Solubility product and the common-ion effect

Result

Solubility product and the common-ion effect

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Statement

Let the sparingly soluble salt \(\text{M}_a\text{X}_b\) be present as a pure macroscopic solid in equilibrium with its saturated aqueous solution at fixed temperature and pressure, dissolving congruently and completely into \(a\) cations \(\text{M}^{z_+}\) and \(b\) anions \(\text{X}^{z_-}\) with no further reaction. Then the product of the ion activities is a constant of the temperature alone, \(K_{sp}=a_+^{\,a}a_-^{\,b}=\exp(-\Delta G^{\circ}_{\mathrm{sol}}/RT)\); in the ideal-dilute limit this becomes \(K_{sp}=[\text{M}^{z_+}]^a[\text{X}^{z_-}]^b\) with concentrations measured against \(c^{\circ}=1\ \text{mol L}^{-1}\), so the molar solubility in pure solvent is \(s=(K_{sp}/a^ab^b)^{1/(a+b)}\), while an independent source of \(\text{X}^{z_-}\) at concentration \(c\) forces the solubility down to the root \(s'\) of \((as')^a(bs'+c)^b=K_{sp}\) — the common-ion effect, which in the swamping limit \(bs'\ll c\) reduces solubility as \(c^{-b/a}\).

Why it matters

Almost every quantitative statement a chemist makes about a precipitate — how much of it dissolves, whether it will form at all when two solutions are mixed, how completely it can be washed without loss, which of two metal ions can be separated from the other — is a statement about one number, \(K_{sp}\), and about the stoichiometric bookkeeping that converts that number into a concentration. The bookkeeping is exactly the mole-and-concentration arithmetic developed in the mole and Avogadro's number and concentration and dilution; the equilibrium content is the general mass-action expression of the law of mass action, specialised to a two-phase system in which one participant is a pure solid.

The common-ion effect is where the two halves meet and where most errors are made. It is not a separate law: it is the observation that the solubility \(s\) is defined by a mass balance, not by \(K_{sp}\) alone, and that when part of an ion's concentration is supplied from outside, the mass balance changes while \(K_{sp}\) does not. Getting this right is what makes gravimetric analysis quantitative (a wash liquid containing the precipitating reagent loses far less analyte than pure water), what sets the residual metal-ion concentration in a selective precipitation, and what explains why hard water deposits scale, why fluoride protects enamel, and why a kidney stone grows in one patient's urine and not another's at the same total calcium.

It also fails in two instructive ways — an inert electrolyte, which shares no ion at all, increases solubility; and a large excess of the common ion can reverse the suppression entirely through complex formation — and those failures are the cleanest available demonstration that \(K_{sp}\) is a statement about activities and about one particular reaction, not a statement about solubility as such.

Hypotheses

A pure, macroscopic, undissolved solid of the stated stoichiometry is present at equilibrium.The solid's activity is set to one because its standard state is the pure phase at the temperature and pressure of interest; if the last crystal has dissolved the system is unsaturated and \(K_{sp}\) yields only the inequality \([\text{M}^{z_+}]^a[\text{X}^{z_-}]^b\lt K_{sp}\), not an equation. If the solid is a nanocrystalline powder, the extra surface free energy raises its chemical potential above that of the bulk phase and the measured ion product exceeds \(K_{sp}\) (the Ostwald–Freundlich effect), which is why fine precipitates ripen into coarse ones on standing.

Dissolution is congruent and complete: the solid releases only \(\text{M}^{z_+}\) and \(\text{X}^{z_-}\), in the ratio \(a:b\).Any ion pairing (\(\text{CaSO}_4(aq)\) is a real, appreciable species), protonation of the anion, hydrolysis of the cation, or complexation removes ions from the two species that appear in \(K_{sp}\) without removing them from solution, so the measured total solubility exceeds the value computed here — sometimes by orders of magnitude.

The solution is ideal-dilute, so that every activity coefficient may be set to one.This is the weakest hypothesis in the list. At an ionic strength of only \(0.010\ \text{mol L}^{-1}\) the mean ionic activity coefficient of a 1:1 salt is already about \(0.89\), and the measured solubility of AgCl is some \(12\%\) above the ideal prediction; for a 2:2 salt the deviation at the same ionic strength is far larger. The correction is supplied by the Debye–Huckel theory and is developed in Step 5.

Temperature and pressure are fixed, and \(K_{sp}\) is evaluated at that temperature.\(K_{sp}\) is not a material constant but a value of \(\exp(-\Delta G^{\circ}_{\mathrm{sol}}/RT)\); it varies with temperature according to the van 't Hoff relation \(\mathrm{d}\ln K_{sp}/\mathrm{d}T=\Delta H^{\circ}_{\mathrm{sol}}/RT^{2}\), and for salts with exothermic dissolution (\(\text{CaSO}_4\), \(\text{Ca(OH)}_2\)) it decreases as the solution is heated. Tabulated values are almost always for \(298.15\ \text{K}\).

The added common-ion source is a fully dissociated strong electrolyte whose counter-ion is a spectator.If the counter-ion itself reacts — adding \(\text{Na}_2\text{CO}_3\) to suppress \(\text{CaCO}_3\) also raises the pH, and adding \(\text{HCl}\) to suppress \(\text{AgCl}\) also changes the medium — then the single-variable mass balance of Step 8 is incomplete and the full set of simultaneous equilibria must be solved instead.

