The solubility, S, of a salt may be defined as the molality of a saturated solution of that salt, and, for a relatively insoluble salt, may be estimated from the standard potential of the appropriate cell.
For a general salt of low solubility, represented by MX, the solubility may be discussed in terms of the solubility equilibrium:
MX(s) ↔ M+(aq) + X–(aq) Ks = a(M+) a(X–)
where the activities are those of the saturated solution (i.e. at equilibrium).
Note that in the expression for Ks, the activity of MX has been taken as one, as it is a solid. The equilibrium constant Ks is called the solubility constant (or solubility product) of the salt. When the solubility of the salt is so low that the mean activity coefficient, γ± , is approximately equal to one even in the saturated solution, then from the definition of activity we may write a ≈ b/bº.
However, in the saturated solution, b = S (from the definition of S)
Thus, since bº = 1
Ks ≈ S2
( and clearly S ≈ √Ks )
Thus it is possible to estimate S from the standard potential of a cell with an overall cell reaction corresponding to the solubility equilibrium (as Ks may be obtained from the relation ln K = νFEº/RT )
For Example, to estimate the solubility of cadmium hydroxide, Cd(OH)2 , we need to consider the solubility equilibrium
Cd(OH)2 (s) ↔ Cd2+(aq) + 2OH –(aq)
which can be expressed as the difference of the following half equations:
Cd(OH)2 (s) + 2e– → Cd (s) + 2OH –(aq) Eº = – 0.81 V
Cd2+(aq) + 2e– → Cd (s) Eº = – 0.40 V
So the cell potential, Eº, is – 0.81 – (- 0.40) = – 0.41 V. For this reaction, ν = 2. From here, calculation of Ks and then S is a trivial matter of substituting the correct values into the equations given above.