The solubility product

Result

The solubility product

T-048Home CU-203Threads equilibrium

Statement

Equilibria of sparingly soluble ionic solids.

Why it matters

law-of-mass-action established the general equilibrium-constant expression for any reversible reaction; the solubility product \(K_{sp}\) is simply that same expression applied to the specific case of a sparingly soluble ionic solid dissolving into its constituent ions, and it is the standard quantitative tool for predicting whether a precipitate will form, how much of a solid will dissolve, and how that solubility responds to other dissolved species already present.

Because \(K_{sp}\) governs which solid phase is thermodynamically stable in a given solution, it underlies practical problems ranging from water hardness and kidney-stone formation to the selective precipitation used in classical qualitative and quantitative analytical chemistry, making it one of the most directly applied results in the whole equilibrium unit.

Hypotheses

The undissolved solid is genuinely present in excess, in direct contact with its saturated solution.The solubility-product equilibrium describes a two-phase, heterogeneous system; if all of the solid has already dissolved (an unsaturated solution), the system is not at this equilibrium at all, and \(K_{sp}\) does not directly constrain the (sub-equilibrium) ion concentrations present. The pure solid's own concentration (activity) is treated as constant and folded into the equilibrium constant, per law-of-mass-action's convention for heterogeneous equilibria.A pure solid's "concentration" does not meaningfully change as more or less of it is present, so, exactly as for any other pure solid or liquid phase in an equilibrium expression, it is omitted entirely from the \(K_{sp}\) expression, which is written purely in terms of the dissolved ionic species. The concentrations appearing in the \(K_{sp}\) expression are formally activities, assumed here to equal molar concentrations; this ideal-solution approximation can break down measurably in solutions of appreciable ionic strength, where ion–ion interactions make true activities differ from raw concentrations.

Proof

\text{M}_a\text{X}_b(s) \rightleftharpoons a\,\text{M}^{n+}(aq) + b\,\text{X}^{m-}(aq)

A sparingly soluble ionic solid establishes a genuine, dynamic dissolution equilibrium with its own saturated solution, with dissolution and re-precipitation proceeding at equal rates once equilibrium is reached. A

K_{sp}=[\text{M}^{n+}]^a[\text{X}^{m-}]^b

Applying law-of-mass-action's general equilibrium-constant expression to Step 1, and omitting the pure solid per the Hypotheses' heterogeneous-equilibrium convention, gives the solubility-product expression directly, in terms only of the dissolved ion concentrations. A

[\text{M}^{n+}]=as,\ [\text{X}^{m-}]=bs \Rightarrow K_{sp}=(as)^a(bs)^b, \qquad s=\text{molar solubility}

If the only source of both ions is the dissolving solid itself, their concentrations are fixed by the solid's own stoichiometry in terms of a single unknown, the molar solubility \(s\); substituting into Step 2 gives an algebraic equation for \(s\) directly from \(K_{sp}\) alone. A

\text{Common ion effect: adding excess } \text{X}^{m-}\text{ (concentration } c\text{) suppresses } s' \text{ below the value found in Step 3.}

Because \(K_{sp}\) is fixed at a given temperature, raising \([\text{X}^{m-}]\) by adding an external source of that ion forces \([\text{M}^{n+}]\) down to keep the product constant — a direct application of Le Chatelier-type reasoning to this specific heterogeneous equilibrium, reducing the solid's net solubility below its value in pure solvent. A

\text{Precipitation criterion: compare } Q=[\text{M}^{n+}]^a[\text{X}^{m-}]^b \text{ (instantaneous) against } K_{sp}.

If \(QK_{sp}\) the solution is supersaturated and precipitation occurs until \(Q\) falls back to \(K_{sp}\) — the same reaction-quotient-versus-equilibrium-constant logic used generally to predict the direction of any net change. A

Result

K_{sp}=[\text{M}^{n+}]^a[\text{X}^{m-}]^b \qquad\qquad QK_{sp}:\text{precipitates}

Reading. A single, fixed equilibrium constant, formed from the dissolved-ion concentrations raised to their stoichiometric powers, governs both how much of a sparingly soluble solid dissolves and whether mixing two solutions will trigger precipitation.

Scope. Requires excess undissolved solid genuinely present at equilibrium (Hypotheses); ordinary \(K_{sp}\) calculations assume no additional equilibria (complex formation, pH-dependent speciation) compete for the same ions, which can otherwise substantially increase the apparent solubility beyond the simple prediction.

Corollaries & converses

Selective precipitation exploits differing \(K_{sp}\) values among several possible precipitates sharing a common ion: adding the shared reagent gradually precipitates the least soluble salt first, a standard analytical separation technique built directly on Step 5's criterion.
le-chatelier's general principle, applied specifically to this heterogeneous ionic equilibrium, is exactly what underlies the common ion effect (Step 4) — the same qualitative shifting-equilibrium logic used throughout the unit, here made quantitative.
kp-kc-relation applies the identical law-of-mass-action logic to gas-phase equilibria; \(K_{sp}\) is the same general framework specialised instead to heterogeneous solid–solution equilibria.

Fails without

Drop the requirement that undissolved solid genuinely be present (Hypotheses): if all of the solid has already dissolved, the system is not at the \(K_{sp}\) equilibrium at all, and \(K_{sp}\) no longer directly constrains the (sub-equilibrium) ion concentrations actually present in solution.
Include the solid's own concentration in the expression, violating the heterogeneous-equilibrium convention: this produces a "constant" that would depend on how much excess solid happens to be present, rather than the genuine, fixed equilibrium constant \(K_{sp}\) is meant to be.

