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)} K_{s} = *a*(M^{+}) *a*(X^{–})

where the activities are those of the saturated solution (i.e. at equilibrium).

Note that in the expression for K_{s}, the activity of MX has been taken as one, as it is a solid. The equilibrium constant K_{s} 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

**K _{s} ≈ S^{2}**

( and clearly *S *≈* √*K_{s} )

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 K_{s} 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)} ↔ Cd^{2+}_{(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

Cd^{2+}_{(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 K_{s} and then *S* is a trivial matter of substituting the correct values into the equations given above.