Solubility equilibrium

Solubility equilibrium is any type chemical equilibrium between solid and dissolved states of a compound at saturation.

Solubility equilibria involve application of chemical principles and constants to predict solubility of substances under specific conditions (because solubility is sensitive to the conditions, while the constants are less so).

The substance that is dissolved can be an organic solid such as sugar or an ionic solid such as table salt. The main difference is that ionic solids dissociate into constituent ions when they dissolve in water. Most commonly water is the solvent of interest, although the same basic principles apply with any solvent.

In the case of environmental science studies of water quality, the total concentration of dissolved solids (not necessarily at saturation) is referred to as total dissolved solids.

Non-ionic compounds
Dissolution of an organic solid can be described as an equilibrium between the substance in its solid and dissolved forms:


 * $$\mathrm{{C}_{12}{H}_{22}{O}_{11}(s)} \rightleftharpoons \mathrm{{C}_{12}{H}_{22}{O}_{11}(aq)}$$

An equilibrium expression for this reaction can be written, as for any chemical reaction (products over reactants):


 * $$K = \frac{\left\{\mathrm{{C}_{12}{H}_{22}{O}_{11}}(aq)\right\}}{ \left \{\mathrm{{C}_{12}{H}_{22}{O}_{11}}(s)\right\}}$$

where K is called the equilibrium constant (or solubility constant). The curly brackets indicate activity. The activity of a pure solid is, by definition, unity. If the activity of the substance in solution is constant (i.e. not affected by any other solutes that may be present) it may be replaced by the concentration.
 * $$K_s = \left[\mathrm{{C}_{12}{H}_{22}{O}_{11}}(aq)\right]\,$$

The square brackets mean molar concentration in mol dm-3 (sometimes called molarity with symbol M).

This statement says that water at equilibrium with solid sugar contains a concentration equal to K. For table sugar (sucrose) at 25 °C, K = 1.971 mol/L. (This solution is very concentrated; sucrose is extremely soluble in water.) This is the maximum amount of sugar that can dissolve at 25 °C; the solution is saturated. If the concentration is below saturation, more sugar dissolves until the solution reaches saturation, or all the solid is consumed. If more sugar is present than is allowed by the solubility expression then the solution is supersaturated and solid will precipitate until the saturation concentration is reached. This process can be slow; the equilibrium expression describes concentrations when the system reaches equilibrium, not how fast it gets there.

Ionic compounds
Ionic compounds normally dissociate into their constituent ions when they dissolve in water. For example, for calcium sulfate:


 * $$\mathrm{CaSO}_4(s) \rightleftharpoons \mbox{Ca}^{2+}(aq) + \mbox{SO}_4^{2-}(aq)\,$$

As for the previous example, the equilibrium expression is:


 * $$K = \frac{\left\{\mbox{Ca} ^{2+}(aq)\right\}\left\{\mbox{SO}_4^{2-}(aq)\right\}}{ \left\{\mbox{CaSO}_4(s)\right\}}$$

where K is called the equilibrium (or solubility) constant and curly brackets indicate activity.

The activity of a pure solid is, by definition, equal to one. When the solubility of the salt is very low the activity coefficients of the ions in solution will also be equal to one and this expression reduces to the solubility product expression:


 * $$K_{\mathrm{sp}} = \left[\mbox{Ca}^{2+}(aq)\right]\left[\mbox{SO}_4^{2-}(aq)\right].\,$$

This expression says that an aqueous solution in equilibrium with (saturated) solid calcium sulfate has concentrations of these two ions such that their product equals Ksp; for calcium sulfate Ksp = 4.93×10&minus;5. If the solution contains only calcium sulfate, and the conditions are such that dissolved species are only Ca2+ and SO42-, then the concentration of each ion (and the overall solubility of calcium sulfate) is


 * $$\sqrt{ K_{\mathrm{sp}}}=\sqrt{4.93\times10^{-5}}=7.02\times10^{-3}=\left[\mbox{Ca}^{2+}\right]=\left[\mbox{SO}_4^{2-}\right].\,$$

When a solution dissociates into unequal parts as in:


 * $$\mathrm{Ca(OH)_2}(s) \rightleftharpoons \mbox{Ca}^{2+}(aq) + \mbox{2OH}^{-}(aq)\,$$,

then determining the solubility from Ksp is slightly more difficult. Generally, for the dissolution reaction:


 * $$\mathrm{A}(s) \rightleftharpoons \mbox{xB}^{p+}(aq) + \mbox{yC}^{q-}(aq)\,$$

the solubility and solubility product are tied with the equation:

$$\sqrt[n]{K_{\mathrm{sp}} \over {x^x \cdot y^y}} = {C \over M_M}$$


 * where:
 * n is the total number of moles on the right hand side, i.e., x+y, dimensionless
 * x is the number of moles of the cation, dimensionless
 * y is the number of moles of the anion, dimensionless
 * Ksp is the solubility product, (mol/kg)n
 * C is the solubility of A expressed as a mass fraction of the solute A in the solvent (kg of A per kg of solvent)
 * MM is the molecular mass of the compound A, kg/mol.

