Solubility equilibria: Ksp, common-ion effect, and dissolution of sparingly soluble salts

Intuition Beginner

Some salts dissolve completely in water (like NaCl). Others dissolve only a little (like AgCl). When a sparingly soluble salt reaches its maximum solubility, the solution is saturated, and an equilibrium exists between the undissolved solid and the dissolved ions.

For silver chloride:

$$\text{AgCl}(s)\rightleftharpoons {\text{Ag}}^{+}(aq)+{\text{Cl}}^{-}(aq)$$

The equilibrium constant for this process is called the solubility product constant, $K_{sp}$. For AgCl:

$$K_{sp}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=1.8\times 10^{-10}$$

A small $K_{sp}$ means the salt is barely soluble. A large $K_{sp}$ means it dissolves readily. When the product of the ion concentrations (the ion product, $Q$) exceeds $K_{sp}$, precipitation occurs. When $Q<K_{sp}$, the solution is unsaturated and more solid can dissolve. At $Q=K_{sp}$, the solution is saturated and equilibrium is established.

Le Chatelier's principle governs solubility equilibria just as it does gas-phase and acid-base equilibria. Adding a common ion (one already present in the solution) shifts the equilibrium toward the solid, reducing solubility. Removing an ion (by forming a complex or adjusting pH) shifts the equilibrium toward dissolution, increasing solubility.

Visual Beginner

A solubility diagram shows the concentration of dissolved ions on the vertical axis and the concentration of a common ion on the horizontal axis. As the common-ion concentration increases, the molar solubility of the sparingly soluble salt decreases.

At low concentrations of common ion, the solubility decreases sharply (the common-ion effect). At very high concentrations, a secondary effect takes over: the dissolved metal ion forms complex ions with the added anion, actually increasing solubility. This dual behaviour is characteristic of many sparingly soluble salts.

Worked example Beginner

The $K_{sp}$ of $\text{AgCl}$ is $1.8\times 10^{-10}$. Calculate the molar solubility of AgCl in pure water.

Step 1: Set up the equilibrium. Let $s$ = molar solubility (mol/L of AgCl that dissolves).

$$[{\text{Ag}}^{+}]=s,[{\text{Cl}}^{-}]=s$$

Step 2: Write the $K_{sp}$ expression.

$$K_{sp}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=s\times s=s^2=1.8\times 10^{-10}$$

Step 3: Solve for $s$.

$$s=\sqrt{1.8\times 10^{-10}}=1.3\times 10^{-5}\text{M}$$

AgCl dissolves to give $1.3\times 10^{-5}$ M of ${\text{Ag}}^{+}$ and ${\text{Cl}}^{-}$ in pure water. That is $0.0019$ g/L -- very sparingly soluble.

Now calculate the molar solubility of AgCl in $0.10\text{M}$ NaCl. The common ion ${\text{Cl}}^{-}$ is already present at $0.10\text{M}$.

$$[{\text{Ag}}^{+}]=s,[{\text{Cl}}^{-}]=0.10+s\approx 0.10$$ $$K_{sp}=s\times 0.10=1.8\times 10^{-10}$$ $$s=1.8\times 10^{-9}\text{M}$$

The common ion reduces the solubility by a factor of about 7,200 compared to pure water.

Check your understanding Beginner

Formal definition Intermediate+

The solubility product constant

For a generic sparingly soluble salt ${\text{M}}_m{\text{A}}_n$ that dissociates as:

$${\text{M}}_m{\text{A}}_n(s)\rightleftharpoons m{\text{M}}^{n+}(aq)+n{\text{A}}^{m-}(aq)$$

the solubility product constant is:

$$K_{sp}=[{\text{M}}^{n+}{]}^m[{\text{A}}^{m-}{]}^n$$

The solid does not appear in the equilibrium expression because its activity is 1 (it is a pure solid in its standard state). The $K_{sp}$ is the product of the ion concentrations, each raised to the power of its stoichiometric coefficient.

