Solubility Equilibria: Quiz

Solubility Equilibria: Quiz

7 questions

1. Question 1EasyWhat is the Ksp expression for Mg(OH)₂?

   AKsp = [Mg²⁺][OH⁻]BKsp = [Mg²⁺][OH⁻]²CKsp = [Mg²⁺][OH⁻]² / [Mg(OH)₂]DKsp = [Mg²⁺][2OH⁻]
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   Answer: Ksp = [Mg²⁺][OH⁻]²

   Mg(OH)₂(s) ⇌ Mg²⁺ + 2OH⁻. Each ion is raised to its coefficient, and the solid is left out.

2. Question 2EasyWhat is the molar solubility of BaSO₄ (Ksp = 1.1 × 10⁻¹⁰)?

   A1.1 × 10⁻¹⁰ mol/LB1.0 × 10⁻⁵ mol/LC5.5 × 10⁻¹¹ mol/LD3.0 × 10⁻⁴ mol/L
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   Answer: 1.0 × 10⁻⁵ mol/L

   For a 1 : 1 salt, Ksp = s², so s = √(1.1 × 10⁻¹⁰) = 1.0 × 10⁻⁵ mol/L.

3. Question 3MediumWhat is the molar solubility of CaF₂ (Ksp = 3.9 × 10⁻¹¹)?

   A6.2 × 10⁻⁶ mol/LB2.1 × 10⁻⁴ mol/LC3.4 × 10⁻⁴ mol/LD9.8 × 10⁻¹² mol/L
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   Answer: 2.1 × 10⁻⁴ mol/L

   Ksp = s(2s)² = 4s³, so s = ∛(3.9 × 10⁻¹¹ ÷ 4) = 2.1 × 10⁻⁴ mol/L. 6.2 × 10⁻⁶ wrongly uses s².

4. Question 4MediumHow does adding NaCl affect the solubility of AgCl?

   AIt increases itBIt decreases itCNo effectDIt changes Ksp
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   Answer: It decreases it

   Cl⁻ is a common ion; the equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ shifts left, so less AgCl dissolves. Ksp itself does not change.

5. Question 5MediumIn a solution, [Ag⁺] = [Cl⁻] = 5.0 × 10⁻⁴ mol/L. Ksp(AgCl) = 1.8 × 10⁻¹⁰. What happens?

   ANo precipitate: Q is less than KspBA precipitate forms: Q is greater than KspCThe solution is exactly saturatedDAgCl dissolves further
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   Answer: A precipitate forms: Q is greater than Ksp

   Q = (5.0 × 10⁻⁴)² = 2.5 × 10⁻⁷, which is greater than 1.8 × 10⁻¹⁰, so AgCl precipitates.

6. Question 6HardWhich salt is MORE soluble in water: AgCl (Ksp 1.8 × 10⁻¹⁰) or Ag₂CrO₄ (Ksp 1.1 × 10⁻¹²)?

   AAgCl, because its Ksp is largerBAg₂CrO₄, because its molar solubility is largerCThey are equally solubleDIt cannot be decided
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   Answer: Ag₂CrO₄, because its molar solubility is larger

   AgCl: s = √Ksp = 1.3 × 10⁻⁵ mol/L. Ag₂CrO₄: s = ∛(Ksp ÷ 4) = 6.5 × 10⁻⁵ mol/L. Different ion ratios mean Ksp values cannot be compared directly.

7. Question 7HardWhat is the molar solubility of AgCl in 0.10 mol/L NaCl? (Ksp = 1.8 × 10⁻¹⁰)

   A1.3 × 10⁻⁵ mol/LB1.8 × 10⁻⁹ mol/LC1.8 × 10⁻¹¹ mol/LD0.10 mol/L
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   Answer: 1.8 × 10⁻⁹ mol/L

   [Cl⁻] ≈ 0.10 mol/L, so s = Ksp ÷ 0.10 = 1.8 × 10⁻⁹ mol/L, thousands of times less than in pure water.