Proof

The argument is thermodynamic at the start and purely stoichiometric at the end. Steps 1–4 establish what \(K_{sp}\) is and why the solid is absent from it; Steps 5–6 descend from activities to concentrations; Steps 7–11 are mass balance, which is where solubility — and the common-ion effect — actually comes from; Step 12 supplies the precipitation criterion.

\[ \text{M}_a\text{X}_b(s) \rightleftharpoons a\,\text{M}^{z_+}(aq) + b\,\text{X}^{z_-}(aq), \qquad \mu_{\mathrm{s}} = a\,\mu_+ + b\,\mu_- \]

At fixed \(T\) and \(p\) a closed system is at equilibrium when \(G\) is stationary against every allowed change of composition. Advancing the dissolution by \(\mathrm{d}\xi\) changes the amounts by \(-\mathrm{d}\xi,\,+a\,\mathrm{d}\xi,\,+b\,\mathrm{d}\xi\), so \((\partial G/\partial\xi)_{T,p}=a\mu_++b\mu_--\mu_{\mathrm{s}}=0\). A

\[ \mu_i = \mu_i^{\circ} + RT\ln a_i \quad\Longrightarrow\quad \mu_{\mathrm{s}}^{\circ}+RT\ln a_{\mathrm{s}} = a\big(\mu_+^{\circ}+RT\ln a_+\big) + b\big(\mu_-^{\circ}+RT\ln a_-\big) \]

Insert the exact definition of activity for every species. This is a definition, not an approximation: \(a_i\) is whatever makes the equation true, and the content lies entirely in the choice of standard state \(\mu_i^{\circ}\). A

\[ a_{\mathrm{s}} = 1 \quad\Longrightarrow\quad \Delta G^{\circ}_{\mathrm{sol}} \equiv a\mu_+^{\circ}+b\mu_-^{\circ}-\mu_{\mathrm{s}}^{\circ} = -RT\ln\!\big(a_+^{\,a}a_-^{\,b}\big) \]

The standard state of a pure condensed phase is that phase itself, so its activity is exactly one whenever it is present, whatever its amount or its degree of subdivision. This — and not the folk explanation that “the concentration of a solid is constant” — is why the solid never appears in the solubility product; a solid has no concentration in solution to be constant. Rearranging Step 2 and collecting the standard chemical potentials defines the standard Gibbs energy of solution. B

\[ K_{sp} \equiv a_+^{\,a}\,a_-^{\,b} = \exp\!\left(-\frac{\Delta G^{\circ}_{\mathrm{sol}}}{RT}\right) \]

Exponentiating Step 3. Three consequences follow immediately: \(K_{sp}\) is dimensionless (activities are ratios to a standard state), it depends on temperature and pressure but on nothing about the particular solution, and it is a special case of the general mass-action constant of the law of mass action and of \(\Delta G^{\circ}=-RT\ln K\) from free energy and the equilibrium constant. A

\[ a_i = \gamma_i\,\frac{c_i}{c^{\circ}},\qquad \gamma_{\pm}^{\,a+b} \equiv \gamma_+^{\,a}\gamma_-^{\,b} \quad\Longrightarrow\quad K_{sp} = \gamma_{\pm}^{\,a+b}\left(\frac{c_+}{c^{\circ}}\right)^{\!a}\left(\frac{c_-}{c^{\circ}}\right)^{\!b} \]

Adopt the molar-concentration scale with \(c^{\circ}=1\ \text{mol L}^{-1}\). Single-ion activity coefficients are not separately measurable — no experiment can add cations without anions — so they are collected into the mean ionic coefficient \(\gamma_{\pm}\) defined by the geometric mean above, which is measurable. Defining the concentration product \(K_{sp}^{c}=K_{sp}/\gamma_{\pm}^{\,a+b}\) shows that the quantity chemists actually compute with grows as \(\gamma_{\pm}\) falls: raising the ionic strength raises the ion concentrations a saturated solution can hold. C

\[ \gamma_{\pm}\to 1 \ \ (\text{ideal-dilute}) \quad\Longrightarrow\quad K_{sp} = [\text{M}^{z_+}]^a[\text{X}^{z_-}]^b \]

The working form. From here on square brackets denote molarity divided by \(c^{\circ}\), so the expression stays dimensionless; the hypothesis being used is the third one, and Step 5 and the first “fails without” regime say what it costs. A

\[ [\text{M}^{z_+}]=as,\quad [\text{X}^{z_-}]=bs \quad\Longrightarrow\quad K_{sp}=(as)^a(bs)^b=a^ab^b\,s^{a+b} \]

Mass balance for a saturated solution in pure solvent: if \(s\) moles of the salt dissolve per litre and every formula unit releases \(a\) cations and \(b\) anions, the ion concentrations are fixed multiples of the single unknown \(s\). Note that this step, not Step 6, is where solubility enters; \(K_{sp}\) knows nothing about \(s\) until a mass balance is imposed. A

\[ s = \left(\frac{K_{sp}}{a^a b^b}\right)^{1/(a+b)} \]

Solve Step 7 for \(s\), symbols before numbers. The exponent \(1/(a+b)\) is the reason two salts of different stoichiometric type cannot be ranked by \(K_{sp}\): the map from \(K_{sp}\) to \(s\) is a different function for each type, and it is monotonic only within a type. A

\[ \text{add } \text{X}^{z_-} \text{ at concentration } c \ \Longrightarrow\ [\text{M}^{z_+}]=as',\ \ [\text{X}^{z_-}]=bs'+c \ \Longrightarrow\ (as')^a\,(bs'+c)^b = K_{sp} \]

Only the mass balance changes. The anion now has two sources — the dissolving solid, which supplies \(bs'\), and the added strong electrolyte, which supplies \(c\) — while the cation still has only one, so the two ion concentrations are no longer in the ratio \(a:b\). \(K_{sp}\) is untouched, because it is a property of the temperature (Step 4). This polynomial in \(s'\) is exact within Steps 6 and the hypotheses, and it is the whole of the common-ion effect. B

\[ bs' \ll c \quad\Longrightarrow\quad (as')^a c^{\,b} \simeq K_{sp} \quad\Longrightarrow\quad s' \simeq \frac{1}{a}\left(\frac{K_{sp}}{c^{\,b}}\right)^{1/a} \]