For common salts:

• AgCl: $K_{sp}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=1.8\times 10^{-10}$
• ${\text{CaF}}_2$: $K_{sp}=[{\text{Ca}}^{2+}][{\text{F}}^{-}{]}^2=3.9\times 10^{-11}$
• ${\text{Ag}}_2{\text{CrO}}_4$: $K_{sp}=[{\text{Ag}}^{+}{]}^2[{\text{CrO}}_4^{2-}]=1.1\times 10^{-12}$
• ${\text{Fe(OH)}}_3$: $K_{sp}=[{\text{Fe}}^{3+}][{\text{OH}}^{-}{]}^3=2.8\times 10^{-39}$

Molar solubility and $K_{sp}$

The molar solubility $s$ is the number of moles of solute that dissolve per litre of saturated solution. The relationship between $s$ and $K_{sp}$ depends on the stoichiometry:

Salt type	Dissolution	$K_{sp}$ in terms of $s$	$s$ in terms of $K_{sp}$
1:1 (AgCl)	$s$ cation, $s$ anion	$s^2$	$\sqrt{K_{sp}}$
1:2 (${\text{CaF}}_2$)	$s$ cation, $2s$ anion	$4s^3$	$\sqrt[3]{K_{sp}/4}$
2:1 (${\text{Ag}}_2{\text{CrO}}_4$)	$2s$ cation, $s$ anion	$4s^3$	$\sqrt[3]{K_{sp}/4}$
1:3 (${\text{Fe(OH)}}_3$)	$s$ cation, $3s$ anion	$27s^4$	$\sqrt[4]{K_{sp}/27}$

These relationships assume that the salt dissociates completely (which is a good approximation for sparingly soluble salts) and that no side reactions (hydrolysis, complex formation) occur.

The common-ion effect

When a solution already contains one of the ions of the sparingly soluble salt, the solubility is reduced. For AgCl in a solution already containing $[{\text{Cl}}^{-}]=c$:

$$K_{sp}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=s\times (c+s)$$

When $c\gg s$, this simplifies to $s\approx K_{sp}/c$. The solubility is inversely proportional to the common-ion concentration. This is a direct consequence of Le Chatelier's principle: the added ion shifts the dissolution equilibrium toward the solid.

The common-ion effect is quantitatively significant. For ${\text{CaF}}_2$ ($K_{sp}=3.9\times 10^{-11}$), the molar solubility in pure water is $2.1\times 10^{-4}$ M. In $0.10\text{M}$ NaF, the solubility drops to $3.9\times 10^{-9}$ M -- a reduction by a factor of about 54,000.

Selective precipitation

When a solution contains two ions that can form sparingly soluble salts with the same counterion, selective precipitation exploits the difference in their $K_{sp}$ values to separate them. The ion whose salt has the smaller $K_{sp}$ precipitates first.

For a solution containing $0.010\text{M}{\text{Ag}}^{+}$ and $0.010\text{M}{\text{Pb}}^{2+}$, to which ${\text{Cl}}^{-}$ is slowly added: $\text{AgCl}$ ($K_{sp}=1.8\times 10^{-10}$) precipitates first because it requires a lower $[{\text{Cl}}^{-}]$ to exceed $K_{sp}$:

$$[{\text{Cl}}^{-}{]}_{\text{AgCl ppt}}=K_{sp}/[{\text{Ag}}^{+}]=1.8\times 10^{-10}/0.010=1.8\times 10^{-8}\text{M}$$

${\text{PbCl}}_2$ ($K_{sp}=1.6\times 10^{-5}$) requires:

$$[{\text{Cl}}^{-}{]}_{{\text{PbCl}}_2\text{ ppt}}=\sqrt{K_{sp}/[{\text{Pb}}^{2+}]}=\sqrt{1.6\times 10^{-5}/0.010}=0.040\text{M}$$

AgCl precipitates at $[{\text{Cl}}^{-}]=1.8\times 10^{-8}$ M, while ${\text{PbCl}}_2$ does not precipitate until $[{\text{Cl}}^{-}]=0.040$ M. Between these two concentrations, AgCl precipitates selectively and ${\text{Pb}}^{2+}$ remains in solution.

Counterexamples to common slips

• $K_{sp}$ does not directly rank molar solubility across different stoichiometries. $\text{AgCl}$ has $K_{sp}=1.8\times 10^{-10}$ ($s=1.3\times 10^{-5}$ M) and ${\text{Ag}}_2{\text{CrO}}_4$ has $K_{sp}=1.1\times 10^{-12}$ ($s=6.5\times 10^{-5}$ M). Despite having a smaller $K_{sp}$, silver chromate is more soluble because the cubic relationship for 2:1 salts produces a larger $s$ for the same $K_{sp}$.
• $K_{sp}$ applies only to saturated solutions. In an unsaturated solution, $Q<K_{sp}$ and no equilibrium with the solid exists. The $K_{sp}$ expression does not constrain the ion concentrations.
• The common ion must actually be present as a free ion. Adding HCl to a solution of AgCl does not produce a simple common-ion effect because ${\text{Cl}}^{-}$ can form the complex $[{\text{AgCl}}_2{]}^{-}$ at high chloride concentrations, increasing rather than decreasing solubility.

