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Quiz: Solubility Equilibria

Test your understanding of the solubility product, molar solubility, the common ion effect, precipitation prediction, complex ion formation, and Henry's law with these questions.

1. Write the correct \(K_{sp}\) expression for the dissolution of \(\ce{Ag2CrO4}\).

\[\ce{Ag2CrO4(s) <=> 2Ag+(aq) + CrO4^{2-}(aq)}\]
  1. \(K_{sp} = [\ce{Ag+}]^2[\ce{CrO4^{2-}}]\)
  2. \(K_{sp} = [\ce{Ag+}][\ce{CrO4^{2-}}]\)
  3. \(K_{sp} = \dfrac{[\ce{Ag+}]^2[\ce{CrO4^{2-}}]}{[\ce{Ag2CrO4}]}\)
  4. \(K_{sp} = 2[\ce{Ag+}][\ce{CrO4^{2-}}]\)
Show Answer

The correct answer is A. The \(K_{sp}\) expression is the product of ion concentrations each raised to the power of their stoichiometric coefficient. For \(\ce{Ag2CrO4}\): \(K_{sp} = [\ce{Ag+}]^2[\ce{CrO4^{2-}}]^1\). The solid \(\ce{Ag2CrO4}\) is omitted because it is a pure solid. Option B omits the exponent of 2 on \([\ce{Ag+}]\). Option C incorrectly includes the solid in the denominator. Option D multiplies by 2 rather than squaring.

Concept Tested: Solubility Product \(K_{sp}\) / Writing Ksp Expressions

2. The \(K_{sp}\) of \(\ce{AgCl}\) is \(1.8 \times 10^{-10}\) at 25 °C. What is the molar solubility of \(\ce{AgCl}\) in pure water?

  1. \(1.8 \times 10^{-10}\,\text{M}\)
  2. \(1.3 \times 10^{-5}\,\text{M}\)
  3. \(3.6 \times 10^{-10}\,\text{M}\)
  4. \(9.0 \times 10^{-6}\,\text{M}\)
Show Answer

The correct answer is B. Setting up the ICE table: \([\ce{Ag+}] = [\ce{Cl-}] = s\), so \(K_{sp} = s^2 = 1.8 \times 10^{-10}\). Therefore \(s = \sqrt{1.8 \times 10^{-10}} = 1.34 \times 10^{-5}\,\text{M} \approx 1.3 \times 10^{-5}\,\text{M}\). Option A is the \(K_{sp}\) itself, not the square root. Option C doubles \(K_{sp}\) instead of taking the square root. Option D results from an arithmetic error in the square root calculation.

Concept Tested: Molar Solubility from \(K_{sp}\)

3. The \(K_{sp}\) of \(\ce{PbI2}\) is \(7.1 \times 10^{-9}\). Which expression correctly gives the molar solubility \(s\) of \(\ce{PbI2}\) in pure water?

\[\ce{PbI2(s) <=> Pb^{2+}(aq) + 2I-(aq)}\]
  1. \(s = \sqrt{7.1 \times 10^{-9}}\)
  2. \(s^2 = 7.1 \times 10^{-9}\), so \(s = 8.4 \times 10^{-5}\,\text{M}\)
  3. \(4s^3 = 7.1 \times 10^{-9}\), so \(s = \left(\dfrac{7.1 \times 10^{-9}}{4}\right)^{1/3}\)
  4. \(2s^2 = 7.1 \times 10^{-9}\), so \(s = 5.96 \times 10^{-5}\,\text{M}\)
Show Answer

The correct answer is C. For \(\ce{PbI2}\), \([\ce{Pb^{2+}}] = s\) and \([\ce{I-}] = 2s\), so \(K_{sp} = (s)(2s)^2 = 4s^3 = 7.1 \times 10^{-9}\), giving \(s = (7.1 \times 10^{-9}/4)^{1/3} = (1.775 \times 10^{-9})^{1/3} \approx 1.2 \times 10^{-3}\,\text{M}\). Options A and B apply the 1:1 formula incorrectly (missing the factor of 4 from stoichiometry). Option D uses the coefficient 2 as a multiplier instead of squaring the iodide term.

Concept Tested: Ksp Calculations / Stoichiometry in Molar Solubility

4. What is the molar solubility of \(\ce{AgCl}\) (\(K_{sp} = 1.8 \times 10^{-10}\)) in a solution that already contains 0.100 M \(\ce{NaCl}\)?

