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**Molar solubility** is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated. It can be calculated from a substance's solubility product constant (K_{sp}) and stoichiometry. The units are mol/L, sometimes written as M.

Given excess of a simple salt A_{x}B_{y} in an aqueous solution of no common ions (A or B already present in the solution), the amount of it which enters solution can be calculated as follows:

The Chemical equation for this salt would be:

where A, B are ions and x, y are coefficients...

1. The relationship of the changes in amount (of which mole is a unit), represented as N_{(∆)}, between the species is given by Stoichiometry as follows:

which, when rearranged for ∆A and ∆B yields:

2. The Deltas(∆), Initials(i) and Finals(f) relate very simply since, in this case, Molar solubility is defined assuming no common ions are already present in the solution.

The Difference law

Which condense to the identities

3. In these variables (with V for volume), the Molar solubility would be written as:

4. The Solubility Product expression is defined as:

These four sets of equations are enough to solve for S_{0} algebraically:

Hence;

If the solubility product constant (K_{sp}) and dissociation product ions are known, the molar solubility can be computed without the aforementioned equation.

Ionic substance AB_{2} dissociates into into A and 2B, or one mole of ion A and two moles of ion B. The soluble ion dissociation equation can thus be written as:

AB_{2} --> A + 2B

where the corresponding solubility product equation is:

K_{sp} = [A][B]^2

If the initial concentration of A is x, then that of B must be 2x. Inserting these initial concentrations into the solubility product equation gives

K_{sp} = (x)(2x)^2

If Ksp is known, x can be computed, which is the molar solubility.