Questions: Precipitation Reactions and the Common Ion Effect

Question 1

A saturated solution of AgCl is prepared in pure water. You then dissolve NaCl (a soluble salt) into this solution. What happens and why?

• A. More AgCl dissolves because the additional Na⁺ ions disrupt the existing equilibrium
• B. The AgCl solubility is unaffected because Cl⁻ is already present in the solution
• C. AgCl precipitates further because the increased [Cl⁻] pushes Q above Ksp, forcing the equilibrium left
• D. The Ksp of AgCl increases to accommodate the extra Cl⁻ ions

Answer: (C)

Adding NaCl introduces Cl⁻ into a solution already at equilibrium for AgCl. This raises [Cl⁻], making Q = [Ag⁺][Cl⁻] exceed Ksp. The system restores equilibrium by shifting left — Ag⁺ and Cl⁻ ions combine to form more solid AgCl, reducing [Ag⁺] until Q returns to Ksp. The AgCl is now less soluble than in pure water. Note that Ksp is a constant at fixed temperature and cannot change — only the equilibrium position shifts.

Question 2

You mix 100 mL of 0.001 M AgNO₃ with 100 mL of 0.001 M NaCl. The Ksp of AgCl is 1.8 × 10⁻¹⁰. What do you predict?

• A. No precipitate forms because both solutions are dilute
• B. A precipitate forms because Q > Ksp after mixing
• C. No precipitate forms because Q = Ksp after mixing
• D. The solution becomes supersaturated but no precipitate forms until heated

Answer: (B)

After mixing, concentrations are halved: [Ag⁺] = 5 × 10⁻⁴ M and [Cl⁻] = 5 × 10⁻⁴ M. Q = (5 × 10⁻⁴)(5 × 10⁻⁴) = 2.5 × 10⁻⁷, which far exceeds Ksp = 1.8 × 10⁻¹⁰. Since Q > Ksp, the solution is supersaturated and AgCl precipitates until the ion product drops to Ksp. The diluteness of the starting solutions is irrelevant — what matters is whether the product of the mixed concentrations exceeds Ksp.

Question 3

The common ion effect decreases the solubility of a sparingly soluble salt by lowering the Ksp of that salt.

• T. True
• F. False

Answer: False

Ksp is a thermodynamic equilibrium constant that depends only on temperature — it cannot be changed by adding ions to the solution. The common ion effect decreases solubility by shifting the dissolution equilibrium to the left (toward more solid), not by altering Ksp itself. The same Ksp is maintained; what changes is the equilibrium position between dissolved ions and solid. This is a Le Chatelier's principle effect on the position of equilibrium, not a change in the equilibrium constant.

Question 4

If Q < Ksp for a sparingly soluble salt, more of that salt will dissolve into the solution if added.

• T. True
• F. False

Answer: True

Q < Ksp means the solution is unsaturated — the ion product is below the equilibrium value, so the system has not yet reached saturation. The dissolution reaction proceeds forward (dissolving more solid) to increase ion concentrations until Q = Ksp. Only when Q = Ksp does the rate of dissolution equal the rate of precipitation and net dissolution ceases. If Q > Ksp, the system is supersaturated and precipitation occurs instead.

Question 5

Explain the conceptual role of Q (the reaction quotient) in predicting whether a precipitate will form when two solutions are mixed.

Model answer: Q is calculated from the actual ion concentrations immediately after mixing, using the same mathematical form as Ksp. Comparing Q to Ksp indicates the system's direction of change: if Q < Ksp the solution is unsaturated and no precipitate forms; if Q = Ksp the solution is at equilibrium; if Q > Ksp the ion product exceeds what equilibrium can sustain, and solid precipitates until ion concentrations decrease enough that Q equals Ksp. Q is the snapshot of the system's current state, while Ksp is the target equilibrium state.

The Q-vs-Ksp comparison generalizes across all equilibria, not just precipitation. In the context of sparingly soluble salts, it provides a quantitative prediction that memorized solubility rules cannot. Any mixing problem — whether two solutions are combined or a common ion is added to a saturated solution — can be handled by calculating Q and comparing it to Ksp.