The swamping limit: when the added ion dominates its own mass balance, the anion concentration is a known constant and the equation for \(s'\) collapses to one term. Suppression is now explicit — \(s'\propto c^{-b/a}\) — and comparing with Step 8 shows the effect is large because \(K_{sp}\) is small: replacing the self-supplied \(bs\) by a much larger \(c\) forces \([\text{M}^{z_+}]\) down by the same factor raised to \(b/a\). B

\[ \theta \equiv \frac{bs'}{bs'+c}\ \in(0,1) \quad\Longrightarrow\quad \frac{\mathrm{d}\ln s'}{\mathrm{d}\ln c} = -\frac{b(1-\theta)}{a+b\theta}\ \xrightarrow[\ \theta\to0\ ]{}\ -\frac{b}{a}, \qquad \frac{s'_{\text{approx}}}{s'_{\text{exact}}} = \left(1+\frac{bs'}{c}\right)^{b/a} \]

Take logarithms of Step 9, \(a\ln(as')+b\ln(bs'+c)=\ln K_{sp}\), and differentiate with respect to \(\ln c\); solving for \(\mathrm{d}\ln s'/\mathrm{d}\ln c\) gives the exact sensitivity in terms of the single dimensionless ratio \(\theta\), the fraction of the anion that the solid itself supplies. The second identity follows by writing Step 9 as Step 10 with \(c\) replaced by \(c+bs'\); it turns the vague instruction “check that \(bs'\) is negligible” into a bound, since the fractional error of the swamping formula is \((1+bs'/c)^{b/a}-1\approx(b/a)(bs'/c)\). Every worked example below quotes this number. C

\[ Q \equiv [\text{M}^{z_+}]^a[\text{X}^{z_-}]^b \ \text{(instantaneous)}, \qquad \Delta G = \Delta G^{\circ}_{\mathrm{sol}} + RT\ln Q = RT\ln\!\frac{Q}{K_{sp}} \]

The same ion product evaluated away from equilibrium is the reaction quotient of the reaction quotient. Its sign relative to \(K_{sp}\) fixes the sign of \(\Delta G\) for dissolving one more formula unit: \(Q\lt K_{sp}\) gives \(\Delta G\lt0\) and the solid dissolves; \(Q=K_{sp}\) is saturation; \(Q\gt K_{sp}\) gives \(\Delta G\gt0\) for dissolution, so the reverse process — precipitation — is spontaneous. This is the precipitation criterion, and it is a thermodynamic statement about direction, not a kinetic one about rate. B

Result

\[ \begin{aligned} K_{sp} &= a^a b^b\,s^{a+b} &&\text{pure solvent}\\ K_{sp} &= (as')^a\,(bs'+c)^b \ \ \simeq\ (as')^a c^{\,b} &&\text{common ion at }c\gg bs'\\ Q\lt K_{sp}\ \big|\ Q=K_{sp}\ \big|\ Q\gt K_{sp} &\ \Longleftrightarrow\ \text{dissolves}\ \big|\ \text{saturated}\ \big|\ \text{precipitates} && \end{aligned} \]

Reading. One temperature-dependent number, \(K_{sp}=\exp(-\Delta G^{\circ}_{\mathrm{sol}}/RT)\), fixes the product of ion activities that a saturated solution can support. The solubility that results is not that number but the solution of a mass balance, so it depends on what else is in the solution: with no other source of either ion, \(s=(K_{sp}/a^ab^b)^{1/(a+b)}\); with the anion also supplied at concentration \(c\), the solubility falls as \(c^{-b/a}\).

Scope. Requires excess pure macroscopic solid, congruent and complete dissociation into exactly the two ions named, and ideal-dilute behaviour (\(\gamma_{\pm}\to1\)), all at the temperature for which \(K_{sp}\) is tabulated. Outside those conditions the correct statement is the activity form of Step 4 together with the full set of competing equilibria; both corrections raise the measured solubility above the value predicted here, and they can do so by many orders of magnitude.

Corollaries & converses

Solubility ranks by \(s\), not by \(K_{sp}\). Within one stoichiometric type, \(s\) is a strictly increasing function of \(K_{sp}\) and the ranking is safe. Across types it is not: \(\text{Ag}_2\text{CrO}_4\) has the smaller solubility product (\(1.1\times10^{-12}\) against \(1.8\times10^{-10}\) for AgCl) and yet the larger solubility (\(6.5\times10^{-5}\ \text{M}\) against \(1.3\times10^{-5}\ \text{M}\)), because \(s\) is a cube root of \(K_{sp}/4\) in the first case and a square root of \(K_{sp}\) in the second.
Suppression exponent. \(\mathrm{d}\ln s'/\mathrm{d}\ln c\to-b/a\) in the swamping limit (Step 11), where \(c\) is an excess of the anion; if instead the cation is supplied in excess the two roles swap and the exponent is \(-a/b\). A tenfold excess of iodide suppresses \(\text{PbI}_2\) a hundredfold (\(b/a=2\)). For \(\text{Ag}_2\text{CrO}_4\) (\(a=2,\ b=1\)) the same salt has two different exponents: a tenfold excess of chromate suppresses it only \(\sqrt{10}\approx3.2\)-fold (\(b/a=1/2\)), while a tenfold excess of silver suppresses it a hundredfold (\(a/b=2\)), since then \(s'=K_{sp}/c^{2}\).
Selective precipitation. For two salts of the same stoichiometric type sharing the precipitant, once both solids are present the ratio of the dissolved cations is pinned at \([\text{M}_1]/[\text{M}_2]=K_{sp,1}/K_{sp,2}\), however much further precipitant is added; before the second solid appears the ratio is still falling, because only the first cation is buffered by a solid. Starting from equal cation concentrations, the first is therefore driven down to the fraction \(K_{sp,1}/K_{sp,2}\) of its initial value before the second begins to come down, so the separation is quantitative (better than \(0.1\%\) carry-over) whenever the \(K_{sp}\) values differ by more than about \(10^3\).
Conditional solubility product. If side reactions leave only fractions \(\alpha_{\text{M}}\) and \(\alpha_{\text{X}}\) of each dissolved ion in the free form, then \(K'_{sp}=K_{sp}/(\alpha_{\text{M}}^{\,a}\alpha_{\text{X}}^{\,b})\) plays the role of \(K_{sp}\) for total dissolved amounts, and every formula above holds with \(K_{sp}\to K'_{sp}\). Since \(\alpha\le1\), side reactions can only increase solubility.
Le Chatelier as a special case. The qualitative statement of Le Chatelier's principle that adding a product shifts an equilibrium back towards reactants is recovered exactly from Step 9; the quantitative content that the principle cannot supply is the exponent \(b/a\).
Thermodynamic reading. \(\Delta G^{\circ}_{\mathrm{sol}}=-RT\ln K_{sp}\) means a \(K_{sp}\) of \(10^{-10}\) corresponds to \(+57\ \text{kJ mol}^{-1}\) at \(298\ \text{K}\), and the van 't Hoff relation converts the temperature dependence of \(K_{sp}\) into \(\Delta H^{\circ}_{\mathrm{sol}}\) — a solubility curve is a calorimeter.

