Questions: Solubility Product Constant (Ksp) and Equilibrium

Question 1

For PbCl₂ dissolving as PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq), if the molar solubility is s, which Ksp expression is correct?

• A. Ksp = s × s = s²
• B. Ksp = s × 2s = 2s²
• C. Ksp = s × (2s)² = 4s³
• D. Ksp = (2s)³ = 8s³

Answer: (C)

When s moles of PbCl₂ dissolve, [Pb²⁺] = s and [Cl⁻] = 2s (twice as much Cl⁻ because of the 2:1 stoichiometry). The Ksp expression is [Pb²⁺][Cl⁻]², giving s × (2s)² = s × 4s² = 4s³. The most common error is forgetting to square the chloride concentration — writing s × 2s = 2s² misses that the exponent in the Ksp expression equals the stoichiometric coefficient, not a multiplier.

Question 2

A student mixes solutions of AgNO₃ and NaCl. The actual ion concentrations at the moment of mixing give an ion product Q = 5.0 × 10⁻⁸ for AgCl (Ksp = 1.8 × 10⁻¹⁰). What will happen?

• A. No precipitate forms because Q is a small number
• B. No precipitate forms because Q > Ksp means the solution is unsaturated
• C. AgCl precipitates because Q > Ksp — the solution is supersaturated
• D. AgCl precipitates because Q < Ksp — the solution is unsaturated

Answer: (C)

When Q > Ksp, the ion concentrations exceed equilibrium values and the system drives toward precipitation to reduce Q back to Ksp. Option B reverses the logic: Q > Ksp signals supersaturation (too many ions), not unsaturation. Option D has both the comparison and conclusion wrong. The ion product Q is calculated identically to Ksp but uses actual concentrations rather than equilibrium values — this comparison is the core tool for predicting whether precipitation occurs.

Question 3

A salt with Ksp = 1.0 × 10⁻¹² (1:1 stoichiometry) is generally more soluble than a salt with Ksp = 4.0 × 10⁻¹² (1:2 stoichiometry, like AB₂).

• T. True
• F. False

Answer: False

You cannot compare molar solubilities by comparing Ksp values alone when stoichiometries differ. For the 1:1 salt: Ksp = s², so s = 1.0 × 10⁻⁶ M. For the 1:2 salt: Ksp = 4s³, so s³ = 1.0 × 10⁻¹², giving s = 1.0 × 10⁻⁴ M — one hundred times more soluble despite its Ksp being only four times larger. The stoichiometric coefficient raises both the concentration and the exponent, making higher-stoichiometry salts relatively more soluble than their Ksp values suggest.

Question 4

When Q < Ksp for a sparingly soluble salt, more solid will dissolve if present.

• T. True
• F. False

Answer: True

Q < Ksp means the ion concentrations are below their equilibrium values — the solution is unsaturated. If undissolved solid is present, dissolution will continue (the forward reaction dominates) until ion concentrations rise to the point where Q = Ksp and saturation is reached. This is the direct application of Le Chatelier's principle to dissolution equilibria.

Question 5

Why can you not determine which of two ionic salts is more soluble simply by comparing their Ksp values?

Model answer: Ksp values are only directly comparable when the salts have the same stoichiometry (same ion ratio). When stoichiometries differ, the relationship between Ksp and molar solubility s involves different algebraic forms — for a 1:1 salt, Ksp = s²; for a 1:2 salt, Ksp = 4s³. A salt with a larger Ksp can therefore have lower actual solubility than a salt with a smaller Ksp if the stoichiometry raises both the concentration and the exponent in the Ksp expression.

The key is that Ksp is the product of ion concentrations raised to their stoichiometric powers — not just concentrations. A 1:2 salt (AB₂) has Ksp = [A²⁺][B⁻]² = 4s³, which grows much faster with s than does Ksp = s² for a 1:1 salt. So even a modest s value gives a large Ksp for a 1:2 salt. Always solve for s from the Ksp expression; never rank solubilities by Ksp alone unless stoichiometries match.