Key result Intermediate+

Theorem (Common-ion effect on molar solubility). For a 1:1 salt MX with $K_{sp}$ in a solution containing a common ion X at concentration $c\gg s$:

$$s=\frac{K_{sp}}{c}$$

For a 1:2 salt ${\text{MX}}_2$ with $K_{sp}$ in a solution containing a common ion X at concentration $c\gg 2s$:

$$s=\frac{K_{sp}}{4c^2}$$

Proof. For the 1:1 salt, $[{\text{M}}^{+}]=s$ and $[{\text{X}}^{-}]=c+s\approx c$ when $c\gg s$.

$$K_{sp}=s\times c,s=K_{sp}/c$$

For the 1:2 salt, $[{\text{M}}^{2+}]=s$ and $[{\text{X}}^{-}]=c+2s\approx c$ when $c\gg 2s$.

$$K_{sp}=s\times c^2,s=K_{sp}/c^2$$

The factor of 4 in the earlier table arises in pure water where $c=2s$; when an external common ion dominates, $c$ is fixed and the relationship simplifies. $\square$

Corollary (Quantitative separation condition). Two ions ${\text{M}}_1$ and ${\text{M}}_2$ can be quantitatively separated by selective precipitation with counterion X if the ratio of their $K_{sp}$ values (adjusted for stoichiometry) exceeds $10^4$ and the initial concentrations are comparable.

Worked example: selective precipitation

A solution contains $0.020\text{M}{\text{Ba}}^{2+}$ and $0.020\text{M}{\text{Ca}}^{2+}$. Sodium sulfate is slowly added. $K_{sp}({\text{BaSO}}_4)=1.1\times 10^{-10}$, $K_{sp}({\text{CaSO}}_4)=2.4\times 10^{-5}$. Can the ions be separated?

Step 1: $[{\text{SO}}_4^{2-}]$ to begin precipitating BaSO${}_4$:

$$[{\text{SO}}_4^{2-}]=K_{sp}/[{\text{Ba}}^{2+}]=1.1\times 10^{-10}/0.020=5.5\times 10^{-9}\text{M}$$

Step 2: $[{\text{SO}}_4^{2-}]$ to begin precipitating CaSO${}_4$:

$$[{\text{SO}}_4^{2-}]=K_{sp}/[{\text{Ca}}^{2+}]=2.4\times 10^{-5}/0.020=1.2\times 10^{-3}\text{M}$$

Step 3: $[{\text{Ba}}^{2+}]$ remaining when CaSO${}_4$ begins to precipitate:

$$[{\text{Ba}}^{2+}]=K_{sp}/[{\text{SO}}_4^{2-}]=1.1\times 10^{-10}/1.2\times 10^{-3}=9.2\times 10^{-8}\text{M}$$

Over 99.999% of the ${\text{Ba}}^{2+}$ has precipitated before CaSO${}_4$ begins to form. The separation is quantitative.

Bridge. Selective precipitation is the basis of the classical qualitative analysis scheme, where cations are separated into groups by sequential addition of precipitating agents. The common-ion effect also connects to 14.10.01 acid-base chemistry: many metal hydroxides are sparingly soluble, and their solubility depends on pH through the $[{\text{OH}}^{-}]$ term in $K_{sp}$. This pH-dependent solubility links directly to 14.11.01 electrochemistry, where the Nernst equation for metal-ion electrodes depends on the free metal-ion concentration, which in turn is governed by solubility equilibria.

Exercises Intermediate+

pH-dependent solubility Master

Metal hydroxides

For a metal hydroxide ${\text{M(OH)}}_n$:

$$K_{sp}=[{\text{M}}^{n+}][{\text{OH}}^{-}{]}^n$$

The molar solubility in a buffer at fixed pH is:

$$s=[{\text{M}}^{n+}]=\frac{K_{sp}}{[{\text{OH}}^{-}{]}^n}=\frac{K_{sp}}{(10^{-(14-\text{pH})}{)}^n}=K_{sp}\times 10^{n(14-\text{pH})}$$

Taking the logarithm:

$$\log s=\log K_{sp}+n(14-\text{pH})$$

A plot of $\log s$ versus pH is a straight line with slope $-n$ and intercept $\log K_{sp}+14n$. For ${\text{Fe(OH)}}_3$ ($K_{sp}=2.8\times 10^{-39}$, $n=3$):

$$\log s=-38.55+3(14-\text{pH})=3.45-3\text{pH}$$

At pH 2: $\log s=3.45-6=-2.55$, $s=2.8\times 10^{-3}$ M (moderately soluble). At pH 7: $\log s=3.45-21=-17.55$, $s=2.8\times 10^{-18}$ M (essentially insoluble). A 5-unit pH increase reduces the solubility by 15 orders of magnitude.

