$\mathrm{M}_{x}^{p+} \mathrm{X}_{y}^{q-}$ is a sparingly soluble salt and solubility is $\mathrm{S} \mathrm{mol} \mathrm{L}^{-1}$. The amount of this salt is dissolve in solution is totally in the form of ions. In solution 1 molecule $x \mathrm{M}^{p+}$ positive ion and $y \mathrm{X}^{q-}$ negative ion and form, from this salt and ionic equation is as follows.

$\mathrm{M}_{x}^{p+} \mathrm{X}_{y(\mathrm{~s})}^{q} \square \quad x \mathrm{M}_{(\mathrm{aq})}^{p+}+y \mathrm{X}_{(\mathrm{aq})}^{q-} \quad \ldots . .(\mathrm{Eq} .-\mathrm{i})$

Where $\left(x \times p^{+}=y \times q^{-}\right)$

The concentration of solid salt mix with $\mathrm{K}_{s p}$, So solubility product is as follows. $\mathrm{K}_{s p}=\left[\mathrm{M}^{\mathrm{p}+}\right]^{x}\left[\mathrm{X}^{\mathrm{q}-}\right]^{y}$ But applying stoichiometry and $\mathrm{S}$

$\mathrm{K}_{s p}=(x \mathrm{~S})^{x}(\mathrm{~S})^{y}$

$=x^{x} \times y^{y}(\mathrm{~S})^{(x+y)} \ldots(\text { Eq.-ii })$

$\therefore(\mathrm{S})^{(x+y)}=\frac{\mathrm{K}_{s p}}{x^{x} y^{y}}$

$\therefore \mathrm{S}=\frac{\mathrm{K}_{s p}}{\left(x^{x} \times y^{y}\right)^{1 /(x+y)}} \ldots(\text { Eq.-iii) }$