U7.11 · Introduction to Solubility Equilibria (Ksp)Practice · 溶解平衡与 Ksp
Lessons covered:U7.11 Introduction to Solubility Equilibria (Ksp)
- 1Multiple choice
What is the Ksp expression for ?
ExplanationAg₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq). The solid is left out, and the coefficient 3 on Ag⁺ becomes an exponent, not a multiplier: Ksp = [Ag⁺]³[PO₄³⁻].
- 2Numeric
At 25 °C, Ksp = 5.0 × 10−13 for AgBr. What is the molar solubility of AgBr in water?
× 10MExplanationAgBr(s) ⇌ Ag⁺ + Br⁻. If s mol/L dissolves, [Ag⁺] = [Br⁻] = s, so Ksp = s² and s = √(5.0 × 10−13) = 7.1 × 10−7 M.
- 3Numeric
At 25 °C, 1.0 L of saturated solution is found to contain 0.0168 g of dissolved CaF₂ (M = 78.07 g/mol). Calculate Ksp for CaF₂.
× 10Explanations = 0.0168 ÷ 78.07 = 2.15 × 10−4 M. CaF₂(s) ⇌ Ca²⁺ + 2F⁻, so [Ca²⁺] = s and [F⁻] = 2s. Ksp = s(2s)² = 4s³ = 4 × (2.15 × 10−4)³ = 4.0 × 10−11. Forgetting to double [F⁻] gives s³ = 1.0 × 10−11.
- 4Multiple choice
At 25 °C, Ksp = 5.0 × 10−13 for AgBr and 1.8 × 10−18 for Ag₃PO₄. Which has the greater molar solubility in water?
ExplanationWith different formula types, Ksp values can’t be compared directly. AgBr: s = √Ksp = 7.1 × 10−7. Ag₃PO₄: Ksp = (3s)³·s = 27s⁴, so s = (1.8 × 10−18 ÷ 27)1/4 = 1.6 × 10−5. Despite its much smaller Ksp, Ag₃PO₄ is about 23 times more soluble.
- 5Numeric
At 25 °C, Ksp = 1.6 × 10−10 for AgCl (M = 143.3 g/mol). What is the greatest mass of AgCl that can dissolve in 2.0 L of water?
× 10gExplanations = √(1.6 × 10−10) = 1.26 × 10−5 M; in 2.0 L that is 1.26 × 10−5 × 2.0 = 2.53 × 10−5 mol, with a mass of 2.53 × 10−5 × 143.3 = 3.6 × 10−3 g, just a few milligrams.
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