Q. 53

A sparingly solub

Answer :

A relationship between the solubility and solubility product for such salt can be derived as follows-

Let the solubility be S.

Initially, assume 1 mole of AxBy was present. From which S moles of it was dissolved to give xS and yS moles of Ap+ and Bq- respectively. This can be understood from the following equation.

Ap+x Bq-y xA+p + yB-q

At t=0, 1 0 0

At Equilibrium, 1-S xS yS

Solubility product is the multiplication product of concentration of products each raise to their stoichiometric coefficients.

Hence, Solubility product (Ksp) = [A+p]x [B-q]y

= (xS)x(yS)y

= xxyySx+y

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