Question

Calculate the molar solubility $(S)$ of a salt like zirconium phosphate of molecular formula. $\left( Zr ^{4+}\right)_{3}\left( PO _{4}^{3-}\right)_{4}$.

• 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 $\left( Zr ^{4+}\right)_{3}\left( PO _{4}^{3-}\right)_{4}$. It dissociates into $3$
zirconium cations of charge $+4$ and $4$ phosphate anions of charge $-3$.
$\left[ Zr ^{4+}\right]=3 S $ and $\left[ PO _{4}^{3-}\right]=4 S$
and $K_{ sp }=(3 S)^{3}(4 S)^{4}=6912(S)^{7} $
or $S =\left\{K_{ sp } /\left(3^{3} \times 4^{4}\right)\right\}^{1 / 7}=\left(K_{ sp } / 6912\right)^{1 / 7}$