Deriving Ksp from Molar Solubility

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Solubility (s) and the solubility product constant (Ksp) describe how much of a sparingly soluble ionic salt dissolves in water at a given temperature. For a salt $A_xB_y$ in equilibrium with its saturated solution, $A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq)$, the equilibrium expression is Ksp = [A^(y+)]^x · [B^(x-)]^y, and the pure solid and water are omitted. If s mol/L of salt dissolves, then [A^(y+)] = xs and [B^(x-)] = ys, giving Ksp = x^x · y^y · s^(x+y). The ion product Q uses actual ion concentrations: when Q > Ksp precipitation occurs, Q = Ksp the solution is saturated, Q < Ksp it is unsaturated. A common ion (senama) shifts the equilibrium left and lowers s; for example, adding Cl⁻ to a saturated AgCl solution drops s = Ksp/[Cl⁻]. Lowering pH dissolves salts whose anions are conjugate bases of weak acids (CO₃²⁻, S²⁻, OH⁻ from weak bases). UTBK frequently tests stoichiometric conversion between s and Ksp, the Q-versus-Ksp rule, and pH effects on carbonate/sulfide/hydroxide precipitates.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Deriving Ksp from Molar Solubility

For any sparingly soluble electrolyte $A_xB_y(s)$ the heterogeneous equilibrium has only aqueous species in the mass-action expression:

$$K_{sp} = [A^{y+}]^x [B^{x-}]^y$$

If s mol/L dissolves, the stoichiometry gives [A^(y+)] = xs and [B^(x-)] = ys, so:

$$K_{sp} = (xs)^x (ys)^y = x^x y^y s^{(x+y)}$$

Worked examples:

AgCl (1:1): Ksp = s² → s = √Ksp
Ag₂CrO₄ (2:1): Ksp = (2s)²(s) = 4s³ → s = ∛(Ksp/4)
Ca₃(PO₄)₂ (3:2): Ksp = (3s)³(2s)² = 108 s⁵ → s = ⁵√(Ksp/108)

This shows the inverse relationship: the larger the stoichiometric exponents, the smaller s for the same Ksp.

Predicting Precipitation: Q vs Ksp

Compute the ion product Q with actual concentrations (not equilibrium ones). The decision rule:

Condition	State	Observation
Q < Ksp	Unsaturated	No precipitate
Q = Ksp	Saturated	Equilibrium; incipient precipitate
Q > Ksp	Supersaturated	Precipitate forms until Q = Ksp

Common-Ion Effect and pH Influence

Adding a senama ion (e.g., NaCl into a saturated AgCl solution) raises one ion’s concentration, forcing Q above Ksp, so the salt precipitates until s falls. Approximate: s’ ≈ s / (1 + [common ion]/Ksp). Acidic pH raises the solubility of salts whose anions are conjugate bases of weak acids: H⁺ protonates CO₃²⁻ → HCO₃⁻ → H₂CO₃, removing anion from solution and pulling the dissolution equilibrium rightward. The same logic applies to S²⁻, OH⁻ from weak bases, and PO₄³⁻.

Typical UTBK Question Types

Calculate s from a given Ksp, or vice versa (multiple-choice with three stoichiometries).
Decide whether mixing two solutions forms a precipitate by comparing Q with Ksp.
Determine the minimum [anion] (often from a weak acid’s Kₐ and pH) needed to start precipitation of a metal hydroxide or sulfide — the basis of qualitative cation group analysis.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Subtler Mechanisms

1. Hydroxide precipitation requires pH control. To precipitate M(OH)ₙ fully, the hydroxide concentration must reach [OH⁻] = ⁿ√(Ksp / [M^n⁺]). Converting this to pH lets you compute the minimum pH for quantitative removal of a metal ion in analytical chemistry — a classic UTBK two-step problem combining Ksp and Kᵥ of water.

2. Selective precipitation. When two cations share the same anion (e.g., both form sulfides), you can separate them by controlling [S²⁻]. The ion that precipitates first is the one whose required [S²⁻] = Ksp / [M^n⁺] is reached first as the precipitant is added — provided the two Ksp values differ by at least 10⁴. This is the principle behind group I–V cation analysis.

3. Simultaneous equilibria and complex ions. AgCl dissolves in NH₃ because Ag⁺ forms [Ag(NH₃)₂]⁺, removing Ag⁺ from solution and forcing the AgCl equilibrium right. The “effective” solubility becomes s = √(Ksp · (1 + β₂[NH₃]²)) where β₂ is the formation constant. Conversely, adding CN⁻ dissolves AgCN, while adding Cl⁻ to Ag₂CrO₄ reduces s without forming complexes.

4. Temperature dependence. Dissolution of most salts is endothermic, so Ksp and s rise with T; a few (e.g., Ca(OH)₂) are exothermic and become less soluble when heated.

Common Mistakes Examined

Forgetting the stoichiometric exponents: writing Ksp = s² for Ag₂CrO₄ is the single most common UTBK error.
Including [AgCl(s)] or [H₂O] in the equilibrium expression — both are constant or pure and must be omitted.
Confusing molar solubility s with ionic concentration: in a saturated Ag₂CrO₄ solution, [Ag⁺] = 2s while s itself is the moles of salt per litre.
Treating pH as irrelevant for hydroxides, sulfides, and carbonates; the UTBK routinely exploits this by giving pH data and asking for conditional solubility.

Practice Prompts

Ksp(AgCl) = 1.8 × 10⁻¹⁰ at 25 °C. Find s in pure water and in 0.10 M NaCl. (Answer: s ≈ 1.34 × 10⁻⁵ M pure; 1.8 × 10⁻⁹ M with common ion.)
500 mL of 0.0010 M Ca(NO₃)₂ is mixed with 500 mL of 0.0020 M NaF. Ksp(CaF₂) = 1.5 × 10⁻¹⁰. Does a precipitate form? (Final [Ca²⁺] = 5.0 × 10⁻⁴ M, [F⁻] = 1.0 × 10⁻³ M; Q = 5.0 × 10⁻¹⁰ > Ksp → yes, CaF₂ precipitates.)

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Deriving Ksp from Molar Solubility