Solubility Product ($K_{sp}$)

When a sparingly soluble ionic compound is placed in water, a dynamic equilibrium is established between the undissolved solid and the ions in solution. Even substances we commonly call 'insoluble' dissolve to a tiny extent, releasing ions into the aqueous phase until the rate of dissolution equals the rate of recrystallisation. This equilibrium is described quantitatively by the solubility product constant, KspK_{sp}, which is the product of the ionic concentrations at saturation, each raised to the power of its stoichiometric coefficient. The concept of KspK_{sp} is enormously important in analytical chemistry, environmental science, and medicine. It allows chemists to predict whether a precipitate will form when two solutions are mixed, to understand why kidney stones develop, and to design processes for removing toxic heavy-metal ions from wastewater. Because KspK_{sp} is a true equilibrium constant, it depends only on temperature — not on the amounts of solid present or the volume of solution. Understanding KspK_{sp} also requires mastering the common ion effect: adding a soluble salt that shares an ion with a sparingly soluble compound shifts the solubility equilibrium, dramatically reducing how much of the sparingly soluble compound dissolves. This principle underpins gravimetric analysis and many industrial precipitation processes.

Key takeaways
  • Writing the Dissolution Equilibrium
  • Deriving the General KspK_{sp} Expression
  • Relating Molar Solubility ss to KspK_{sp}
  • The Common Ion Effect
  • Precipitation Criterion: The Reaction Quotient QspQ_{sp}

Writing the Dissolution Equilibrium

For a generic sparingly soluble salt \ceMaXb\ce{M_aX_b}, the dissolution equilibrium is: \ceMaXb(s)<=>aMn+(aq)+bXm−(aq)\ce{M_aX_b(s) <=> a M^{n+}(aq) + b X^{m-}(aq)}. The solid is in its standard state so it does not appear in the equilibrium expression. For example, silver chloride dissolves as \ceAgCl(s)<=>Ag+(aq)+Cl−(aq)\ce{AgCl(s) <=> Ag^+(aq) + Cl^-(aq)}, giving Ksp=[Ag+][Cl−]K_{sp} = [\text{Ag}^+][\text{Cl}^-].

Deriving the General KspK_{sp} Expression

For \ceMaXb(s)<=>aMn+(aq)+bXm−(aq)\ce{M_aX_b(s) <=> a M^{n+}(aq) + b X^{m-}(aq)}, the solubility product is Ksp=[Mn+]a[Xm−]bK_{sp} = [\text{M}^{n+}]^a [\text{X}^{m-}]^b. For calcium phosphate, \ceCa3(PO4)2(s)<=>3Ca2+(aq)+2PO43−(aq)\ce{Ca3(PO4)2(s) <=> 3 Ca^{2+}(aq) + 2 PO4^{3-}(aq)}, so Ksp=[Ca2+]3[PO43−]2K_{sp} = [\text{Ca}^{2+}]^3 [\text{PO}_4^{3-}]^2. Note that the exponents come from the stoichiometry, not from the charges.

Relating Molar Solubility ss to KspK_{sp}

The molar solubility ss is the number of moles of solute that dissolve per litre of solution to form a saturated solution. For \ceAgCl\ce{AgCl}, if ss mol/L dissolves, then [Ag+]=s[\text{Ag}^+] = s and [Cl−]=s[\text{Cl}^-] = s, so Ksp=s2K_{sp} = s^2, giving s=Ksps = \sqrt{K_{sp}}. For \ceAg2CrO4(s)<=>2Ag+(aq)+CrO42−(aq)\ce{Ag2CrO4(s) <=> 2 Ag^+(aq) + CrO4^{2-}(aq)}: [Ag+]=2s[\text{Ag}^+] = 2s, [CrO42−]=s[\text{CrO}_4^{2-}] = s, so Ksp=(2s)2(s)=4s3K_{sp} = (2s)^2(s) = 4s^3, giving s=(Ksp4)1/3s = \left(\frac{K_{sp}}{4}\right)^{1/3}.

The Common Ion Effect

If a common ion is already present in solution, Le Chatelier's principle predicts that the equilibrium shifts left, reducing solubility. For \ceAgCl\ce{AgCl} dissolving in 0.10 M \ceNaCl0.10\text{ M }\ce{NaCl}: let s′s' be the new solubility. Then [Ag+]=s′[\text{Ag}^+] = s' and [Cl−]=0.10+s′≈0.10[\text{Cl}^-] = 0.10 + s' \approx 0.10 (since s′≪0.10s' \ll 0.10). So Ksp=s′(0.10)K_{sp} = s'(0.10), giving s′=Ksp/0.10s' = K_{sp}/0.10. With Ksp(\ceAgCl)=1.8×10−10K_{sp}(\ce{AgCl}) = 1.8 \times 10^{-10}, s′=1.8×10−9s' = 1.8 \times 10^{-9} mol/L — about 100 times smaller than in pure water.