This extreme pH dependence is the basis for many industrial and environmental processes. Water treatment plants adjust pH to precipitate heavy metals as hydroxides. A pH above 9 removes ${\text{Fe}}^{3+}$, ${\text{Al}}^{3+}$, and ${\text{Cr}}^{3+}$ to below $10^{-6}$ M. Conversely, acid mine drainage (pH < 3) dissolves metal hydroxides and releases toxic metals into waterways.

Salts of weak acids

For a sparingly soluble salt of a weak acid, the anion undergoes protonation at low pH, reducing its concentration and increasing solubility. For ${\text{CaCO}}_3$ ($K_{sp}=3.4\times 10^{-9}$):

$${\text{CaCO}}_3(s)\rightleftharpoons {\text{Ca}}^{2+}+{\text{CO}}_3^{2-}$$

The carbonate ion is a weak base: ${\text{CO}}_3^{2-}+{\text{H}}^{+}\rightleftharpoons {\text{HCO}}_3^{-}$ ($K_{a2}=4.7\times 10^{-11}$, $K_b=2.1\times 10^{-4}$). At pH below about 8, a significant fraction of ${\text{CO}}_3^{2-}$ is protonated to ${\text{HCO}}_3^{-}$, reducing $[{\text{CO}}_3^{2-}]$ below the value predicted by $K_{sp}$ alone and driving further dissolution.

The total dissolved calcium in a saturated ${\text{CaCO}}_3$ solution at pH $\le 8$ is:

$$[{\text{Ca}}^{2+}]=\frac{K_{sp}}{[{\text{CO}}_3^{2-}]}$$

where $[{\text{CO}}_3^{2-}]$ is determined by the fractional composition of the carbonate system at the given pH. Using ${\alpha }_2=K_{a1}{K}_{a2}/D$ (where $D$ is the carbonate system denominator):

$$[{\text{Ca}}^{2+}]=\frac{K_{sp}}{{\alpha }_2\cdot c_T}$$

where $c_T=[{\text{CO}}_3^{2-}]+[{\text{HCO}}_3^{-}]+[{\text{H}}_2{\text{CO}}_3]=[{\text{Ca}}^{2+}]$ (from the stoichiometry of dissolution in a closed system). This gives:

$$[{\text{Ca}}^{2+}{]}^2=K_{sp}/{\alpha }_2$$ $$[{\text{Ca}}^{2+}]=\sqrt{K_{sp}/{\alpha }_2}$$

At pH 7, ${\alpha }_2\approx 3.5\times 10^{-4}$, so $[{\text{Ca}}^{2+}]=\sqrt{3.4\times 10^{-9}/3.5\times 10^{-4}}=\sqrt{9.7\times 10^{-6}}=3.1\times 10^{-3}$ M. Compare with pure water at pH ~10 (where ${\alpha }_2\approx 0.5$): $[{\text{Ca}}^{2+}]=\sqrt{3.4\times 10^{-9}/0.5}=8.2\times 10^{-5}$ M. The lower pH increases solubility by a factor of about 38.

This is why acid rain dissolves limestone (${\text{CaCO}}_3$) statues and buildings, and why the ocean acidification (decreasing pH of surface seawater) threatens coral reefs and shell-forming organisms that rely on ${\text{CaCO}}_3$ precipitation.

Complex ion formation and solubility Master

Ligand-enhanced dissolution

When a sparingly soluble salt contains a metal ion that can form complex ions with a ligand in solution, the complex formation removes the free metal ion from solution, shifting the dissolution equilibrium to the right and increasing solubility.

