# Molar solubility

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 (Ksp) and stoichiometry. The units are mol/L, sometimes written as M.

## Derivation

Given excess of a simple salt AxBy 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:
${\text{A}_x \text{B}_y}_{(s)} \Longleftrightarrow x \text{A}_{(aq)} + y \text{B}_{(aq)}\,$
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:
$-\frac{N_{ AxBy(\Delta)}}{1} = \frac{N_{A(\Delta)}}{x} = \frac{N_{B(\Delta)}}{y}\,$
which, when rearranged for ∆A and ∆B yields:
$N_{A(\Delta)} = -xN_{AxBy(\Delta)}\,$
$N_{B(\Delta)} = -yN_{AxBy(\Delta)}\,$

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.
$N_{(i)} + N_{(\Delta)} = N_{(f)}\,$ The Difference law

$N_{A(i)} = 0\,$
$N_{B(i)} = 0\,$

Which condense to the identities
$N_{A(f)} = N_{A(\Delta)}\,$
$N_{B(f)} = N_{B(\Delta)}\,$

3. In these variables (with V for volume), the Molar solubility would be written as:
$S_0 = -\frac{N_{ AxBy(\Delta)}}{V}\,$

4. The Solubility Product expression is defined as:
$K_{sp} = [A]^x[B]^y\,$

These four sets of equations are enough to solve for S0 algebraically:

$K_{sp} = {\left(\frac{N_{A(f)}}{V}\right)}^x {\left(\frac{N_{B(f)}}{V}\right)}^y\,$

$K_{sp} = {\left(\frac{N_{A(\Delta)}}{V}\right)}^x {\left(\frac{N_{B(\Delta)}}{V}\right)}^y\,$

$K_{sp} = \frac{{(N_{A(\Delta)})}^x {(N_{B(\Delta)})}^y}{V^{(x+y)}}\,$

$K_{sp} = \frac{{(-xN_{AxBy(\Delta)})}^x {(-yN_{AxBy(\Delta)})}^y}{V^{(x+y)}}\,$

$K_{sp} = \frac{{(-1)}^x {(x)}^x {(N_{AxBy(\Delta)})}^x {(-1)}^y {(y)}^y {(N_{AxBy(\Delta)})}^y}{V^{(x+y)}}\,$

$K_{sp} = x^x y^y \frac{{(-1)}^x {(-1)}^y {(N_{AxBy(\Delta)})}^x {(N_{AxBy(\Delta)})}^y}{V^{(x+y)}}\,$

$K_{sp} = x^x y^y \frac{{(-1)}^{(x+y)}{(N_{AxBy(\Delta)})}^{(x+y)}}{V^{(x+y)}}\,$

$\frac{K_{sp}}{x^x y^y} = {\left(-\frac{N_{AxBy(\Delta)}}{V}\right)}^{(x+y)}\,$

$\frac{K_{sp}}{x^x y^y} = {\left(S_0\right)}^{(x+y)}\,$

Hence;

$S_0 = \sqrt[(x+y)]{\frac{K_{sp}}{x^x y^y}}\,$

## Simple calculation

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

### Example

Ionic substance AB2 dissociates 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:

$AB_{2(s)} \Longleftrightarrow A_{aq} + 2B_{aq}$

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:

$K_{sp} = (x)(2x)^2$
$K_{sp} = (x)(4x^2)$
$K_{sp} = 4x^3$

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