Question

Calculate solubility of $\ce{Zr_3(PO_4)_4(s)}$ 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)

$\begin{matrix}&Zr_{3}\left(PO_{4}\right)_{4}\left(s\right)& \ce{<=>}&3Zr^{+4}\left(aq\right)&+ &4PO_{4}^{-3}\left(aq\right)\\ t=0&a&&0&&0\\ at \; eqm&a-S&&3S&&4S\end{matrix}$
$\ce{K_{sp} = (Zr^{+4})^3 (PO_4{^{-3}})^4}$
$K_{sp} = (3S)^3 (4S)^4$
$K_{sp} = 6912 \; S^7$
$S = \left(\frac{K_{sp}}{6912}\right)^{1/7}$