for `Zr_(3)(PO_(4))_(4)` the solubility product is `K_(sp)` and solubility is S. find the correct relation

Options

A. `S=(K_(sp)/(6912))^(1/7)`
B. `S=(K_(sp)/(929))^(1/9)`
C. `S=(K_(sp)/(216))^(1/7)`
D. `S=(K_(sp)/(144))^(1/6)`

Explanation

To solve the problem regarding the solubility product \( K_{sp} \) and solubility \( S \) of zirconium phosphate \( Zr_3(PO_4)_4 \), we will follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of zirconium phosphate in water can be represented as:
\[ Zr_3(PO_4)_4 (s) \rightleftharpoons 3 Zr^{4+} (aq) + 4 PO_4^{3-} (aq) \]

### Step 2: Define Solubility
Let the solubility of \( Zr_3(PO_4)_4 \) be \( S \) mol/L. This means that when \( Zr_3(PO_4)_4 \) dissolves, it produces:
- 3 moles of \( Zr^{4+} \) ions for every mole of \( Zr_3(PO_4)_4 \) that dissolves, which gives \( 3S \) for \( Zr^{4+} \).
- 4 moles of \( PO_4^{3-} \) ions for every mole of \( Zr_3(PO_4)_4 \) that dissolves, which gives \( 4S \) for \( PO_4^{3-} \).

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (3S)^3 (4S)^4 \]

### Step 4: Simplify the Expression
Now, we can simplify the expression:
\[ K_{sp} = (27S^3)(256S^4) \]
\[ K_{sp} = 6912 S^{7} \]

### Step 5: Relate \( K_{sp} \) to \( S \)
From the equation above, we can express \( S \) in terms of \( K_{sp} \):
\[ S^{7} = \frac{K_{sp}}{6912} \]
Taking the seventh root of both sides gives:
\[ S = \left( \frac{K_{sp}}{6912} \right)^{\frac{1}{7}} \]

### Conclusion
Thus, the correct relation between the solubility product \( K_{sp} \) and the solubility \( S \) of zirconium phosphate \( Zr_3(PO_4)_4 \) is:
\[ S = \left( \frac{K_{sp}}{6912} \right)^{\frac{1}{7}} \]