For PbCl₂ ⇌ Pb²⁺ + 2 Cl⁻, Ksp =

Answer: [Pb²⁺][Cl⁻]² Solid omitted; product of ion concentrations to coefficients.

A saturated solution in contact with undissolved solute is

Answer: At dissolution-precipitation equilibrium Saturation = dynamic equilibrium between dissolved and solid phases.

Answer: 1.7×10⁻¹⁰ Ksp = [Ag⁺][Cl⁻] = (1.3×10⁻⁵)² = 1.69×10⁻¹⁰ ≈ 1.7×10⁻¹⁰.

AP Chemistry · Topic 7.11

Introduction to Solubility Equilibria Practice

Part of Equilibrium.(SPQ-5.A)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$For PbCl₂ ⇌ Pb²⁺ + 2 Cl⁻, K_sp =$
- A
  $[Pb²⁺]²[Cl⁻]$
- B
  $[PbCl₂]/[Pb²⁺][Cl⁻]$
- C
  $[Pb²⁺][Cl⁻]²$
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- D
  $[Pb²⁺][Cl⁻]$
Why
$Solid omitted; product of ion concentrations to coefficients.$
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Sample 2difficulty 2/5
$A saturated solution in contact with undissolved solute is$
- A
  $Not at equilibrium$
- B
  $Subsaturated$
- C
  $Pure solute$
- D
  $At dissolution-precipitation equilibrium$
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Why
$Saturation = dynamic equilibrium between dissolved and solid phases.$
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Sample 3difficulty 2/5
$A saturated solution of AgCl gives [Ag⁺] = 1.3\times10⁻⁵ M.$
$What is Ksp of AgCl?$
- A
  $1.7\times10⁻⁵$
- B
  $1.7\times10⁻¹⁰$
  check_circle
- C
  $1.3\times10⁻⁵$
- D
  $2.6\times10⁻⁵$
Why
$Ksp = [Ag⁺][Cl⁻] = (1.3\times10⁻⁵)² = 1.69\times10⁻¹⁰ \approx 1.7\times10⁻¹⁰.$
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Sample 4difficulty 3/5
$Ksp(PbI2) = 7.1e-9.$
$What is the molar solubility s in pure water (approximately)?$
- A
  $7.1e-9 M$
- B
  $2.7e-3 M$
- C
  $8.4e-5 M$
- D
  $1.2e-3 M$
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Why
$s^3 = Ksp/4 = 1.78e-9; s = cube root ~ 1.2e-3 M.$
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Sample 5difficulty 3/5
$The Ksp values above are tabulated at 25 C in pure water.$
$Which silver halide is least soluble in pure water?$
- A
  $All have equal molar solubility$
- B
  $AgI$
  check_circle
- C
  $AgBr$
- D
  $AgCl$
Why
$Among the 1:1 salts (AgX), the smallest Ksp gives the smallest molar solubility. AgI has Ksp = 8.5e-17, the smallest.$
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