Question

The solubility of $M{L}_2$ (formula weight = 60 g/mol) in water is $2.4\times {10}^{-5}$ g/100 mL solution. Calculate the solubility product constant for $M{L}_2$

Answer 1

Molecular weight of ${ML}_2=60g/mol$

$\Rightarrow 1mole=60g$ of ${ML}_2$

Solubility of ${ML}_2=2.4\times {10}^{-5}\frac{g}{100ml}$

$=2.4\times {10}^{-5}\times 10\times \frac{60g}{60}\times \frac{1}{1000ml}$

$S=\frac{24\times {10}^{-5}}{60}\frac{mol}{litre}$

${ML}_2\rightleftharpoons M^{2+}+2L^{-}$

$K_{sp}={4S}^3$

$=4{(\frac{24}{60}\times {10}^{-5})}^3=2.56\times {10}^{-16}$

$\Rightarrow$ Solubility product constant is $2.56\times {10}^{-16}$.