Fails without

Ideal-dilute activities dropped (the inert-salt regime): add \(0.010\ \text{mol L}^{-1}\) \(\text{KNO}_3\), which shares no ion with AgCl at all. The ionic strength rises to \(0.010\ \text{mol L}^{-1}\), the Debye–Huckel limiting law gives \(\log_{10}\gamma_{\pm}=-0.509\sqrt{0.010}=-0.0509\), so \(\gamma_{\pm}=0.889\) and \(s=\sqrt{K_{sp}}/\gamma_{\pm}=1.51\times10^{-5}\ \text{M}\) — \(12\%\) more soluble, not less. The ideal treatment predicts no effect whatever. The common-ion effect and this salt effect act in opposite directions and are routinely confused.
Congruent, complete dissolution dropped (the competing-equilibrium regime): at \(\text{pH}\ 3.00\) only \(40\%\) of the dissolved fluoride survives as free \(\text{F}^-\), the other \(60\%\) being HF (\(K_a(\text{HF})=6.8\times10^{-4}\)), so only part of the dissolved fluoride counts towards \(K_{sp}\) and \(\text{CaF}_2\) is \(1.8\) times more soluble than in neutral water; in \(1\ \text{mol L}^{-1}\) chloride the complex \(\text{AgCl}_2^-\) makes AgCl more soluble than in pure water, so the common ion, past a threshold, reverses its own effect. Problems 4 and 5 work both cases through.
Excess solid dropped: in an unsaturated solution there is no equilibrium to apply. \(K_{sp}\) then supplies only the inequality \(Q\lt K_{sp}\), and any attempt to compute “the” ion concentrations from it is meaningless — they are set by how much salt was weighed out, not by \(K_{sp}\).
Macroscopic pure solid dropped: a freshly formed precipitate of \(10\ \text{nm}\) crystallites carries enough surface free energy that its effective \(K_{sp}\) is measurably larger than the bulk value, and a solid solution (\(\text{Ba}_{1-x}\text{Sr}_x\text{SO}_4\)) has solid activity \(\ne1\) for each component. Both raise the apparent solubility, and the first is time-dependent, since Ostwald ripening drives the system to the bulk value.
Thermodynamics mistaken for kinetics: \(Q\gt K_{sp}\) makes precipitation spontaneous but does not make it happen. Nucleation requires a critical cluster, so solutions routinely persist metastably at \(Q/K_{sp}\) of \(10\) or more (the metastable zone exploited in every crystallisation), and a “no precipitate observed” result never by itself refutes \(Q\gt K_{sp}\).

Common errors

“\(K_{sp}(\text{PbI}_2)=s^2\).” The exponents are the stoichiometric coefficients and the concentrations are multiples of \(s\): \(K_{sp}=(s)(2s)^2=4s^3\). Omitting either the coefficient inside the bracket or the exponent outside it is the single most common numerical mistake in the topic, and it is off by a factor of \(4\) even before the wrong root is taken.
“This salt has the smaller \(K_{sp}\), so it is the less soluble.” Only true within one stoichiometric type (Step 8). \(\text{Ag}_2\text{CrO}_4\) has a \(K_{sp}\) \(160\) times smaller than AgCl and a solubility five times larger.
“\(K_{sp}\) has units of \(\text{M}^3\).” It does not: activities are ratios to a standard state, so \(K_{sp}\) is dimensionless (Step 4). Quoting units is harmless as bookkeeping for a fixed \(c^{\circ}\), but it hides the fact that the number depends on the chosen standard state, and it makes \(\exp(-\Delta G^{\circ}/RT)\) look dimensionally wrong.
“\([\text{X}^{z_-}]=c\), because the added salt swamps the solid.” True only when \(bs'\ll c\), which must be verified, not assumed. Step 11 gives the fractional error as \((1+bs'/c)^{b/a}-1\); for \(\text{PbI}_2\) in \(0.010\ \text{mol L}^{-1}\) KI it is \(2.8\%\), small but not zero, and for a more soluble salt or a more dilute common ion it can exceed \(100\%\).
“Adding any electrolyte reduces solubility.” Only a genuine common ion does. An inert electrolyte raises solubility by lowering \(\gamma_{\pm}\) (first “fails without” regime), and even a common ion raises it once complexation takes over (Problem 5).
“The solid is left out because its concentration is constant.” A solid has no concentration in the solution. It is left out because a pure condensed phase is its own standard state and its activity is exactly one (Step 3) — which is also why the amount of excess solid, and its state of subdivision, cannot shift the equilibrium.
Mixing two solutions and forgetting the dilution. Combining equal volumes halves both concentrations before \(Q\) is computed. Since \(Q\) carries the exponents \(a\) and \(b\), a factor of \(2\) in each concentration is a factor of \(2^{a+b}\) in \(Q\) — an eightfold error for a 1:2 salt.

