If solubility product of `Zr_(3)(PO_(4))_(4)` is denotes by `K_(sp)` and its molar solubility is denoted by S, then which of the following relation between S and `K_(sp)`­ is correct ?

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 find the relationship between the solubility product \( K_{sp} \) of \( Zr_3(PO_4)_4 \) and its molar solubility \( S \), we can follow these steps:

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

### Step 2: Define Molar Solubility
Let the molar solubility of \( Zr_3(PO_4)_4 \) be \( S \). This means that when \( Zr_3(PO_4)_4 \) dissolves, it produces:
- \( 3S \) moles of \( Zr^{3+} \)
- \( 4S \) moles of \( PO_4^{3-} \)

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

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

### Step 5: Solve for \( S \)
Rearranging the equation to solve for \( S \):
\[ S^7 = \frac{K_{sp}}{6912} \]
Taking the seventh root of both sides:
\[ S = \left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}} \]

### Conclusion
Thus, the relationship between \( S \) and \( K_{sp} \) is:
\[ S = \left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}} \]