Precipitation Criterion: The Reaction Quotient QspQ_{sp}

To predict whether precipitation occurs, calculate the ion product QspQ_{sp} (same formula as KspK_{sp} but using initial concentrations before equilibrium is reached). Three outcomes: (1) Qsp<KspQ_{sp} < K_{sp}: solution is unsaturated, no precipitate forms; (2) Qsp=KspQ_{sp} = K_{sp}: solution is exactly at saturation; (3) Qsp>KspQ_{sp} > K_{sp}: solution is supersaturated and precipitation occurs until QspQ_{sp} drops to KspK_{sp}. For example, mixing 50 mL50\text{ mL} of 2×10−4 M AgNO32 \times 10^{-4}\text{ M AgNO}_3 with 50 mL50\text{ mL} of 2×10−4 M NaCl2 \times 10^{-4}\text{ M NaCl} dilutes both concentrations to 1×10−4 M1 \times 10^{-4}\text{ M}, so Qsp=(10−4)2=10−8>Ksp(\ceAgCl)=1.8×10−10Q_{sp} = (10^{-4})^2 = 10^{-8} > K_{sp}(\ce{AgCl}) = 1.8 \times 10^{-10}: a white precipitate forms.

The Key Equation

Ksp=[Mn+]a [Xm−]bK_{sp} = [\text{M}^{n+}]^{a}\,[\text{X}^{m-}]^{b}
  • K_{sp} — Solubility product constant (dimensionless (formally mol^{a+b} L^{-(a+b)})): Equilibrium constant for the dissolution of a sparingly soluble ionic compound; depends only on temperature.
  • [\text{M}^{n+}] — Molar concentration of cation (mol L^{-1}): Equilibrium concentration of the cation in the saturated solution.
  • [\text{X}^{m-}] — Molar concentration of anion (mol L^{-1}): Equilibrium concentration of the anion in the saturated solution.
  • a,\, b — Stoichiometric coefficients (dimensionless): The number of moles of each ion produced per formula unit dissolved; these become the exponents in the KspK_{sp} expression.

Worked Example

Problem. The KspK_{sp} of lead(II) iodide, \cePbI2\ce{PbI2}, is 9.8×10−99.8 \times 10^{-9} at 25 °C. (a) Calculate the molar solubility of \cePbI2\ce{PbI2} in pure water. (b) Calculate the molar solubility in a 0.0500.050 M \ceKI\ce{KI} solution.

s=1.35×10−3 mol L−1;s′=3.92×10−6 mol L−1s = 1.35 \times 10^{-3}\text{ mol L}^{-1};\quad s' = 3.92 \times 10^{-6}\text{ mol L}^{-1}

Answer. Molar solubility in pure water ≈ 1.35 × 10⁻³ mol/L; in 0.050 M KI ≈ 3.92 × 10⁻⁶ mol/L.

Figure
Diagram
Solubility Product: Common Ion Effect on [Pb²⁺][I⁻] added (common ion concentration, mol/L)Solubility s (mol/L)00.010.020.030.040.050.0605×10⁻⁷1×10⁻⁶2×10⁻⁶4×10⁻⁶s ≈ 3.9×10⁻⁶at 0.05 M I⁻s = 1.35×10⁻³(pure water)Ksp = [Pb²⁺][I⁻]² = 9.8×10⁻⁹Dissolution equilibrium:PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq)Common I⁻ suppresses solubility
Origins · 1 minJacobus Henricus van 't Hoff and how this was discovered
JH
Jacobus Henricus van 't Hoff
1852–1911 · Dutch

Van 't Hoff was the first winner of the Nobel Prize in Chemistry (1901). He developed the modern theory of chemical equilibrium and showed how equilibrium constants relate to thermodynamic quantities such as Gibbs free energy. His equation ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln K directly connects KspK_{sp} to the free energy of dissolution, providing the theoretical foundation for all solubility equilibrium calculations. Van 't Hoff originally trained as an engineer before switching to chemistry. His landmark paper on three-dimensional carbon bonding was initially ridiculed by some senior chemists but later proved completely correct and underpins all of modern stereochemistry.

  • Le Chatelier's Principle — The common ion effect is a direct application of Le Chatelier's principle: adding a common ion stresses the solubility equilibrium and shifts it toward the solid, reducing solubility.
  • Equilibrium Constant (Kc) — Ksp is a specific form of Kc for dissolution reactions; understanding how to write equilibrium expressions and interpret their magnitude is essential before studying Ksp.
  • Acid Dissociation Constant (Ka) — Ka is another ionic equilibrium constant describing partial dissociation in solution; the mathematical treatment of Ka problems (ICE tables, approximations) is directly analogous to Ksp calculations.
  • Gibbs Free Energy and Equilibrium — The relationship ΔG° = −RT ln K links Ksp to thermodynamic spontaneity, explaining why Ksp changes with temperature and why some salts dissolve endothermically.