Solubility product

Solubility is a measure of a solute’s ability to dissolve in a solvent. More specifically, it is the concentration of a saturated solution of a solute when dissolved in a particular solvent at a particular temperature, measured in grams per hundred millilitres (g/100 mL) or, for molar solubility, moles per litre (mol/L).

A solution that has the maximum amount of solute dissolved in it is saturated. If there is less, the solution is unsaturated. If there is more, the solution is supersaturated—this is unstable, and the extra solute will usually precipitate.

The solubility product constant (`K_"sp"`) is a special case of the equilibrium constant. It applies only to saturated solutions of solutes that have very low solubilities. Here is an example of one such solubility equilibrium:

BaSO4(s) ⇌ Ba2+(aq) + SO42−(aq).

The fact that this system is at equilibrium means that barium sulfate is dissolving at the same rate that its ions are precipitating. These processes cancel each other out, so there is no overall change in the concentrations of the ions. Since we don’t include solids or liquids in the equilibrium law expression, the solubility product constant for this system is

`K_"sp" = ["Ba"^(2+)]\["SO"_4^(2-)]`.

The name solubility product constant should help you remember to only include the products (the ions). Smaller values of `K_"sp"` indicate lower solubilities because the position of the equilibrium is further left. Greater values indicate higher solubilities.

What if you have an extremely dilute solution of barium sulfate—so much water and so little solute that it all dissolves? This means you have an unsaturated solution; the equilibrium equation doesn’t apply (and neither do `K_"sp"` or LCP) because there is no dynamic equilibrium: the solid is not in equilibrium with the ions because there is no solid. What if you increase the concentration of the ions in a saturated solution? Some ions will precipitate, just like you would expect using LCP.

Sometimes it will be necessary to convert between mol/L and g/100 mL. To convert the numerator of the unit between g and mol, multiply or divide by molar mass. To convert the denominator of the unit from 100 mL to L, multiply or divide by 0.1. It’s easier to visualize when you write each one as a fraction and cross out the units that cancel.

Example

What is the solubility of silver carbonate in g/100 mL if the solubility product constant is known to be 8.8 × 10−12?

First, write the solubility equilibrium equation:

Ag2CO3(s) ⇌ 2 Ag+(aq) + CO32−(aq).

Now we can write the equilibrium law expression:

`K_"sp" = ["Ag"^+]^2["CO"_3^(2-)]`.

Let `x` represent the equilibrium concentration of Ag2CO3(aq). Note that Ag2CO3(aq) is not in the equilibrium equation above, but it represents the entire right side, which is just the dissociated version of the aqueous compound. You can use the coefficients in the equation to write the concentration of each ion in terms of `x`.

`8.8xx10^-12 = (2x)^2(x) = 4x^3`.

Now we can solve for `x`, which represents the molar solubility of silver carbonate:

`x = root3((8.8xx10^-12)/4) = 1.3xx10^-4\ "mol/L"`.