A student dissolves NaCl in water to give a solution containing 0.050 M Cl⁻. She then adds AgNO₃ until [Ag⁺] = 2.0 × 10⁻⁹ M. The Ksp of AgCl is 1.8 × 10⁻¹⁰. Does a precipitate form?
AYes, because Q = [Ag⁺][Cl⁻] = (2.0×10⁻⁹)(0.050) = 1.0×10⁻¹⁰, which is less than Ksp
BNo, because Q = 1.0×10⁻¹⁰ < Ksp = 1.8×10⁻¹⁰, so the solution is still unsaturated
DNo, because the common ion (Cl⁻) from NaCl prevents any AgCl from precipitating
Q = [Ag⁺][Cl⁻] = (2.0×10⁻⁹)(0.050) = 1.0×10⁻¹⁰. Since Q = 1.0×10⁻¹⁰ < Ksp = 1.8×10⁻¹⁰, the solution is undersaturated and no precipitate forms. Precipitation only occurs when Q > Ksp (supersaturated). Option 4 is a common misconception: the common ion effect does not prevent precipitation — it shifts the equilibrium toward solid, which means a common ion makes precipitation more likely (lowering the threshold for Q > Ksp), not less.
Question 2 Multiple Choice
A student looks up Ksp values: AgCl has Ksp = 1.8×10⁻¹⁰ and PbI₂ has Ksp = 9.8×10⁻⁹. She concludes that PbI₂ is more soluble because its Ksp is larger. Is her reasoning correct?
AYes — a larger Ksp always means greater molar solubility, regardless of salt formula
BNo — Ksp values can only be directly compared for salts with the same stoichiometric formula type; she must calculate molar solubility (x) for each salt individually
CYes — both are sparingly soluble salts, so their Ksp values are directly comparable
DNo — Ksp comparisons are only meaningful for salts that share a common ion
For AgCl (1:1 formula): Ksp = x², so x = √(1.8×10⁻¹⁰) ≈ 1.3×10⁻⁵ M. For PbI₂ (1:2 formula): Ksp = (x)(2x)² = 4x³, so x = (Ksp/4)^(1/3) = (9.8×10⁻⁹/4)^(1/3) ≈ 1.3×10⁻³ M. PbI₂ is indeed about 100 times more soluble — but this is because the different stoichiometry changes the relationship between Ksp and x, not simply because the Ksp is larger. A 1:2 salt can have a larger Ksp than a 1:1 salt yet still be less soluble, depending on the specific values. You must always solve for molar solubility.
Question 3 True / False
Adding NaCl to a saturated AgCl solution at equilibrium will cause additional AgCl to precipitate out of solution.
TTrue
FFalse
Answer: True
This is the common ion effect, a direct application of Le Chatelier's principle. The dissolution equilibrium is AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). Adding NaCl introduces additional Cl⁻ ions, increasing [Cl⁻] above the equilibrium value and making Q > Ksp. The system responds by shifting left — toward the solid — until equilibrium is re-established at a lower [Ag⁺]. The Ksp does not change, but the solubility of AgCl (the amount that dissolves per liter) decreases substantially. This is why ionic compounds are far less soluble in solutions containing a common ion than in pure water.
Question 4 True / False
A compound with a very small Ksp (e.g., 10⁻³⁰) releases no detectable ions into solution — it is effectively insoluble.
TTrue
FFalse
Answer: False
No ionic compound is truly insoluble — equilibrium always requires some dissolved ions to be present. For AgI with Ksp ≈ 8×10⁻¹⁷, the molar solubility in pure water is √(8×10⁻¹⁷) ≈ 9×10⁻⁹ M — extremely small but not zero. Even compounds with Ksp ~ 10⁻³⁰ have some ions in solution at equilibrium, just at concentrations far below what is analytically detectable. 'Sparingly soluble' and 'slightly soluble' are the correct technical terms; 'insoluble' is a practical simplification, not a physical truth. This distinction matters in contexts like selective precipitation, where tiny differences in solubility are exploited analytically.
Question 5 Short Answer
Explain why AgCl dissolves to a much smaller extent in a 0.10 M NaCl solution than in pure water, even though the Ksp of AgCl has not changed.
Think about your answer, then reveal below.
Model answer: Ksp is a constant: [Ag⁺][Cl⁻] must equal Ksp = 1.8×10⁻¹⁰ at equilibrium. In pure water, both ions come only from AgCl dissolution, so [Ag⁺] = [Cl⁻] = x ≈ 1.3×10⁻⁵ M. In 0.10 M NaCl, the solution already contains [Cl⁻] = 0.10 M from the salt. To satisfy Ksp, [Ag⁺] = Ksp/[Cl⁻] = 1.8×10⁻¹⁰/0.10 = 1.8×10⁻⁹ M. The solubility of AgCl has dropped from 1.3×10⁻⁵ M to 1.8×10⁻⁹ M — about 7,000-fold reduction. The Ksp is unchanged; what changes is how much AgCl needs to dissolve to reach equilibrium when one ion is already present at high concentration.
This is the common ion effect expressed quantitatively. Le Chatelier's principle describes the direction (equilibrium shifts toward solid), and the Ksp expression gives the magnitude (the ion product must still equal Ksp). The key conceptual point is that Ksp constrains the product of ion concentrations, so if one concentration is forced high by an external source, the other must be proportionally low — meaning less of the compound can dissolve.