  1. \(1.8 \times 10^{-9}\,\text{M}\) — greatly reduced by the common ion effect
  2. \(1.3 \times 10^{-5}\,\text{M}\) — the same as in pure water
  3. \(1.8 \times 10^{-11}\,\text{M}\) — reduced because \(K_{sp}\) itself decreases
  4. \(1.8 \times 10^{-8}\,\text{M}\) — slightly reduced due to ionic strength effects
Show Answer

The correct answer is A. In the presence of 0.100 M \(\ce{Cl-}\) (from \(\ce{NaCl}\)), the ICE table gives \(K_{sp} = s(0.100 + s) \approx s(0.100) = 1.8 \times 10^{-10}\), so \(s = 1.8 \times 10^{-9}\,\text{M}\). This is over 7,000 times less soluble than in pure water — the common ion effect. Option B ignores the common ion entirely. Option C incorrectly states \(K_{sp}\) changes (\(K_{sp}\) is constant at fixed temperature). Option D uses incorrect arithmetic.

Concept Tested: Common Ion Effect / Molar Solubility

5. When 100 mL of 0.200 M \(\ce{Pb(NO3)2}\) is mixed with 100 mL of 0.200 M \(\ce{NaI}\), will \(\ce{PbI2}\) (\(K_{sp} = 7.1 \times 10^{-9}\)) precipitate?

  1. No, because after mixing \([\ce{Pb^{2+}}] = [\ce{I-}] = 0.100\,\text{M}\) and \(Q_{sp} < K_{sp}\).
  2. Yes, because after mixing \(Q_{sp} = 1.0 \times 10^{-3} \gg K_{sp} = 7.1 \times 10^{-9}\).
  3. No, because \(\ce{PbI2}\) is soluble in water at room temperature.
  4. Yes, but only if the temperature is raised above 25 °C to exceed the solubility limit.
Show Answer

The correct answer is B. After mixing equal volumes, concentrations are halved: \([\ce{Pb^{2+}}] = [\ce{I-}] = 0.100\,\text{M}\). Then \(Q_{sp} = (0.100)(0.100)^2 = 1.0 \times 10^{-3}\). Since \(Q_{sp} = 1.0 \times 10^{-3} \gg K_{sp} = 7.1 \times 10^{-9}\), the solution is supersaturated and \(\ce{PbI2}\) precipitates. Option A incorrectly evaluates the comparison. Options C and D mischaracterize the chemistry.

Concept Tested: Ion Product / Predicting Precipitation

6. A solution contains both \(\ce{Ag+}\) and \(\ce{Pb^{2+}}\) each at 0.100 M. Chloride ion is slowly added. Which ion precipitates first, and why?

\(K_{sp}(\ce{AgCl}) = 1.8 \times 10^{-10}\); \(K_{sp}(\ce{PbCl2}) = 1.7 \times 10^{-5}\)

  1. \(\ce{Pb^{2+}}\) precipitates first because \(K_{sp}(\ce{PbCl2})\) is larger.
  2. \(\ce{Ag+}\) precipitates first because \(\ce{AgCl}\) requires a much lower \([\ce{Cl-}]\) to begin precipitation.
  3. Both ions precipitate simultaneously because they are at the same initial concentration.
  4. \(\ce{Ag+}\) precipitates first because silver has a higher molar mass than lead.
Show Answer

The correct answer is B. \(\ce{AgCl}\) begins to precipitate when \([\ce{Cl-}] > K_{sp}/[\ce{Ag+}] = 1.8 \times 10^{-10}/0.100 = 1.8 \times 10^{-9}\,\text{M}\). \(\ce{PbCl2}\) begins to precipitate when \([\ce{Cl-}] > \sqrt{K_{sp}/[\ce{Pb^{2+}}]} = \sqrt{1.7 \times 10^{-4}} = 0.013\,\text{M}\). Because \(\ce{AgCl}\) requires a far smaller \([\ce{Cl-}]\) to start precipitating, \(\ce{Ag+}\) is removed first. Option A confuses larger \(K_{sp}\) with precipitating first. Molar mass is irrelevant (option D).