For $\text{AgCl}$ in a solution containing ${\text{NH}}_3$:

$$\text{AgCl}(s)\rightleftharpoons {\text{Ag}}^{+}+{\text{Cl}}^{-}{K}_{sp}=1.8\times 10^{-10}$$ $${\text{Ag}}^{+}+2{\text{NH}}_3\rightleftharpoons [{\text{Ag(NH}}_3{)}_2{]}^{+}{K}_f=1.7\times 10^7$$

The overall dissolution reaction is:

$$\text{AgCl}(s)+2{\text{NH}}_3\rightleftharpoons [{\text{Ag(NH}}_3{)}_2{]}^{+}+{\text{Cl}}^{-}K=K_{sp}\times K_f=3.1\times 10^{-3}$$

The solubility in $1.0\text{M}{\text{NH}}_3$ is much greater than in pure water. Let $s$ = molar solubility. Then $[[{\text{Ag(NH}}_3{)}_2{]}^{+}]=s$, $[{\text{Cl}}^{-}]=s$, and $[{\text{NH}}_3]=1.0-2s\approx 1.0$:

$$K=\frac{s\times s}{(1.0{)}^2}=s^2=3.1\times 10^{-3}$$ $$s=0.056\text{M}$$

Compare with pure water: $s=1.3\times 10^{-5}$ M. Ammonia increases the solubility of AgCl by a factor of about 4,300.

The combined equilibrium constant

The overall dissolution constant for a salt $\text{MX}$ forming a complex $[{\text{ML}}_n{]}^{+}$ with $n$ ligands L is:

$$K_{\text{overall}}=K_{sp}\times K_f$$

where $K_f$ is the formation constant of the complex. For the dissolution to be significant, $K_{\text{overall}}$ must not be too small. A large $K_f$ can compensate for a small $K_{sp}$, which is why strong complexing agents (EDTA, cyanide, ammonia) effectively dissolve otherwise insoluble salts.

The quantitative treatment requires tracking all species simultaneously. For $\text{AgCl}$ in ${\text{NH}}_3$, the free $[{\text{Ag}}^{+}]$ is determined by the complex equilibrium:

$$[{\text{Ag}}^{+}]=\frac{[{\text{Ag(NH}}_3{)}_2{]}^{+}}{K_f\cdot [{\text{NH}}_3{]}^2}$$

Combined with $K_{sp}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]$ and mass balance on silver ($[{\text{Ag}}^{+}]+[{\text{Ag(NH}}_3{)}_2^{+}]=s$) and ammonia, the complete system of equations is solved simultaneously. When $K_f$ is large, virtually all dissolved silver exists as the complex, and the simplified treatment above is accurate.

Amphoteric hydroxides

Metal hydroxides that are also amphoteric (such as ${\text{Al(OH)}}_3$, ${\text{Zn(OH)}}_2$, ${\text{Cr(OH)}}_3$) dissolve both at low pH (where ${\text{OH}}^{-}$ is consumed by protonation) and at high pH (where the metal ion forms hydroxo complexes):

Low pH: ${\text{Al(OH)}}_3(s)+3{\text{H}}^{+}\rightleftharpoons {\text{Al}}^{3+}+3{\text{H}}_2\text{O}$

High pH: ${\text{Al(OH)}}_3(s)+{\text{OH}}^{-}\rightleftharpoons [{\text{Al(OH)}}_4{]}^{-}$

The solubility of ${\text{Al(OH)}}_3$ is therefore a U-shaped function of pH: high at low pH (acidic dissolution), minimum near neutral pH, and high again at high pH (alkaline dissolution via complex formation). The minimum solubility occurs near the isoelectric point of the hydroxide.

This amphoteric behaviour is exploited in the Bayer process for refining bauxite (${\text{Al}}_2{\text{O}}_3$). The bauxite ore is dissolved in hot concentrated NaOH, which selectively dissolves aluminium as $[{\text{Al(OH)}}_4{]}^{-}$ while leaving iron and silicon impurities as insoluble residues. The solution is then cooled and seeded to precipitate pure ${\text{Al(OH)}}_3$.

Ksp and the Gibbs energy of dissolution Master

Thermodynamic relationship

The standard Gibbs energy change for the dissolution of a salt is related to $K_{sp}$ by:

$$\Delta G_{\text{diss}}^{\circ }=-RT\ln K_{sp}$$

For $\text{AgCl}$ at 298 K: $\Delta G^{\circ }=-8.314\times 298\times \ln (1.8\times 10^{-10})=-2477\times (-22.23)=+55.0\text{kJ/mol}$. The positive $\Delta G^{\circ }$ confirms that dissolution is non-spontaneous under standard conditions (unit activities of all species). The salt dissolves only to the extent that the ion product reaches $K_{sp}$, at which point $\Delta G=0$.

The enthalpy and entropy of dissolution decompose the Gibbs energy:

$$\Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }$$

For most sparingly soluble salts, $\Delta H_{\text{diss}}^{\circ }$ is modest (slightly positive or negative, typically $\pm 20\text{kJ/mol}$) and $\Delta S_{\text{diss}}^{\circ }$ is negative (the ions are strongly ordered by hydration shells in solution). The negative entropy contribution makes $\Delta G^{\circ }$ positive even when $\Delta H^{\circ }$ is slightly favourable.

