How will you calculate solubility product from molar solubility ?

Explanation

Solubility can be calculated from the molar solubility i.e., the maximum number of moles of solute that can be dissolved in one litre of the solution.
For a solute `X_m Y_n`
`X_(m)Y_(n(s)) hArr mX_((aq))^(n+)+nY_((aq))^(m-)`
From the above stoichiometrically balanced equation we have come to know that 1 mole of `X_(m)Y_(n(s))` dissociated to furnish 'm' moles of `X^(n+)` and 'n' mole of `Y^(m-)` if 's' is molar solubility of `X_(m)Y_(n)`, then
`[X^(n+)]=ms and [Y^(m-)]=ns`
`therefore K_(sp)=[X^(n+)]^(m)[Y^(m-)]^n`
`K_(sp)=(ms)^m (ns)^n`
`K_(sp)=(m)^m (n)^n (s)^(m+n)`