Discussion

The historical order was the reverse of the logical one. Nernst's 1889 paper on the mutual influence of the solubilities of salts (“Über gegenseitige Beeinflussung der Löslichkeit von Salzen”) introduced the solubility product as an empirical mass-action rule for precipitates, and it was immediately useful in analytical chemistry long before activity was a well-defined concept; the thermodynamic derivation in Steps 1–4, which explains both why the solid is absent and why the “constant” is only constant in dilute solution, came a generation later with Lewis's formulation of activity and then Debye and Huckel's 1923 calculation of the coefficients. The residue of that history is the persistent textbook explanation that the solid is omitted because “its concentration does not change”, which is not a statement about anything real.

The practical importance of the common-ion effect is that it is the cheapest available control on a solubility. In gravimetric analysis a precipitate is washed with dilute precipitating reagent rather than water precisely because Step 10 says the loss scales as \(c^{-b/a}\): washing \(\text{BaSO}_4\) with \(0.01\ \text{mol L}^{-1}\) sulfate rather than pure water reduces dissolution loss by three orders of magnitude. The same reasoning, run backwards, is why fractional precipitation works, why a fluoridated water supply drives enamel hydroxyapatite towards the far less soluble fluorapatite, and why boiler scale forms where the local carbonate concentration is highest rather than where the calcium is.

The deepest subtlety is that a tabulated \(K_{sp}\) is not an experimental datum but an extrapolation. What is measured is a solubility, hence a concentration product, at some finite ionic strength; the thermodynamic \(K_{sp}\) is obtained by measuring at several ionic strengths and extrapolating \(\log_{10}K_{sp}^{c}\) against \(\sqrt{I}\) to \(I=0\), using the limiting-law slope of Step 5 as the guide. Different tables therefore disagree in the second or third significant figure — \(K_{sp}(\text{CaF}_2)\) is quoted variously as \(3.45\times10^{-11}\) and \(3.9\times10^{-11}\) — and the disagreement is dominated not by measurement error but by which extrapolation, and which set of competing equilibria (ion pairing above all), the compiler assumed. A calculation carried to four significant figures from a two-figure \(K_{sp}\) is arithmetic, not chemistry. It also follows that using a thermodynamic \(K_{sp}\) with raw concentrations in a solution of appreciable ionic strength double-counts nothing but omits a correction that is often larger than any other term in the problem.

Common misconceptions. That \(K_{sp}\) measures solubility (it measures an ion-activity product; solubility additionally requires a mass balance, and the two differ whenever anything else is in the solution). That the common-ion effect is a distinct physical law (it is Step 9's mass balance, nothing more). That a larger \(K_{sp}\) always means a larger solubility (only within one stoichiometric type). And that a precipitate “does not dissolve”: every precipitate has a finite \(s\), and in analysis the question is never whether analyte is lost but whether the loss is below the tolerance, which is exactly what Step 10 lets one compute.

Worked examples

Example 1. Calcium fluoride, \(K_{sp}=3.9\times10^{-11}\) at \(298\ \text{K}\) (a widely tabulated value; sources differ in the second figure). Find (i) the molar solubility in pure water and its equivalent in \(\text{mg L}^{-1}\), taking \(M(\text{CaF}_2)=78.08\ \text{g mol}^{-1}\); (ii) the solubility in \(0.010\ \text{mol L}^{-1}\) NaF; and (iii) the error committed by the swamping approximation in (ii).

\[ \text{CaF}_2(s)\rightleftharpoons\text{Ca}^{2+}+2\,\text{F}^-,\qquad a=1,\ b=2 \ \Longrightarrow\ K_{sp}=(s)(2s)^2=4s^3 \]

Step 7 with \(a=1,b=2\). Each formula unit releases one calcium and two fluorides, so \([\text{Ca}^{2+}]=s\) and \([\text{F}^-]=2s\); the coefficient \(2\) appears once inside the bracket and once as the exponent. A

\[ s=\left(\frac{K_{sp}}{4}\right)^{1/3}=\left(\frac{3.9\times10^{-11}}{4}\right)^{1/3}=\big(9.75\times10^{-12}\big)^{1/3}=2.14\times10^{-4}\ \text{mol L}^{-1} \]

Step 8, symbols rearranged before any number is substituted. In mass terms \(2.14\times10^{-4}\times78.08=1.67\times10^{-2}\ \text{g L}^{-1}=16.7\ \text{mg L}^{-1}\), which is why fluorite is a stable mineral in contact with groundwater yet not entirely inert to it. A

\[ (s')\,(2s'+0.010)^2 = 3.9\times10^{-11} \]

Step 9 with \(c=0.010\). Sodium fluoride is a strong electrolyte and \(\text{Na}^+\) is a spectator, so the only change from Step 1 is that fluoride now has a second source. This cubic is exact within the hypotheses. A

\[ 2s'\ll0.010 \ \Longrightarrow\ s'\simeq\frac{K_{sp}}{c^{2}}=\frac{3.9\times10^{-11}}{(0.010)^2}=\frac{3.9\times10^{-11}}{1.0\times10^{-4}}=3.9\times10^{-7}\ \text{mol L}^{-1} \]

