Chemistry · Ch 3 — Ionic Equilibria

Relationship between solubility and solubility product

3.9.2

The solubility of a compound is the amount in grams that dissolves per unit volume (which may be 100 mL or 1 L) of its saturated solution.

Molar solubility : The number of moles of a compound that dissolve to give one litre of saturated solution is called its molar solubility.

\text{molar solubility (mol/L)} = \frac{\text{solubility in g/L}}{\text{molar mass in g/mol}}

Consider once again the solubility equilibrium for $\mathrm{B}_x\mathrm{A}_y$ ,

\mathrm{B}_x\mathrm{A}_y\mathrm{(s)} \rightleftharpoons x\,\mathrm{B^{y+}(aq)} + y\,\mathrm{A^{x-}(aq)}

The solubility product is given by Eq. (3.28) :

K_{sp} = [\mathrm{B^{y+}}]^x\,[\mathrm{A^{x-}}]^y

If $S$ is the molar solubility of the compound, the equilibrium concentrations of the ions in the saturated solution will be

[\mathrm{B^{y+}}] = xS \ \text{mol/L} \quad \text{and} \quad [\mathrm{A^{x-}}] = yS \ \text{mol/L}

From Eq. (3.28)

K_{sp} = [xS]^x\,[yS]^y = x^x y^y S^{x+y} \qquad \text{...(3.29)}

For example :

i. For AgBr, $\mathrm{AgBr(s)} \rightleftharpoons \mathrm{Ag^{+}(aq)} + \mathrm{Br^{-}(aq)}$ . Here, $x = 1$ , $y = 1$ :

$\therefore K_{sp} = S \times S = S^2$ …