Question

Zirconium phosphate $[{\left(Zr\right)}_3({\left(PO\right)}_4{)}_4]$ 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. $\text{S}={\left\{{\text{K}}_{\text{sp}}/6912\right\}}^7$
• B. $\text{S}={\left\{{\text{K}}_{\text{sp}}/144\right\}}^{1/7}$
• C. $\text{S}=\left\{{\left(\text{K}\right)}_{\text{sp}}/{\left(6912\right)}^{1/7}\right\}$
• D. $\text{S}={\left({\left(\text{K}\right)}_{sp}/6912\right)}^{1/7}$

Answer 1

Answer: (D)

In the saturated solution of
${\left(\text{Zr}\right)}_3{\left({\left(\text{PO}\right)}_4\right)}_4+\text{aq}\rightleftharpoons 3\underset{3\text{S molar}}{{\left(\text{Zr}\right)}^{4+}}+4\underset{4\text{S molar}}{{\text{PO}}_4^{3-}}$
${\left(\text{K}\right)}_{\text{sp}}={\left(\left[{\left(\text{Zr}\right)}^{4+}\right]\right)}^3{\left(\left[{\text{PO}}_4^{3-}\right]\right)}^4={\left(3\text{S}\right)}^3{\left(4\text{S}\right)}^4$
$=27\times 256\times {\text{s}}^7=6912{\text{ S}}^7$
$\text{S}={\left(\frac{{\left(\text{K}\right)}_{\text{sp}}}{6912}\right)}^{1/7}$