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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