Zirconium phosphate [Zr3(PO4)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 Ksp, then which of the following relationships between S and Ksp is correct?

A. S = (Ksp / (6912))^(1/7)

B. S = (Ksp / 144)^(1/7)

C. S = (Ksp / 6912)^(1/7)

D. S = (Ksp / 6912)^7

To solve this problem, let's first write down the dissociation equation for zirconium phosphate, \([Zr_3(PO_4)_4]\).

1. **Dissociation of Zirconium Phosphate:**

\[
Zr_3(PO_4)_4 \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}
\]

2. **Molar Solubility:**

If the molar solubility of zirconium phosphate is \(S\), then in solution:
- The concentration of \(Zr^{4+}\) will be \(3S\).
- The concentration of \(PO_4^{3-}\) will be \(4S\).

3. **Expression for Solubility Product \({K_{sp}}\):**

The solubility product (\(K_{sp}\)) is given by:

\[
K_{sp} = [Zr^{4+}]^3 \times [PO_4^{3-}]^4
\]

Substitute the values of \([Zr^{4+}] = 3S\) and \([PO_4^{3-}] = 4S\) into the equation:

\[
K_{sp} = (3S)^3 \times (4S)^4
\]

\[
K_{sp} = 27S^3 \times 256S^4
\]

\[
K_{sp} = 6912S^7
\]

4. **Relationship between \(S\) and \({K_{sp}}\):**

From the equation above, we get:

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

5. **Conclusion:**

The correct relationship is option (C):

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