Ncert Exercises & Solutions

A sparingly soluble salt having general formula $A_{x}^{p +} B_{y}^{q -}$ and molar solubility S is in equilibrium with its saturated solution. Derive a relationship between the solubility and solubility product for such salt.

A sparingly soluble salt having general formula

A_{x}^{p +} B_{y}^{q -}

. Its molar solubility is S mol L^-1.

Then,

A_{x}^{p +} B_{y}^{q -} ⇌ \underset{x}{{xA}^{p +}} (aq) + \underset{y}{{yB}^{q -}} (aq)

S moles of

A_{x} B_{y}

dissolve to give x moles of

A^{p +}

and y moles of

B^{q -}

Therefore, solubility product

(K_{sp}) = {[A^{p +}]}^{x} {[B^{q -}]}^{y}

= {[xS]}^{x} {[yS]}^{y}

= x^{x} y^{y} S^{x + y}

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