Solubility Product (Ksp) & Precipitation Checker

If you're running a solubility product Ksp and precipitation check and the tool says "supersaturated," it means the ion concentrations in your mixture exceed what the solid's dissolution equilibrium can sustain. The most common mistake at this step is forgetting dilution. When you mix two solutions, the total volume increases and every concentration drops. Students who plug in the original concentrations instead of the diluted ones overestimate Qsp and predict precipitation where none occurs.

The comparison itself is straightforward. Calculate Qsp the same way you'd calculate Ksp—products of ion concentrations raised to their stoichiometric coefficients—but use current concentrations, not equilibrium values. If Qsp < Ksp, the solution can hold more dissolved ions. No precipitate forms. If Qsp > Ksp, the solution is supersaturated and precipitation is thermodynamically favored. If Qsp = Ksp, you're sitting right at saturation.

"Thermodynamically favored" doesn't mean it happens instantly. Precipitation requires nucleation—tiny seed crystals must form before bulk solid can grow. A supersaturated sugar solution can sit for hours without crystallizing until you scratch the glass or drop in a seed crystal. The same kinetic barrier applies to ionic precipitates, though most common ones (AgCl, BaSO₄) nucleate fast enough that you see a cloud within seconds of mixing.

Dissolving AgCl in pure water gives a molar solubility of about 1.3 × 10⁻⁵ M (from Ksp = 1.8 × 10⁻¹⁰). But dissolve it in 0.10 M NaCl instead, and solubility drops to 1.8 × 10⁻⁹ M—roughly 7000 times less. The Cl⁻ already in solution from NaCl pushes the dissolution equilibrium AgCl(s) ⇌ Ag⁺ + Cl⁻ back toward the solid. That's the common ion effect.

Mathematically, Ksp doesn't change—it's still 1.8 × 10⁻¹⁰ at 25 °C regardless of what else is dissolved. What changes is how much additional AgCl can dissolve. If [Cl⁻] is already 0.10 M from NaCl, then at equilibrium [Ag⁺] = Ksp / [Cl⁻] = 1.8 × 10⁻¹⁰ / 0.10 = 1.8 × 10⁻⁹ M. You treat the common ion concentration as fixed (it's in huge excess compared to what AgCl contributes) and solve for the other ion.

The error students make is setting up an ICE table where both ions start at zero, ignoring the common ion already present. If [Cl⁻]₀ = 0.10 M, the change column for Cl⁻ is +S where S is the molar solubility, but since S ≈ 10⁻⁹ and [Cl⁻]₀ = 0.10, the change is negligible. You can skip the ICE table entirely and just use [Cl⁻] = 0.10 M directly. Only set up a full ICE table when the common ion concentration is comparable to S.

Molar solubility S is the number of moles of salt that dissolve per liter to reach saturation. For a 1:1 salt like AgCl (→ Ag⁺ + Cl⁻), each formula unit produces one of each ion, so [Ag⁺] = S and [Cl⁻] = S. Ksp = S × S = S². Solving: S = √Ksp. For AgCl, S = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ M.

For salts with unequal stoichiometry, the exponents matter. PbCl₂ dissociates as PbCl₂ → Pb²⁺ + 2 Cl⁻. If S moles dissolve, [Pb²⁺] = S and [Cl⁻] = 2S. Ksp = (S)(2S)² = 4S³. Solving: S = (Ksp / 4)^1/3. For Ksp = 1.7 × 10⁻⁵, S = (4.25 × 10⁻⁶)^1/3 = 0.016 M. Students who write Ksp = S³ (forgetting the coefficient 2 in front of Cl⁻) get S = 0.026 M—off by 60%.

The general formula for M_aX_b(s) ⇌ aM⁺(aq) + bX⁻(aq) is Ksp = (aS)^a(bS)^b = a^ab^bS^(a+b). Solve for S = (Ksp / a^ab^b)^1/(a+b). This formula assumes pure water with no common ion. In the presence of a common ion, you can't use this shortcut—you need to set [common ion] = known value and solve the Ksp expression algebraically.

A saturated solution is at equilibrium with undissolved solid—Qsp = Ksp exactly. If you add more solid to a saturated solution, it just sits there undissolved. If you remove some solid, nothing changes in solution because Qsp is still at Ksp. The system is stable.

A supersaturated solution has Qsp > Ksp—more ions dissolved than equilibrium allows. This state is metastable. It persists only because precipitation hasn't started yet (nucleation barrier). Once a seed crystal forms or is introduced, ions crash out of solution rapidly until Qsp drops to Ksp. The classic demo is sodium acetate "hot ice": a supersaturated solution crystallizes spectacularly on contact with a single crystal.

An unsaturated solution has Qsp < Ksp. More solid can dissolve. If you drop a crystal of the salt into an unsaturated solution, it dissolves. No precipitation occurs. The ratio Qsp/Ksp quantifies how far you are from saturation: 0.01 means very unsaturated, 0.99 means nearly saturated, 1.0 is saturated, 1.5 means 50% above Ksp. On an exam, always state which comparison you made and what it means—just writing "Qsp > Ksp" without saying "therefore precipitation occurs" may cost you the interpretation mark.

Can you compare Ksp values directly to rank solubility? Only for salts with the same stoichiometry. AgCl (Ksp = 1.8 × 10⁻¹⁰) is less soluble than AgBr (Ksp = 5.0 × 10⁻¹³)? Actually no—AgBr has a smaller Ksp, so it's less soluble. But you can't compare AgCl (1:1 salt) to PbCl₂ (1:2 salt) by Ksp alone. You need to compute molar solubility for each and compare those numbers.

Does temperature affect Ksp? Yes. Ksp is an equilibrium constant, so it depends on temperature only. Most salts have larger Ksp at higher temperatures (dissolution is usually endothermic), meaning they become more soluble when heated. A few, like Ca(OH)₂, have smaller Ksp at higher temperatures. Always check that your Ksp value matches the temperature of your problem.

Why does pH affect metal hydroxide solubility? For a salt like Fe(OH)₃, Ksp = [Fe³⁺][OH⁻]³. Lowering pH reduces [OH⁻], which reduces Qsp, pushing the system below Ksp and dissolving the solid. Raising pH increases [OH⁻], raises Qsp above Ksp, and drives precipitation. This is why metal hydroxides dissolve in acid and precipitate in base—it's all Le Chatelier via the [OH⁻] term in the Ksp expression.

What about complex ion formation? Ksp assumes the ions stay as free ions in solution. If one of them forms a complex (like Ag⁺ + 2 NH₃ → Ag(NH₃)₂⁺), the free [Ag⁺] drops, Qsp drops below Ksp, and more solid dissolves. This is why AgCl dissolves in concentrated ammonia despite having a tiny Ksp. The simple Ksp model doesn't account for this—you need to combine Ksp with the formation constant Kf.

Problem: Mix 25.0 mL of 0.0020 M AgNO₃ with 75.0 mL of 0.0040 M NaCl. Will AgCl precipitate? Ksp(AgCl) = 1.8 × 10⁻¹⁰ at 25 °C.

Qsp exceeds Ksp by over four orders of magnitude—AgCl will absolutely precipitate. If you had used the undiluted concentrations (0.0020 M and 0.0040 M), you'd get Qsp = 8.0 × 10⁻⁶, which still predicts precipitation but overstates Qsp by about 5×. On a quantitative exam, that arithmetic error costs points even though the qualitative answer happens to be the same.