Temperature dependence

The van 't Hoff equation gives the temperature dependence of $K_{sp}$:

$$\frac{d\ln K_{sp}}{dT}=\frac{\Delta H^{\circ }}{R{T}^2}$$

For most sparingly soluble salts, dissolution is endothermic ($\Delta H^{\circ }>0$), so $K_{sp}$ increases with temperature and solubility increases. This is the basis of recrystallisation as a purification technique: dissolve the salt in hot solvent and cool to precipitate the purified solid.

Exceptions exist. ${\text{Ca(OH)}}_2$ and ${\text{Li}}_2{\text{CO}}_3$ have exothermic dissolution ($\Delta H^{\circ }<0$) and become less soluble at higher temperature. ${\text{Ca(OH)}}_2$ is used in the sugar-refining industry because its solubility decreases at elevated temperature, facilitating precipitation of impurities.

Lattice energy and solvation energy

The Gibbs energy of dissolution can be analysed as the sum of two large opposing contributions:

$$\Delta G_{\text{diss}}^{\circ }=\Delta G_{\text{lattice}}^{\circ }+\Delta G_{\text{solvation}}^{\circ }$$

The lattice energy (energy required to separate the solid into gas-phase ions) is always positive and opposes dissolution. The solvation energy (energy released when gas-phase ions are hydrated) is always negative and favours dissolution. For sparingly soluble salts, the lattice energy slightly exceeds the solvation energy, giving a small positive $\Delta G^{\circ }$ and hence a small $K_{sp}$.

The Born-Lande equation estimates the lattice energy:

$$U=\frac{N_{A}M{z}^{+}{z}^{-}{e}^2}{4\pi {ϵ}_0{r}_0}\left(1-\frac{1}{n}\right)$$

where $M$ is the Madelung constant, $z$ are the ion charges, $r_0$ is the interionic distance, and $n$ is the Born exponent. Small, highly charged ions (like ${\text{Al}}^{3+}$ and ${\text{O}}^{2-}$ in ${\text{Al}}_2{\text{O}}_3$) produce very large lattice energies, explaining the extreme insolubility of many oxides and phosphates.

Qualitative analysis scheme Master

Group separation by selective precipitation

The classical qualitative analysis scheme separates cations into five groups by sequential precipitation. Each group is defined by the precipitating reagent and the $K_{sp}$ values of the resulting salts.

Group I: Ag${}^{+}$, Pb${}^{2+}$, Hg${}_2^{2+}$. Precipitated as chlorides by adding dilute HCl. $K_{sp}$ values: $\text{AgCl}=1.8\times 10^{-10}$, ${\text{PbCl}}_2=1.6\times 10^{-5}$, ${\text{Hg}}_2{\text{Cl}}_2=1.4\times 10^{-18}$. The very low $K_{sp}$ values of AgCl and ${\text{Hg}}_2{\text{Cl}}_2$ ensure quantitative precipitation. ${\text{PbCl}}_2$ is partially precipitated (its higher $K_{sp}$ means some remains in solution) and is confirmed in subsequent groups.

Group II: Cu${}^{2+}$, Bi${}^{3+}$, Cd${}^{2+}$, Hg${}^{2+}$, As(III/V), Sb(III/V), Sn(II/IV). Precipitated as sulfides in acidic solution (pH ~0.5) by adding ${\text{H}}_2\text{S}$. The very low $K_{sp}$ values of these sulfides ($10^{-20}$ to $10^{-50}$) ensure precipitation even at the low $[{\text{S}}^{2-}]$ available in acidic solution (where most ${\text{H}}_2\text{S}$ is protonated). The higher-$K_{sp}$ sulfides (MnS, FeS, ZnS, with $K_{sp}$ values of $10^{-10}$ to $10^{-24}$) remain in solution at this pH because the available $[{\text{S}}^{2-}]$ is too low.

Group III: Al${}^{3+}$, Cr${}^{3+}$, Fe${}^{2+}$, Fe${}^{3+}$, Zn${}^{2+}$, Ni${}^{2+}$, Co${}^{2+}$, Mn${}^{2+}$. Precipitated as hydroxides or sulfides in basic solution (pH ~9) with ${\text{NH}}_3$/${\text{NH}}_4\text{Cl}$ buffer and $({\text{NH}}_4{)}_2\text{S}$. The basic pH raises $[{\text{OH}}^{-}]$ and $[{\text{S}}^{2-}]$ enough to precipitate the remaining cations. The ${\text{NH}}_4\text{Cl}$ buffer prevents the pH from becoming so high that amphoteric hydroxides (${\text{Al(OH)}}_3$, ${\text{Zn(OH)}}_2$) redissolve.

