If solubility product of Zr$$_3$$(PO$$_4$$)$$_4$$ is denoted by K$$_{sp}$$ and its molar solubility is denoted by S, then which of the following relation between S and K$$_{sp}$$ is correct?

First, we recall the definition of the solubility product. For a salt that dissociates according to

$$\text{Salt (s)} \rightleftharpoons a\,\text{C}^{m+} + b\,\text{A}^{n-},$$

the solubility product is stated as

$$K_{sp} = [\text{C}^{m+}]^{\,a}\,[\text{A}^{n-}]^{\,b}.$$

Now we apply this to zirconium phosphate, $$\text{Zr}_3(\text{PO}_4)_4.$$ Its dissolution in water is

$$\text{Zr}_3(\text{PO}_4)_4 (s) \rightleftharpoons 3\,\text{Zr}^{4+} + 4\,\text{PO}_4^{3-}.$$

Let the molar solubility be $$S\ \text{mol L}^{-1}.$$ That means one mole of solid gives:

$$[\text{Zr}^{4+}] = 3S \quad\text{and}\quad [\text{PO}_4^{3-}] = 4S.$$

We substitute these concentrations into the expression for $$K_{sp}$$:

$$\begin{aligned} K_{sp} &= [\text{Zr}^{4+}]^{\,3}\,[\text{PO}_4^{3-}]^{\,4} \\ &= (3S)^{3}\,(4S)^{4}. \end{aligned}$$

We now separate numerical factors from the power of $$S$$:

$$\begin{aligned} (3S)^{3} &= 3^{3}\,S^{3} = 27\,S^{3},\\ (4S)^{4} &= 4^{4}\,S^{4} = 256\,S^{4}. \end{aligned}$$

Multiplying these two results, we obtain

$$K_{sp} = 27\,S^{3} \times 256\,S^{4} = (27 \times 256)\,S^{7}.$$

Next we evaluate the product of the two numbers:

$$27 \times 256 = 6912.$$

So the solubility product becomes

$$K_{sp} = 6912\,S^{7}.$$

We rearrange this equation to solve for $$S$$ in terms of $$K_{sp}$$:

$$S^{7} = \frac{K_{sp}}{6912},$$

and therefore

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

Hence, the correct answer is Option B.