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

Derivation

Given excess of a simple salt A_xB_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:

${\displaystyle {\ce {{{A}_{\mathit {x}}{B}_{\mathit {y}}}{(s)}->{\mathit {x}}{A}{(aq)}+{\mathit {y}}{B}{(aq)}}}}$

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:

   ${\displaystyle -{\frac {N_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )}}{1}}={\frac {N_{\ce {A(\Delta )}}}{x}}={\frac {N_{\ce {B(\Delta )}}}{y}}\,}$

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

   ${\displaystyle N_{\ce {A(\Delta )}}=-xN_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )}\,}$

   ${\displaystyle N_{\ce {B(\Delta )}}=-yN_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )}\,}$

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

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

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

   Which condense to the identities

   ${\displaystyle N_{{\ce {A}}(f)}=N_{{\ce {A}}(\Delta )}\,}$

   ${\displaystyle N_{{\ce {B}}(f)}=N_{{\ce {B}}(\Delta )}\,}$

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

   ${\displaystyle S_{0}=-{\frac {N_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )}}{V}}\,}$

4. The Solubility Product expression is defined as:

   ${\displaystyle K_{sp}=[{\ce {A}}]^{x}[{\ce {B}}]^{y}\,}$

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

   ${\displaystyle {\begin{aligned}K_{sp}&={\left({\frac {N_{{\ce {A}}(f)}}{V}}\right)}^{x}{\left({\frac {N_{{\ce {B}}(f)}}{V}}\right)}^{y}\\&={\left({\frac {N_{\ce {A(\Delta )}}}{V}}\right)}^{x}{\left({\frac {N_{\ce {B(\Delta )}}}{V}}\right)}^{y}\\&={\frac {{(N_{\ce {A(\Delta )}})}^{x}{(N_{\ce {B(\Delta )}})}^{y}}{V^{(x+y)}}}\\&={\frac {{(-xN_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )})}^{x}{(-yN_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )})}^{y}}{V^{(x+y)}}}\\&={\frac {{(-1)}^{x}{(x)}^{x}{(N_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )})}^{x}{(-1)}^{y}{(y)}^{y}{(N_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )})}^{y}}{V^{(x+y)}}}\\&=x^{x}y^{y}{\frac {{(-1)}^{x}{(-1)}^{y}{(N_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )})}^{x}{(N_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )})}^{y}}{V^{(x+y)}}}\\&=x^{x}y^{y}{\frac {{(-1)}^{(x+y)}{(N_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )})}^{(x+y)}}{V^{(x+y)}}}\end{aligned}}}$

   ${\displaystyle {\begin{aligned}{\frac {K_{sp}}{x^{x}y^{y}}}&={\left(-{\frac {N_{{\ce {A}}_{x}{\ce {B}}_{y}(\Delta )}}{V}}\right)}^{(x+y)}\\&={\left(S_{0}\right)}^{(x+y)}\end{aligned}}}$

   Hence;

   ${\displaystyle S_{0}={\sqrt[{(x+y)}]{\frac {K_{sp}}{x^{x}y^{y}}}}}$

Simple calculation

If the solubility product constant (K_sp) 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:

${\displaystyle {\ce {AB2(s)->A{(aq)}+2B{(aq)}}}}$

where the corresponding solubility product equation is:

${\displaystyle K_{sp}={\ce {[A][B]^{2}}}}$

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:

${\displaystyle {\begin{aligned}K_{sp}&=(x)(2x)^{2}\\&=(x)(4x^{2})\\&=4x^{3}\end{aligned}}}$

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