By what factor is the molar solubility of PbI2 reduced when going from pure water to 0.10 M KI?

Answer: About 1700-fold lower Solubility (= [Pb^2+]) drops from 1.3e-3 in pure water to 7.9e-7 in 0.10 M KI. Ratio = 1.3e-3 / 7.9e-7 1.6e3.

What value of Ksp is calculated from the 0.10 M KI trial, and how does it compare to the pure-water value?

Answer: 7.9e-9; values agree within experimental error Ksp = [Pb^2+][I-]^2 = (7.9e-7)(0.10)^2 = 7.9e-9. This is consistent with 8.8e-9 from the first trial.

What is Ksp of PbI2 from the pure-water trial?

Answer: 8.8e-9 PbI2 - Pb^2+ + 2 I-. [Pb^2+] = 1.3e-3, [I-] = 2.6e-3. Ksp = (1.3e-3)(2.6e-3)^2 = 8.8e-9.

AP Chemistry · Topic 7.12

Common-Ion Effect Practice

Part of Equilibrium.(SPQ-5.B)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$A student measures the solubility of PbI2 in pure water and in 0.10 M KI by saturating each solvent at 25 C and titrating the I- in the saturated supernatant. In pure water, [I-] = 2.6e-3 M at saturation. In 0.10 M KI, the additional I- from PbI2 dissolution is negligible, and [Pb^2+] is found to be 7.9e-7 M.$
$By what factor is the molar solubility of PbI2 reduced when going from pure water to 0.10 M KI?$
- A
  $About 130-fold higher$
- B
  $About 10-fold lower$
- C
  $About 1700-fold lower$
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- D
  $Unchanged$
Why
$Solubility (= [Pb^2+]) drops from 1.3e-3 in pure water to 7.9e-7 in 0.10 M KI. Ratio = 1.3e-3 / 7.9e-7 ~ 1.6e3.$
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Sample 2difficulty 2/5
$A student measures the solubility of PbI2 in pure water and in 0.10 M KI by saturating each solvent at 25 C and titrating the I- in the saturated supernatant. In pure water, [I-] = 2.6e-3 M at saturation. In 0.10 M KI, the additional I- from PbI2 dissolution is negligible, and [Pb^2+] is found to be 7.9e-7 M.$
$What value of Ksp is calculated from the 0.10 M KI trial, and how does it compare to the pure-water value?$
- A
  $1.6e-7; common ion increases Ksp$
- B
  $0.10; Ksp depends on [KI]$
- C
  $7.9e-9; values agree within experimental error$
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- D
  $7.9e-7; KI dissolved Pb^2+$
Why
$Ksp = [Pb^2+][I-]^2 = (7.9e-7)(0.10)^2 = 7.9e-9. This is consistent with 8.8e-9 from the first trial.$
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Sample 3difficulty 2/5
$A student measures the solubility of PbI2 in pure water and in 0.10 M KI by saturating each solvent at 25 C and titrating the I- in the saturated supernatant. In pure water, [I-] = 2.6e-3 M at saturation. In 0.10 M KI, the additional I- from PbI2 dissolution is negligible, and [Pb^2+] is found to be 7.9e-7 M.$
$What is Ksp of PbI2 from the pure-water trial?$
- A
  $2.6e-3$
- B
  $1.7e-9$
- C
  $6.8e-6$
- D
  $8.8e-9$
  check_circle
Why
$PbI2 -> Pb^2+ + 2 I-. [Pb^2+] = 1.3e-3, [I-] = 2.6e-3. Ksp = (1.3e-3)(2.6e-3)^2 = 8.8e-9.$
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Sample 4difficulty 3/5
$A saturated solution of AgCl is in contact with solid AgCl. Solid NaCl is added.$
$What happens to the solubility of AgCl when NaCl is added?$
- A
  $Decreases because Ksp decreases$
- B
  $Decreases due to common-ion effect$
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- C
  $Unchanged because Ksp is constant$
- D
  $Increases$
Why
$Adding Cl- (common ion) shifts AgCl(s) <-> Ag+ + Cl- to the left, decreasing the solubility of AgCl. Ksp is constant.$
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Sample 5difficulty 4/5
$The common-ion effect on molar solubility$
- A
  $Increases K_sp$
- B
  $Increases solubility$
- C
  $Decreases solubility$
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- D
  $No effect$
Why
$Adding one of the ions reduces how much of the salt can dissolve.$
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