Step 10 with \(a=1,\ b=2\). The assumption is provisional and is tested in the next step. A

\[ \frac{bs'}{c}=\frac{2(3.9\times10^{-7})}{0.010}=7.8\times10^{-5} \ \Longrightarrow\ \frac{s'_{\text{approx}}}{s'_{\text{exact}}}=\big(1+7.8\times10^{-5}\big)^{2}=1.00016 \]

Step 11's bound: the swamping formula overestimates \(s'\) by \(0.016\%\), far below the uncertainty in \(K_{sp}\) itself, so the approximation is not merely acceptable but wasteful to improve on. The suppression factor is \(s/s'=2.14\times10^{-4}/3.9\times10^{-7}=5.5\times10^{2}\). B

\[ s(\text{pure water})=2.14\times10^{-4}\ \text{mol L}^{-1}=16.7\ \text{mg L}^{-1};\qquad s'(0.010\ \text{M NaF})=3.9\times10^{-7}\ \text{mol L}^{-1};\qquad \frac{s}{s'}=5.5\times10^{2} \]

Reading. A fluoride concentration of only \(0.010\ \text{mol L}^{-1}\) — about \(190\ \text{mg L}^{-1}\), a chemically unremarkable amount — suppresses the solubility of calcium fluoride by a factor of five hundred, because the suppression enters as \(c^{-b/a}=c^{-2}\).

Scope. Ideal-dilute throughout: at \(I=0.010\ \text{mol L}^{-1}\) the limiting law gives \(\gamma_{\pm}=0.79\) for a \(1{:}2\) salt, so \(s'=K_{sp}/(\gamma_{\pm}^{3}c^{2})\) is about twice the value quoted here, and at \(\text{pH}\) below about \(4\) the protonation of fluoride (Problem 4) matters more than either correction.

Example 2. A solution is \(0.010\ \text{mol L}^{-1}\) in both \(\text{Ba}^{2+}\) and \(\text{Sr}^{2+}\). Sulfate is added slowly, with negligible change in volume. Given \(K_{sp}(\text{BaSO}_4)=1.1\times10^{-10}\) and \(K_{sp}(\text{SrSO}_4)=3.4\times10^{-7}\) at \(298\ \text{K}\), find the sulfate concentration at which each solid first appears, and the fraction of the barium still in solution at the moment strontium sulfate begins to precipitate.

\[ Q_{\text{M}}=[\text{M}^{2+}][\text{SO}_4^{2-}] \ \ \text{first reaches}\ \ K_{sp,\text{M}} \ \ \text{at}\ \ [\text{SO}_4^{2-}]_{\text{onset}}=\frac{K_{sp,\text{M}}}{[\text{M}^{2+}]_0} \]

Step 12: precipitation begins at the instant \(Q=K_{sp}\). Both salts are 1:1, so \(a=b=1\) and the criterion is linear in the sulfate concentration. A

\[ [\text{SO}_4^{2-}]_{\text{Ba}}=\frac{1.1\times10^{-10}}{0.010}=1.1\times10^{-8}\ \text{mol L}^{-1}, \qquad [\text{SO}_4^{2-}]_{\text{Sr}}=\frac{3.4\times10^{-7}}{0.010}=3.4\times10^{-5}\ \text{mol L}^{-1} \]

Barium sulfate appears first, at a sulfate concentration about \(3.1\times10^{3}\) times lower — exactly the ratio of the two \(K_{sp}\) values, because the two cations started at equal concentration. A

\[ [\text{Ba}^{2+}] = \frac{K_{sp,\text{Ba}}}{[\text{SO}_4^{2-}]} = \frac{1.1\times10^{-10}}{3.4\times10^{-5}} = 3.2\times10^{-6}\ \text{mol L}^{-1} \]

While solid \(\text{BaSO}_4\) is present the barium in solution is pinned by its own \(K_{sp}\), and the added sulfate is now the common ion suppressing it (Step 10 with \(a=b=1\), so \(s'\propto c^{-1}\)). Evaluating at the sulfate concentration found in Step 2. B

\[ \frac{[\text{Ba}^{2+}]}{[\text{Ba}^{2+}]_0}=\frac{3.2\times10^{-6}}{0.010}=3.2\times10^{-4}, \qquad \frac{[\text{Ba}^{2+}]}{[\text{Sr}^{2+}]}=\frac{K_{sp,\text{Ba}}}{K_{sp,\text{Sr}}}=3.2\times10^{-4} \]

Only \(0.032\%\) of the barium remains when strontium begins to come down, so the separation is quantitative by any analytical standard. The second equality holds because the two cations began at the same concentration and the barium is now pinned by its own solid while the strontium has only just reached saturation; from this instant on, with both solids present, the ratio of the dissolved cations stays pinned at \(K_{sp,\text{Ba}}/K_{sp,\text{Sr}}\) however much more sulfate is added. Earlier in the titration, with only \(\text{BaSO}_4\) down, the ratio was larger and still falling. B

\[ [\text{SO}_4^{2-}]_{\text{Ba}}=1.1\times10^{-8}\ \text{M} \ \lt\ [\text{SO}_4^{2-}]_{\text{Sr}}=3.4\times10^{-5}\ \text{M}; \qquad \text{Ba}^{2+}\ \text{remaining}=3.2\times10^{-4}=0.032\% \]

Reading. Two cations whose solubility products differ by three orders of magnitude can be separated essentially completely by slow addition of a shared precipitant: the first salt is driven to \(0.03\%\) of its initial concentration before the second one starts.

Scope. Assumes slow addition with no local supersaturation, no coprecipitation or solid-solution formation between the two sulfates (which in the \(\text{Ba/Sr}\) system is a real limitation in practice), and ideal-dilute activities at the working ionic strength.