Again, the above equation assumes that the dissolution takes place in pure solvent (no common ion effect), that there is no complexation or hydrolysis (i.e., only ions Bp+ and Cq- are present in the solution), and that the concentrations are sufficiently low for the activity coefficients to be taken as unity.

Common ion effect
The common-ion effect refers to the fact that solubility equilibria shift in accordance with Le Chatelier's Principle. In the above example, addition of sulfate ions to a saturated solution of calcium sulfate causes CaSO4 to precipitate until the concentration of the ions in solution are such that they again satisfy the solubility product. (Addition of sulfate ions can, for example, be accomplished by adding a very soluble salt, such as Na2SO4.)

Salt effect
The salt effect refers to the fact that the presence of another salt, even though there is no common ion, has an effect on the ionic strength of the solution and hence on the activity coefficients of the ions, so that solubility changes even though Ksp remains constant (assuming that the activity of the solid remains unity).

Speciation effect
On dissolution, ionic salts typically dissociate into their constituent ions, but the ions may speciate in the solution. On speciation, the solubility will always increase although the solubility product does not change. For example, solubility equilibrium for calcium carbonate may be expressed by:


 * $$\mathrm{CaCO}_3(s) \rightleftharpoons \mbox{Ca}^{2+}(aq) + \mbox{CO}_3^{2-}(aq)\,$$
 * $$K_{\mathrm{sp}} = \left[\mbox{Ca}^{2+}(aq)\right]\left[\mbox{CO}_3^{2-}(aq)\right].\,$$

Now, if the conditions (e.g., pH) are such that other carbonate (or calcium) species appear in the solution (for example, bicarbonate ion HCO3-), then the solubility of the solid will increase so that the solubility product remains constant.

Similarly, if a complexing agent, for example EDTA, was present in the solution, solubility will increase because of the complexation of calcium (a complex has a different chemical identity than uncomplexed Ca2+ and therefore does not enter the solubility equilibrium).

To correctly predict solubility from a given solubility product, the speciation need to be known (or evaluated, at least approximately). A failure to do so is a common problem and can lead to large errors.

Phase effect
Equilibria are defined for specific crystal phases. Therefore, the solubility product is expected to be different depending on the phase of the solid. For example, aragonite and calcite will have different solubility products even though they have both the same chemical identity (calcium carbonate). Nevertheless, under given conditions, most likely only one phase is thermodynamically stable and therefore this phase enters a true equilibrium.

Particle size effect
The thermodynamic solubility constant is defined for large monocrystals. Solubility will increase with decreasing size of solute particle (or droplet) because of the additional surface energy. This effect is generally small unless particles become very small, typically smaller than 1 μm. The effect of the particle size on solubility constant can be quantified as follows:


 * $$\log(^*K_{A}) = \log(^*K_{A \to 0}) + \frac{2 \gamma A_m} {3\ln(10)RT}$$

where $$^*K_{A}$$ is the solublity constant for the solute particles with the molar surface area A, $$^*K_{A \to 0}$$ is the solubility constant for substance with molar surface area tending to zero (i.e., when the particles are large), γ is the surface tension of the solute particle in the solvent, Am is the molar surface area of the solute (in m2/mol), R is the universal gas constant, and T is the absolute temperature.

Temperature effects
Solubility is sensitive to changes in temperature. For example, sugar is more soluble in hot water than cool water. It occurs because solubility constants, like other types of equilibrium constant, are functions of temperature. In accordance with Le Chatelier's Principle, when the dissolution process is endothermic (heat is absorbed,) solubility increases with rising temperature, but when the process is exothermic (heat is released) solubility decreases with rising temperature.

Solubility constants
Solubility constants have been experimentally determined for a large number of compounds and tables are readily available. For ionic compounds the constants are called solubility products. Concentration units are assumed to be molar unless otherwise stated. Solubility is sometimes listed in units of grams dissolved per liter of water.

Some values at 25°C:
 * Barium carbonate: 2.60
 * Copper(I) chloride: 1.72
 * Lead(II) sulfate: 1.81
 * Magnesium carbonate: 1.15
 * Silver chloride: 1.70
 * Silver bromide: 7.7
 * Calcium hydroxide: 8.0

See also
 * IUPAC-NIST solubility database
 * Solubility products of simple inorganic compounds