+}(aq) + 2\text{OH}^{-}(aq) shows that dissolving s moles produces s moles of Fe2+\text{Fe}^{2+} and 2s moles of OH−\text{OH}^{-}. The KspK_{sp} expression is Ksp=[Fe2+][OH−]2=s⋅(2s)2=4s3K_{sp} = [\text{Fe}^{2+}][\text{OH}^{-}]^2 = s \cdot (2s)^2 = 4s^3. Solving: 4s3=4.0×10−144s^3 = 4.0 \times 10^{-14}, so s3=1.0×10−14s^3 = 1.0 \times 10^{-14}, and s=1.0×10−5 Ms = 1.0 \times 10^{-5} \, \text{M}, making [OH−]=2s=2.0×10−5 M[\text{OH}^{-}] = 2s = 2.0 \times 10^{-5} \, \text{M}. Choice C (4.0×10−144.0 \times 10^{-14}) incorrectly uses the Ksp value as the concentration. For hydroxide salts, remember to multiply the molar solubility by 2 to get the OH⁻ concentration.

Question 10

Excess solid zinc sulfide, ZnS(s)\text{ZnS}(s), is added to pure water at 25∘C25^\circ\text{C} and allowed to reach equilibrium. The KspK_{sp} of ZnS\text{ZnS} is 1.0×10−241.0\times10^{-24}.

Dissolution: ZnS(s)⇌Zn2+(aq)+S2−(aq)\text{ZnS}(s)\rightleftharpoons \text{Zn}^{2+}(aq)+\text{S}^{2-}(aq)

Assuming initial ion concentrations are zero, what is the molar solubility of ZnS\text{ZnS}?

ICE table (in M):

Zn2+\text{Zn}^{2+}S2−\text{S}^{2-}
I00
C+ss+ss
Essss
  1. 1.0×10−12 M1.0\times10^{-12}\,\text{M} (correct answer)
  2. 1.0×10−24 M1.0\times10^{-24}\,\text{M}
  3. 5.0×10−13 M5.0\times10^{-13}\,\text{M}
  4. 1.0×10−8 M1.0\times10^{-8}\,\text{M}
  5. 2.0×10−12 M2.0\times10^{-12}\,\text{M}

Explanation: This problem tests solubility equilibria (quantitative). The dissolution equation ZnS(s) ⇌ Zn²⁺(aq) + S²⁻(aq) shows a 1:1 stoichiometry, so dissolving s moles produces s moles each of Zn²⁺ and S²⁻. The Ksp expression is Ksp = [Zn²⁺][S²⁻] = (s)(s) = s². Solving: s² = 1.0×10⁻²⁴, so s = √(1.0×10⁻²⁴) = 1.0×10⁻¹² M. Choice B (1.0×10⁻²⁴) incorrectly uses the Ksp value as the molar solubility without taking the square root. For extremely insoluble salts like ZnS, the molar solubility is still found by taking the square root of Ksp.