Concept Tested: Selective Precipitation

7. \(\ce{AgCl}\) barely dissolves in pure water, but dissolves readily in concentrated ammonia solution. The best explanation is:

  1. \(\ce{NH3}\) reacts with \(\ce{Cl-}\) to form a new precipitate, removing it from solution.
  2. \(\ce{NH3}\) increases the temperature of the solution, raising \(K_{sp}\).
  3. \(\ce{NH3}\) forms the complex ion \(\ce{Ag(NH3)2+}\), removing \(\ce{Ag+}\) from solution and shifting dissolution equilibrium to the right.
  4. \(\ce{NH3}\) neutralizes the \(\ce{Cl-}\) ions, reducing \(Q_{sp}\) below \(K_{sp}\).
Show Answer

The correct answer is C. Ammonia acts as a ligand and reacts with \(\ce{Ag+}\) to form the stable complex ion \(\ce{Ag(NH3)2+}\) (\(K_f = 1.7 \times 10^7\)). This removes free \(\ce{Ag+}\) from solution, lowering \(Q_{sp}\) below \(K_{sp}\) and shifting the dissolution equilibrium to the right (Le Chatelier's principle). The net \(K = K_{sp} \times K_f = 3.1 \times 10^{-3}\) is much larger than \(K_{sp}\) alone. The other options misidentify the chemical mechanism.

Concept Tested: Complex Ion Formation / Effect on Solubility

8. The dissolution of most ionic solids in water is endothermic. Applying Le Chatelier's principle, what happens to solubility as temperature increases?

  1. Solubility decreases because higher temperature reduces the value of \(K_{sp}\).
  2. Solubility is unaffected because \(K_{sp}\) is a constant.
  3. Solubility decreases because ions move faster and recombine more readily at higher temperature.
  4. Solubility increases because heat shifts the dissolution equilibrium to the right.
Show Answer

The correct answer is D. For an endothermic dissolution, heat can be thought of as a reactant. Raising the temperature shifts the equilibrium to the right (more dissolution), increasing \(K_{sp}\) and therefore increasing molar solubility. Option A states the opposite trend. Option B is wrong because \(K_{sp}\) varies with temperature. Option C incorrectly describes the kinetics and thermodynamics of the process.

Concept Tested: Temperature and Solubility

9. A soda bottle is pressurized with \(\ce{CO2}\) at 3.0 atm. The Henry's law constant for \(\ce{CO2}\) is \(k_H = 3.4 \times 10^{-2}\,\text{mol/L·atm}\). What is the concentration of dissolved \(\ce{CO2}\) in the sealed bottle?

  1. \(1.1 \times 10^{-2}\,\text{mol/L}\)
  2. \(8.8\,\text{mol/L}\)
  3. \(3.4 \times 10^{-2}\,\text{mol/L}\)
  4. \(1.0 \times 10^{-1}\,\text{mol/L}\)
Show Answer

The correct answer is D. By Henry's law: \(C = k_H \times P = (3.4 \times 10^{-2}\,\text{mol/L·atm})(3.0\,\text{atm}) = 0.102\,\text{mol/L} \approx 1.0 \times 10^{-1}\,\text{mol/L}\). Option A divides instead of multiplies. Option C uses only the Henry's constant without multiplying by pressure. Option B incorrectly uses pressure in a reciprocal calculation. Henry's law is a direct proportion: higher pressure means more gas dissolves.

Concept Tested: Henry's Law / Gas Solubility

10. Two equilibria are combined to find the net \(K\) for \(\ce{AgCl(s) + 2NH3(aq) <=> Ag(NH3)2+(aq) + Cl-(aq)}\):

\[K_{sp}(\ce{AgCl}) = 1.8 \times 10^{-10}; \quad K_f(\ce{Ag(NH3)2+}) = 1.7 \times 10^{7}\]

Which mathematical operation and result are correct?

  1. \(K_\text{net} = K_{sp} + K_f = 1.7 \times 10^{7}\)
  2. \(K_\text{net} = K_{sp} / K_f = 1.1 \times 10^{-17}\)
  3. \(K_\text{net} = K_{sp} \times K_f = 3.1 \times 10^{-3}\)
  4. \(K_\text{net} = \sqrt{K_{sp} \times K_f} = 5.5 \times 10^{-2}\)
Show Answer

The correct answer is C. When two reactions are added together, their equilibrium constants are multiplied: \(K_\text{net} = K_1 \times K_2 = (1.8 \times 10^{-10})(1.7 \times 10^7) = 3.1 \times 10^{-3}\). This value, much larger than \(K_{sp}\) alone, explains why \(\ce{AgCl}\) dissolves readily in ammonia. Option A adds the constants (incorrect for combining equilibria). Option B divides (this would apply if one reaction were the reverse of the other). Option D takes the square root for no valid reason.

Concept Tested: Manipulating K Expressions / Adding Reactions