Question

Calculate solubility of $\mathrm{Zr}{}_3(\mathrm{PO}{}_4){}_4(\mathrm{s})$ in terms of $K_{sp}$ ?

• A. $S={\left(\frac{K_{sp}}{6912}\right)}^{1/7}$
• B. $S={\left(\frac{K_{sp}}{6912}\right)}^{1/6}$
• C. $S={\left(\frac{K_{sp}}{1728}\right)}^{1/6}$
• D. $S={\left(\frac{K_{sp}}{1728}\right)}^{1/7}$

Answer 1

Answer: (A)

$\begin{array}{cccccc}& Z{r}_3{\left(P{O}_4\right)}_4\left(s\right)& \rightleftharpoons & 3Z{r}^{+4}\left(aq\right)& +& 4P{O}_4^{-3}\left(aq\right)\\ t=0& a& & 0& & 0\\ ateqm& a-S& & 3S& & 4S\end{array}$
$\mathrm{K}{}_{\mathrm{sp}}=(\mathrm{Zr}{}^{+4}){}^3(\mathrm{PO}{}_4{}^{\text{-}3}){}^4$
$K_{sp}=(3S{)}^3(4S{)}^4$
$K_{sp}=6912S^7$
$S={\left(\frac{K_{sp}}{6912}\right)}^{1/7}$