Question

Zirconium phosphate $\left[\right.\left(Zr\right)_{3}\left(\right. \left(PO\right)_{4} \left.\right)_{4}\left]\right.$ dissociates into three zirconium cations of charge​ $+4$ and four phosphate anions of charge $-3$ . If molar solubility of zirconium phosphate is denoted by $s$ and its solubility product by $K_{sp}$ then which of the following relationship between $s$ and $K_{sp}$ is correct ​?

• A. $\text{S} = \left\{\text{K}_{\text{sp}} / 6 9 1 2\right\}^{7}$
• B. $\text{S} = \left\{\text{K}_{\text{sp}} / 1 4 4\right\}^{1 / 7}$
• C. $\text{S} = \left\{\left(\text{K}\right)_{\text{sp}} / \left(6 9 1 2\right)^{1 / 7}\right\}$
• D. $\text{S} = \left(\left(\text{K}\right)_{sp} / 6 9 1 2\right)^{1 / 7}$

Answer 1

Answer: (D)

In the saturated solution of
$\left(\text{Zr}\right)_{3} \left(\left(\text{PO}\right)_{4}\right)_{4} + \text{aq} \rightleftharpoons 3 \underset{3 \text{S molar}}{\left(\text{Zr}\right)^{4 +}} + 4 \underset{4 \text{S molar}}{\text{PO}_{4}^{3 -}}$
$\left(\text{K}\right)_{\text{sp}} = \left(\left[\left(\text{Zr}\right)^{4 +}\right]\right)^{3} \left(\left[\text{PO}_{4}^{3 -}\right]\right)^{4} = \left(3 \text{S}\right)^{3} \left(4 \text{S}\right)^{4}$
$= 2 7 \times 2 5 6 \times \text{s}^{7} = 6 9 1 2 \text{ S}^{7}$
$\text{S} = \left(\frac{\left(\text{K}\right)_{\text{sp}}}{6 9 1 2}\right)^{1 / 7}$