Group IV: Ba${}^{2+}$, Ca${}^{2+}$, Sr${}^{2+}$. Precipitated as carbonates by adding $({\text{NH}}_4{)}_2{\text{CO}}_3$ at pH ~10. The sulfates and phosphates of these ions are also sparingly soluble, but carbonate precipitation is used because the earlier groups have already removed all other cations.

Group V: Na${}^{+}$, K${}^{+}$, NH${}_4^{+}$. Remain in solution after Groups I-IV have been removed. Identified by flame tests (Na: yellow, K: violet) and specific chemical tests (Nessler's reagent for ${\text{NH}}_4^{+}$).

The pH window for sulfide precipitation

The separation of Group II from Group III relies on controlling $[{\text{S}}^{2-}]$ through pH. Hydrogen sulfide is a diprotic acid:

$${\text{H}}_2\text{S}\rightleftharpoons {\text{H}}^{+}+{\text{HS}}^{-}{K}_{a1}=9.5\times 10^{-8}$$ $${\text{HS}}^{-}\rightleftharpoons {\text{H}}^{+}+{\text{S}}^{2-}{K}_{a2}=1.0\times 10^{-19}$$

The concentration of ${\text{S}}^{2-}$ depends on pH:

$$[{\text{S}}^{2-}]=\frac{K_{a1}{K}_{a2}[{\text{H}}_2\text{S}]}{[{\text{H}}^{+}{]}^2}$$

In a saturated ${\text{H}}_2\text{S}$ solution ($[{\text{H}}_2\text{S}]\approx 0.10$ M) at pH 0.5:

$$[{\text{S}}^{2-}]=\frac{9.5\times 10^{-8}\times 10^{-19}\times 0.10}{(0.316{)}^2}=9.5\times 10^{-29}/0.100=9.5\times 10^{-28}\text{M}$$

This extremely low $[{\text{S}}^{2-}]$ is sufficient to precipitate CuS ($K_{sp}=8.5\times 10^{-45}$) but not MnS ($K_{sp}=2.3\times 10^{-13}$). At pH 9, $[{\text{S}}^{2-}]$ increases by a factor of $10^{17}$, sufficient to precipitate the Group III sulfides.

Activity corrections in solubility calculations Master

Ionic-strength effects on $K_{sp}$

The thermodynamic $K_{sp}$ is defined in terms of activities:

$$K_{sp}^{\circ }=a_{{\text{M}}^{n+}}^m\cdot a_{{\text{A}}^{m-}}^n$$

The concentration-based apparent $K_{sp}^{\prime }$ is related to the thermodynamic value by:

$$K_{sp}^{\circ }=K_{sp}^{\prime }\cdot {\gamma }_{{\text{M}}^{n+}}^m\cdot {\gamma }_{{\text{A}}^{m-}}^n$$

For a 1:1 salt like AgCl:

$$K_{sp}^{\circ }=K_{sp}^{\prime }\cdot {\gamma }_{\pm }^2$$

In a solution of $0.10\text{M}$ KNO${}_3$ ($I=0.10$ M), the Davies equation gives ${\gamma }_{\pm }\approx 0.38$ for a 1:1 electrolyte. The apparent $K_{sp}^{\prime }=K_{sp}^{\circ }/0.38^2=6.9K_{sp}^{\circ }$. The measured solubility in $0.10\text{M}$ KNO${}_3$ is about 2.6 times the value in pure water -- an increase due entirely to the ionic-strength effect on activity coefficients, not to a chemical reaction.

This "salt effect" (increase in solubility with ionic strength) is distinct from the common-ion effect. An inert electrolyte (no common ions) increases solubility by lowering activity coefficients. A common ion decreases solubility by the mass-action effect. The net effect of a solution containing both a common ion and an inert electrolyte requires combining both corrections.

Seawater solubility

Seawater ($I\approx 0.70$ M) provides a practical example where activity corrections are essential. The solubility of ${\text{CaCO}}_3$ in seawater is several times higher than in pure water because the high ionic strength reduces the activity coefficients of ${\text{Ca}}^{2+}$ and ${\text{CO}}_3^{2-}$. The apparent $K_{sp}^{\prime }$ in seawater is about 100 times larger than the thermodynamic $K_{sp}^{\circ }$, primarily because ${\gamma }_{\pm }$ for a 2:2 electrolyte at $I=0.70$ is very small (${\gamma }_{\pm }\approx 0.10$).

