U7.12 · Common-Ion EffectPractice · 同离子效应
Lessons covered:U7.11 Introduction to Solubility Equilibria (Ksp)U7.12 Common-Ion Effect
- 1Numeric
What is the molar solubility of AgBr (Ksp = 5.0 × 10−13) in 0.020 M NaBr?
× 10MExplanationIf s M dissolves, [Ag⁺] = s and [Br⁻] = 0.020 + s ≈ 0.020. s × 0.020 = 5.0 × 10−13, so s = 2.5 × 10−11 M, about 4 × 10−5 times its solubility in pure water (7.1 × 10−7).
Review: U7.12 Common-Ion Effect
- 2Multiple choice
A little solid NaCl is stirred into a saturated AgCl solution that still has solid AgCl at the bottom, and the temperature returns to its original value. Which statement is correct?
ExplanationThe added Cl⁻ makes Q = [Ag⁺][Cl⁻] > Ksp, so the equilibrium shifts toward precipitating AgCl: Ag⁺ falls and solid builds up until Q equals Ksp again. At the same temperature Ksp is unchanged; what changes is the solubility.
Review: U7.12 Common-Ion Effect
- 3Numeric
What is the molar solubility of Mg(OH)₂ (Ksp = 8.9 × 10−12) in 0.050 M NaOH?
× 10MExplanation[Mg²⁺] = s and [OH⁻] = 0.050 + 2s ≈ 0.050. Ksp = s × 0.050² = 8.9 × 10−12, so s = 8.9 × 10−12 ÷ 2.5 × 10−3 = 3.6 × 10−9 M. Remember to square [OH⁻].
Review: U7.12 Common-Ion Effect
- 4Multiple choice
When finding the solubility of AgBr in 0.020 M NaBr, [Br⁻] can be approximated as 0.020 M. Why?
Explanation[Br⁻] = 0.020 + s, and s is only about 2.5 × 10−11 M, negligible by comparison. After solving, check that s really is far below 0.020. Some AgBr still dissolves, and what dissolves is present as Ag⁺ and Br⁻.
Review: U7.12 Common-Ion Effect
- 5Numeric
Silver is recovered from waste: a solution has [Ag⁺] = 1.0 × 10−3 M. NaCl is added so that after precipitation [Cl⁻] = 0.010 M (ignore volume changes). For AgCl, Ksp = 1.6 × 10−10. What [Ag⁺] remains in solution?
× 10MExplanationAfter precipitation the solution is saturated with AgCl: [Ag⁺] = Ksp ÷ [Cl⁻] = 1.6 × 10−10 ÷ 0.010 = 1.6 × 10−8 M, just 1.6 × 10−5 of the original. Nearly all the silver has precipitated, though it never reaches zero.
Review: U7.12 Common-Ion Effect
- 6Multiple choice
The graph shows the molar solubility of at 25 °C in solutions with different concentrations of . Which best explains why the curve drops so quickly?
ExplanationIn pure water (x = 0) the solubility is about 20.5 × 10⁻⁵ M. When F⁻ is already present, only a little Ca²⁺ is needed for Q = [Ca²⁺][F⁻]² to reach Ksp, so the equilibrium shifts toward the solid. At [NaF] = 0.002 M, for example, the solubility falls to about 0.85 × 10⁻⁵ M. Ksp depends only on temperature, and Na⁺ forms no complex with F⁻ or Ca²⁺.
Review: U7.12 Common-Ion EffectU7.11 Introduction to Solubility Equilibria (Ksp)
Your missed questions were saved automatically to your mistake notebook · 0
Your best score counts only full runs through the set, not retries of missed questions.