Question

Calculate the molar solubility $(S)$ of a salt like zirconium phosphate of molecular formula. ${\left(Z{r}^{4+}\right)}_3{\left(P{O}_4^{3-}\right)}_4$.

• A. ${\left(\frac{K_{sp}}{9612}\right)}^{1/8}$
• B. ${\left(\frac{K_{sp}}{6912}\right)}^{1/7}$
• C. ${\left(\frac{K_{sp}}{5348}\right)}^{1/6}$
• D. ${\left(\frac{K_{sp}}{8435}\right)}^{1/7}$

Answer 1

Answer: (B)

Consider a salt, zirconium phosphate of molecular formula ${\left(Z{r}^{4+}\right)}_3{\left(P{O}_4^{3-}\right)}_4$. It dissociates into $3$
zirconium cations of charge $+4$ and $4$ phosphate anions of charge $-3$.
$\left[Z{r}^{4+}\right]=3S$ and $\left[P{O}_4^{3-}\right]=4S$
and $K_{sp}=(3S{)}^3(4S{)}^4=6912(S{)}^7$
or $S={\left\{K_{sp}/\left(3^3\times 4^4\right)\right\}}^{1/7}={\left(K_{sp}/6912\right)}^{1/7}$