Common errors

Including the solid's own "concentration" in the \(K_{sp}\) expression, rather than omitting it entirely per the heterogeneous-equilibrium convention (Step 2).
Forgetting the stoichiometric coefficients when relating molar solubility \(s\) to \(K_{sp}\) for a salt that is not 1:1, e.g. writing \(K_{sp}=s^2\) for a salt \(\text{MX}_2\) instead of the correct \(K_{sp}=4s^3\) (Step 3).
Assuming any additive that raises ionic strength must reduce solubility exactly like a genuine common ion — the common ion effect specifically requires the added ion to be one of the salt's own dissolved species, not merely any dissolved electrolyte.
Directly comparing raw \(K_{sp}\) values of two salts with different ion-ratio stoichiometries to rank which is "more soluble" — this comparison is only valid between salts of the same stoichiometric type (Problems gives a worked contrast).

Discussion

Solubility-product calculations remain the standard quantitative language for a wide range of practical chemistry: geological mineral formation and dissolution, the precipitation reactions used throughout classical qualitative analysis to separate metal ions into groups, and physiological processes such as kidney-stone formation, where a locally elevated ion product exceeding the relevant \(K_{sp}\) drives crystallisation of a sparingly soluble salt within the body.

Common misconception: that a larger \(K_{sp}\) always means "more soluble," full stop. This comparison is only reliable between salts sharing the same stoichiometric type (both 1:1, or both 1:2, and so on); comparing across different stoichiometries (Step 3's exponents differ) can reverse the apparent ranking, since the relationship between \(K_{sp}\) and \(s\) is not the same functional form in each case.

Worked examples

\text{AgCl(s)} \rightleftharpoons \text{Ag}^+(aq)+\text{Cl}^-(aq), \quad K_{sp}\approx1.8\times10^{-10}\ (\text{a typical, widely tabulated value})

For this 1:1 salt, \([\text{Ag}^+]=[\text{Cl}^-]=s\) (Step 3), so \(K_{sp}=s^2\); solving gives \(s=\sqrt{1.8\times10^{-10}}\approx1.34\times10^{-5}\,\text{M}\), the molar solubility of silver chloride in pure water. A

\text{Same salt, in 0.10 M NaCl: } K_{sp}\approx[\text{Ag}^+](0.10) \Rightarrow s'\approx1.8\times10^{-9}\,\text{M}

With chloride already present at \(0.10\,\text{M}\) from the added common-ion source, and \(s'\ll0.10\), the total \([\text{Cl}^-]\) is dominated by the added NaCl rather than the small additional amount from AgCl itself; solving \(K_{sp}=[\text{Ag}^+](0.10)\) directly for \([\text{Ag}^+]=s'\) shows solubility suppressed by roughly four orders of magnitude relative to pure water (Step 4). A

s(\text{AgCl, pure water})\approx1.34\times10^{-5}\,\text{M} \quad\to\quad s'(\text{AgCl, 0.10 M NaCl})\approx1.8\times10^{-9}\,\text{M}

Reading. A modest concentration of a shared ion, added from an entirely separate source, suppresses a sparingly soluble salt's solubility by several orders of magnitude, a direct, quantitative demonstration of the common ion effect.

Scope. The identical two-step procedure (write \(K_{sp}\), substitute known or unknown ion concentrations) applies to any sparingly soluble salt whose \(K_{sp}\) is known, with or without a common ion present.

Problems

Given \(K_{sp}(\text{PbI}_2)\approx7.1\times10^{-9}\) (a typical, widely tabulated value), find the molar solubility of \(\text{PbI}_2\) in pure water.
Solution
\(\text{PbI}_2\rightleftharpoons\text{Pb}^{2+}+2\text{I}^-\), so \([\text{Pb}^{2+}]=s\), \([\text{I}^-]=2s\), giving \(K_{sp}=s(2s)^2=4s^3\). Solving, \(s^3=7.1\times10^{-9}/4=1.78\times10^{-9}\), \(s\approx1.21\times10^{-3}\,\text{M}\).
Solutions of \(0.010\,\text{M Pb(NO}_3)_2\) and \(0.020\,\text{M NaI}\) are mixed in equal volumes. Determine whether a precipitate of \(\text{PbI}_2\) forms, given \(K_{sp}\approx7.1\times10^{-9}\).
Solution
Mixing equal volumes halves each concentration: \([\text{Pb}^{2+}]=0.0050\,\text{M}\), \([\text{I}^-]=0.010\,\text{M}\). \(Q=[\text{Pb}^{2+}][\text{I}^-]^2=(0.0050)(0.010)^2=5.0\times10^{-7}\), which is far larger than \(K_{sp}=7.1\times10^{-9}\); since \(Q>K_{sp}\) (Step 5), a precipitate of \(\text{PbI}_2\) does form.
Two hypothetical salts, \(\text{AB}\) with \(K_{sp}=1.0\times10^{-8}\) and \(\text{AB}_2\) with \(K_{sp}=1.0\times10^{-8}\) (numerically equal), are compared. Explain why it is incorrect to conclude they have equal molar solubility.
Solution
For \(\text{AB}\), \(K_{sp}=s^2\), giving \(s=\sqrt{10^{-8}}=1.0\times10^{-4}\,\text{M}\). For \(\text{AB}_2\), \(K_{sp}=4s^3\), giving \(s=(10^{-8}/4)^{1/3}\approx1.36\times10^{-3}\,\text{M}\), over an order of magnitude larger despite the identical \(K_{sp}\) value — exactly the Common errors' warning that raw \(K_{sp}\) values are only directly comparable between salts of the same stoichiometric type.