Chemistry:Molar solubility

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• Chemistry:Solubility equilibrium

Molar solubility is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution has achieved saturation 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:

[math]\ce{ {{A}_\mathit{x} {B}_\mathit{y}}{(s)} -> \mathit{x} {A}{(aq)} + \mathit{y} {B}{(aq)} }[/math]

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:

   [math]\displaystyle{ -\frac{N_{\ce{A}_x\ce{B}_y(\Delta)}}{1} = \frac{N_\ce{A(\Delta)}}{x} = \frac{N_\ce{B(\Delta)}}{y}\, }[/math]

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

   [math]\displaystyle{ N_\ce{A(\Delta)} = -xN_{\ce{A}_x\ce{B}_y(\Delta)}\, }[/math]

   [math]\displaystyle{ N_\ce{B(\Delta)} = -yN_{\ce{A}_x\ce{B}_y(\Delta)}\, }[/math]

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.
   [math]\displaystyle{ N_{(i)} + N_{(\Delta)} = N_{(f)}\, }[/math] Difference law

   [math]\displaystyle{ N_{\ce A(i)} = 0\, }[/math]

   [math]\displaystyle{ N_{\ce B(i)} = 0\, }[/math]

   Which condense to the identities

   [math]\displaystyle{ N_{\ce A(f)} = N_{\ce A(\Delta)}\, }[/math]

   [math]\displaystyle{ N_{\ce B(f)} = N_{\ce B(\Delta)}\, }[/math]

3. In these variables (with V for volume), the molar solubility would be written as:
   [math]\displaystyle{ S_0 = -\frac{N_{ \ce{A}_x\ce{B}_y(\Delta)}}{V}\, }[/math]
4. The solubility product expression is defined as:
   [math]\displaystyle{ K_{sp} = [\ce A]^x[\ce B]^y\, }[/math]

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

   [math]\displaystyle{ \begin{align} 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\\ & = x^x y^y \frac{{(-1)}^{(x+y)}{(N_{\ce{A}_x\ce{B}_y(\Delta)})}^{(x+y)}}{V^{(x+y)}} \end{align} }[/math]

   [math]\displaystyle{ \begin{align} \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{align} }[/math]

   Hence;

   [math]\displaystyle{ S_0 = \sqrt[(x+y)]{\frac{K_{sp}}{x^x y^y}} }[/math]

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:

[math]\ce{ AB2(s) -> A{(aq)} + 2B{(aq)} }[/math]

where the corresponding solubility product equation is:

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

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

[math]\displaystyle{ \begin{align} K_{sp} &= (x)(2x)^2\\ & = (x)(4x^2)\\ & = 4x^3 \end{align} }[/math]

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

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