Problems

Silver chromate has \(K_{sp}=1.1\times10^{-12}\) at \(298\ \text{K}\). Find its molar solubility in pure water, and in \(0.10\ \text{mol L}^{-1}\) \(\text{AgNO}_3\); verify the swamping approximation in the second case.
Solution
\(\text{Ag}_2\text{CrO}_4(s)\rightleftharpoons2\,\text{Ag}^++\text{CrO}_4^{2-}\), so \(a=2,\ b=1\) and \(K_{sp}=(2s)^2(s)=4s^3\).
Pure water: \(s=(K_{sp}/4)^{1/3}=(2.75\times10^{-13})^{1/3}=6.5\times10^{-5}\ \text{mol L}^{-1}\).
With \(0.10\ \text{mol L}^{-1}\) silver added, the cation is now the common ion. Mass balance: \([\text{Ag}^+]=2s'+0.10\), \([\text{CrO}_4^{2-}]=s'\), so \((2s'+0.10)^2 s'=K_{sp}\). Swamping (\(2s'\ll0.10\)) gives \(s'\simeq K_{sp}/(0.10)^2=1.1\times10^{-12}/1.0\times10^{-2}=1.1\times10^{-10}\ \text{mol L}^{-1}\).
Check: \(2s'/c=2.2\times10^{-10}/0.10=2.2\times10^{-9}\), so the error is about \(2\times10^{-9}\) — utterly negligible. The suppression factor is \(6.5\times10^{-5}/1.1\times10^{-10}=5.9\times10^{5}\). Note also that \(\text{Ag}_2\text{CrO}_4\) has a \(K_{sp}\) \(160\) times smaller than AgCl (\(1.8\times10^{-10}\)) yet a solubility about five times larger, since \(s\) is a cube root here and a square root there.
Lead(II) iodide has \(K_{sp}=7.1\times10^{-9}\). Find its molar solubility in pure water and in \(0.010\ \text{mol L}^{-1}\) KI, this time solving the exact cubic by iteration, and compare with the swamping estimate.
Solution
\(\text{PbI}_2\rightleftharpoons\text{Pb}^{2+}+2\text{I}^-\): \(a=1,\ b=2,\ K_{sp}=4s^3\). Pure water: \(s=(7.1\times10^{-9}/4)^{1/3}=(1.775\times10^{-9})^{1/3}=1.21\times10^{-3}\ \text{mol L}^{-1}\).
With \(c=0.010\): exact equation \(s'(2s'+0.010)^2=7.1\times10^{-9}\). Swamping estimate \(s'_0=K_{sp}/c^2=7.1\times10^{-9}/1.0\times10^{-4}=7.1\times10^{-5}\ \text{mol L}^{-1}\).
Iterate on \(s'_{n+1}=K_{sp}/(2s'_n+0.010)^2\). With \(s'_0=7.1\times10^{-5}\): \(2s'_0+0.010=0.010142\), squared \(=1.0286\times10^{-4}\), so \(s'_1=6.903\times10^{-5}\). Again: \(2s'_1+0.010=0.0101381\), squared \(=1.02780\times10^{-4}\), \(s'_2=6.908\times10^{-5}\). Converged: \(s'=6.91\times10^{-5}\ \text{mol L}^{-1}\).
The swamping estimate is high by \(7.1/6.91-1=2.8\%\), which is exactly what Step 11 predicts: \((1+bs'/c)^{b/a}-1=(1+2(6.91\times10^{-5})/0.010)^2-1=(1.01382)^2-1=0.0278\). Here the iodide from the solid is no longer wholly negligible, unlike Example 1, because \(K_{sp}(\text{PbI}_2)\) is some \(180\) times larger than \(K_{sp}(\text{CaF}_2)\); both salts are of the \(1{:}2\) type, so at the same common-ion concentration \(s'=K_{sp}/c^{2}\) is larger by that same factor (\(6.9\times10^{-5}\) against \(3.9\times10^{-7}\ \text{mol L}^{-1}\)), even though in pure water \(\text{PbI}_2\) is only about six times the more soluble.
Silver chloride has \(K_{sp}=1.8\times10^{-10}\). Using the Debye–Huckel limiting law \(\log_{10}\gamma_{\pm}=-A|z_+z_-|\sqrt{I}\) with \(A=0.509\) for water at \(298\ \text{K}\), find its solubility (i) in pure water on the ideal assumption, (ii) in \(0.010\ \text{mol L}^{-1}\) \(\text{KNO}_3\), and (iii) in \(0.010\ \text{mol L}^{-1}\) KCl with the activity correction included. Comment on the two additives.
Solution
(i) Ideal, pure water: \(K_{sp}=s^2\), \(s=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}\ \text{mol L}^{-1}\). (The solution's own ionic strength, \(1.3\times10^{-5}\), gives \(\gamma_{\pm}=0.996\), a \(0.4\%\) correction that is negligible here.)
(ii) \(\text{KNO}_3\) shares no ion with AgCl, so the mass balance is unchanged: \([\text{Ag}^+]=[\text{Cl}^-]=s\). But \(I=\tfrac12[(0.010)(1)^2+(0.010)(1)^2]=0.010\ \text{mol L}^{-1}\), so \(\log_{10}\gamma_{\pm}=-0.509(1)\sqrt{0.010}=-0.0509\) and \(\gamma_{\pm}=0.889\). From Step 5, \(K_{sp}=\gamma_{\pm}^2s^2\), hence \(s=\sqrt{K_{sp}}/\gamma_{\pm}=1.34\times10^{-5}/0.889=1.51\times10^{-5}\ \text{mol L}^{-1}\), an increase of \(12\%\).
(iii) \(0.010\ \text{mol L}^{-1}\) KCl has the same ionic strength and hence the same \(\gamma_{\pm}=0.889\), but now supplies a common ion. \(K_{sp}=\gamma_{\pm}^2[\text{Ag}^+][\text{Cl}^-]\) with \([\text{Cl}^-]\simeq0.010\) gives \([\text{Ag}^+]=K_{sp}/(\gamma_{\pm}^2 c)=1.8\times10^{-10}/(0.790\times0.010)=2.28\times10^{-8}\ \text{mol L}^{-1}\), against \(1.8\times10^{-8}\) on the ideal treatment.
