Question

For the reaction ${\mathrm{Zr}}_3{\left({\mathrm{PO}}_4\right)}_4$ the solubility product is ${\mathrm{K}}_{\mathrm{sp}}$ and solubility is $\mathrm{S}$. Find the correct relation.

• A. $S={\left(\frac{K_{sp}}{6912}\right)}^{1/2}$
• B. $S={\left(\frac{K_{sp}}{216}\right)}^{1/7}$
• C. $S={\left(\frac{K_{sp}}{216}\right)}^{1/8}$
• D. $S={\left(\frac{K_{sp}}{912}\right)}^{1/3}$

Answer 1

The correct option is A

$S={\left(\frac{K_{sp}}{6912}\right)}^{1/2}$

Explanation for the correct option:

Step 1:

$S=\left(\frac{K_{sp}}{6912}\right)1/7$

It is of ${\mathrm{A}}_3{\mathrm{B}}_4$type of salt.

Step 2:

${\mathrm{Zr}}_3\left({\mathrm{Po}}_4\right)4\left(\mathrm{s}\right)\rightleftharpoons 3{\mathrm{Zr}}^{4+}\left(\mathrm{aq}\right)+4{\mathrm{Po}}_4^{3-}\left(\mathrm{aq}\right)$

Step 3:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Zr}}^{4+}\right]}^3{\left[{\mathrm{Po}}_4^{3-}\right]}^4 \\ ={\left(3\mathrm{S}\right)}^3.{\left(4\mathrm{S}\right)}^4 \\ {\mathrm{K}}_{\mathrm{sp}}=6912{\mathrm{S}}^7 \end{gathered}$$

$S=\left(\frac{K_{sp}}{6912}\right)1/7$

Therefore, the correct relation between solubility product(${\mathrm{K}}_{\mathrm{sp}}$) and solubility($\mathrm{S}$) is $S={\left(\frac{K_{sp}}{6912}\right)}^{1/2}$. Hence option A is correct.${\mathrm{A}}_3{\mathrm{B}}_4$

Explanation for the incorrect option:

The above explanation shows that the correct relation between solubility product(${\mathrm{K}}_{\mathrm{sp}}$) and solubility($\mathrm{S}$) is $S={\left(\frac{K_{sp}}{6912}\right)}^{1/2}$. Hence option B is incorrect.

The above explanation shows that the correct relation between solubility product(${\mathrm{K}}_{\mathrm{sp}}$) and solubility($\mathrm{S}$) is $S={\left(\frac{K_{sp}}{6912}\right)}^{1/2}$. Hence option C is incorrect.

The above explanation shows that the correct relation between solubility product(${\mathrm{K}}_{\mathrm{sp}}$) and solubility($\mathrm{S}$) is $S={\left(\frac{K_{sp}}{6912}\right)}^{1/2}$. Hence option D is incorrect