Historical notes Master

The concept of solubility equilibrium emerged from the study of precipitation reactions in analytical chemistry. Carl Remigius Fresenius established the qualitative analysis scheme in his 1841 textbook Anleitung zur qualitativen chemischen Analyse, which became the standard instructional method for analytical chemistry throughout the 19th and 20th centuries. Fresenius's scheme organised cations into groups based on their precipitation behaviour, providing a systematic method for identifying the constituents of an unknown sample.

The quantitative treatment of solubility equilibria began with Walther Nernst's 1889 derivation of the solubility product concept. Nernst showed that the product of ion concentrations in a saturated solution is a constant at fixed temperature, providing the mathematical framework for predicting precipitation and dissolution.

Gilbert Newton Lewis and Merle Randall's Thermodynamics and the Free Energy of Chemical Substances (1923, 2nd edition 1961) connected $K_{sp}$ to thermodynamic quantities, showing that $\Delta G^{\circ }=-RT\ln K_{sp}$ and decomposing the Gibbs energy into lattice-energy and solvation-energy contributions ^{[Lewis Randall 1961]}. Their treatment established the modern thermodynamic framework for solubility.

Werner Stumm and James Morgan's Aquatic Chemistry (1st edition 1970, 3rd edition 1996) applied solubility equilibria to environmental systems, developing the framework for understanding mineral dissolution and precipitation in natural waters ^{[Stumm Morgan 1996]}. Their treatment of pH-dependent solubility, complex-ion formation, and the carbonic acid-calcite system remains the foundation of environmental chemistry and geochemistry.

James Newton Butler's Ionic Equilibrium (1964, revised edition 1998) provided the comprehensive systematic treatment of all equilibrium types in aqueous solution, including the full derivation of pH-dependent solubility, the common-ion effect with activity corrections, and the qualitative analysis scheme from a rigorous equilibrium perspective ^{[Butler 1998]}.

Connections Master

• Chemical equilibrium (14.07): Ksp is a special case of the general equilibrium constant, applied to heterogeneous dissolution reactions. The common-ion effect is Le Chatelier's principle applied to solubility, and the relationship $\Delta G^{\circ }=-RT\ln K_{sp}$ connects directly to the thermodynamic treatment of equilibrium.
• Acid-base chemistry (14.10.01-14.10.03): Metal hydroxide solubility is pH-dependent through the $[{\text{OH}}^{-}]$ term in Ksp, making acid-base equilibria inseparable from solubility equilibria for these salts. The qualitative analysis scheme exploits precisely this coupling to separate cation groups by pH control.
• Electrochemistry (14.11): Ion-selective electrodes measure the activity of dissolved ions, enabling potentiometric determination of Ksp. The Nernst equation for metal-ion electrodes depends on the free metal-ion concentration, which in turn is governed by solubility equilibria when sparingly soluble salts are present.
• Coordination chemistry (inorganic): Complex-ion formation enhances solubility by removing free metal ions from solution, shifting the dissolution equilibrium. The formation constant Kf and the solubility product Ksp combine to give the overall dissolution constant, linking complexation equilibria to solubility.

Bibliography Master

@book{LewisRandall1961,
  author = {Lewis, G. N. and Randall, M.},
  title = {Thermodynamics},
  publisher = {McGraw-Hill},
  year = {1961},
  edition = {2nd},
  note = {Revised by K. S. Pitzer and L. Brewer}
}

@book{StummMorgan1996,
  author = {Stumm, W. and Morgan, J. J.},
  title = {Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters},
  publisher = {Wiley},
  year = {1996},
  edition = {3rd}
}

@book{Butler1998,
  author = {Butler, J. N.},
  title = {Ionic Equilibrium: Solubility and pH Calculations},
  publisher = {Wiley},
  year = {1998}
}

@book{Zumdahl2019,
  author = {Zumdahl, S. S. and DeCoste, D. J.},
  title = {Chemistry},
  publisher = {Cengage Learning},
  year = {2019},
  edition = {10th}
}

@book{Harris2015,
  author = {Harris, D. C.},
  title = {Quantitative Chemical Analysis},
  publisher = {W. H. Freeman},
  year = {2015},
  edition = {9th}
}

@book{Atkins2023,
  author = {Atkins, P. and de Paula, J.},
  title = {Physical Chemistry},
  publisher = {Oxford University Press},
  year = {2023},
  edition = {12th}
}