Comment: at identical ionic strength the inert salt raises solubility \(12\%\) while the common-ion salt lowers it by a factor of \(1.51\times10^{-5}/2.28\times10^{-8}=6.6\times10^{2}\). The activity correction and the common-ion effect are independent and act in opposite directions; the second is much the larger here, but the first is the one usually omitted without comment.
A saturated solution of \(\text{CaF}_2\) (\(K_{sp}=3.9\times10^{-11}\)) is held at \(\text{pH}\ 3.00\) by a buffer. Given \(K_a(\text{HF})=6.8\times10^{-4}\), find the molar solubility and compare it with the neutral-water value of \(2.14\times10^{-4}\ \text{mol L}^{-1}\). Assume ideal-dilute behaviour and that the buffer holds the pH fixed.
Solution
Only free \(\text{F}^-\) enters \(K_{sp}\), but the dissolved fluoride is distributed between \(\text{F}^-\) and HF. The fraction free is \(\alpha=K_a/(K_a+[\text{H}^+])=6.8\times10^{-4}/(6.8\times10^{-4}+1.00\times10^{-3})=0.405\).
Mass balance on fluoride: total dissolved fluoride is \(2s\), so \([\text{F}^-]=2s\alpha\), while \([\text{Ca}^{2+}]=s\) (calcium has no side reaction here). Hence
\[ K_{sp}=s\,(2s\alpha)^2=4\alpha^2 s^3 \quad\Longrightarrow\quad s=\left(\frac{K_{sp}}{4\alpha^2}\right)^{1/3}. \]
Numerically \(4\alpha^2=4(0.405)^2=0.655\), so \(s=(3.9\times10^{-11}/0.655)^{1/3}=(5.95\times10^{-11})^{1/3}=3.90\times10^{-4}\ \text{mol L}^{-1}\), a factor \(3.90/2.14=1.83\) above the neutral value.
Equivalently, in the conditional-solubility-product language of the corollaries, \(K'_{sp}=K_{sp}/\alpha^2=3.9\times10^{-11}/0.164=2.38\times10^{-10}\), and \(s=(K'_{sp}/4)^{1/3}\) reproduces the same number. The general lesson is that the protonation removes fluoride from the species that \(K_{sp}\) constrains without removing it from solution, so more solid must dissolve to restore the ion product — the mechanism by which acid rain dissolves carbonate rock, where the effect is far larger because \(K_a(\text{HCO}_3^-)\) is much smaller.
Silver chloride dissolves both as free \(\text{Ag}^+\) and, in chloride-rich media, as the complex \(\text{AgCl}_2^-\) formed by \(\text{Ag}^++2\text{Cl}^-\rightleftharpoons\text{AgCl}_2^-\) with \(\beta_2=1.1\times10^{5}\) (concentration basis). Taking \(K_{sp}=1.8\times10^{-10}\) and ignoring \(\text{AgCl}(aq)\) and \(\text{AgCl}_3^{2-}\), derive the chloride concentration at which the total dissolved silver is a minimum, evaluate that minimum, and find the chloride concentration above which AgCl is more soluble than in pure water.
Solution
Total dissolved silver, with solid present, is
\[ S=[\text{Ag}^+]+[\text{AgCl}_2^-]=\frac{K_{sp}}{[\text{Cl}^-]}+\beta_2K_{sp}[\text{Cl}^-], \]
since \([\text{Ag}^+]=K_{sp}/[\text{Cl}^-]\) from the solubility product and \([\text{AgCl}_2^-]=\beta_2[\text{Ag}^+][\text{Cl}^-]^2=\beta_2K_{sp}[\text{Cl}^-]\). The first term is the common-ion suppression, the second is the complexation enhancement, and they scale oppositely with \([\text{Cl}^-]\).
Minimise: \(\mathrm{d}S/\mathrm{d}[\text{Cl}^-]=-K_{sp}/[\text{Cl}^-]^2+\beta_2K_{sp}=0\) gives \([\text{Cl}^-]_{\min}=\beta_2^{-1/2}=(1.1\times10^{5})^{-1/2}=3.0\times10^{-3}\ \text{mol L}^{-1}\), and the second derivative \(2K_{sp}/[\text{Cl}^-]^3\gt0\) confirms a minimum.
At that point \(S_{\min}=2K_{sp}\sqrt{\beta_2}=2(1.8\times10^{-10})(331.7)=1.2\times10^{-7}\ \text{mol L}^{-1}\), about \(110\) times below the pure-water solubility \(1.34\times10^{-5}\ \text{mol L}^{-1}\).
Above the minimum the second term dominates. Setting \(S=1.34\times10^{-5}\) with the complex term alone, \([\text{Cl}^-]=1.34\times10^{-5}/(\beta_2K_{sp})=1.34\times10^{-5}/(1.98\times10^{-5})=0.68\ \text{mol L}^{-1}\). So beyond roughly \(0.7\ \text{mol L}^{-1}\) chloride the common ion has stopped suppressing and started dissolving the solid, and at \(1.0\ \text{mol L}^{-1}\) the total silver is \(2.0\times10^{-5}\ \text{mol L}^{-1}\), above the pure-water value. This is the second “fails without” regime made quantitative, and it is why a chloride wash for silver chloride must be dilute. At molar chloride the ideal-dilute hypothesis has also collapsed, so the last figure is illustrative rather than accurate.