`K_(sp)` of `Zr_(3)(PO_(4))_(4)` in terms of solubility (S) is :

Options

A. `108 S^(7)`
B. `4 S^(3)`
C. `6912 S^(7)`
D. None of these

Answer

To find the solubility product constant \( K_{sp} \) of zirconium phosphate \( Zr_3(PO_4)_4 \) in terms of its solubility \( S \), we can follow these steps:

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

### Step 2: Define solubility
Let \( S \) be the solubility of \( Zr_3(PO_4)_4 \) in moles per liter. When \( Zr_3(PO_4)_4 \) dissolves, it produces:
- 3 moles of \( Zr^{4+} \) ions for every mole of \( Zr_3(PO_4)_4 \)
- 4 moles of \( PO_4^{3-} \) ions for every mole of \( Zr_3(PO_4)_4 \)

### Step 3: Express concentrations in terms of \( S \)
At equilibrium:
- The concentration of \( Zr^{4+} \) ions will be \( 3S \)
- The concentration of \( PO_4^{3-} \) ions will be \( 4S \)

### Step 4: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) is given by the formula:
\[ K_{sp} = [Zr^{4+}]^3 \cdot [PO_4^{3-}]^4 \]

Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (3S)^3 \cdot (4S)^4 \]

### Step 5: Calculate \( K_{sp} \)
Now, we can calculate \( K_{sp} \):
\[ K_{sp} = (3^3 \cdot S^3) \cdot (4^4 \cdot S^4) \]
Calculating the powers:
\[ 3^3 = 27 \quad \text{and} \quad 4^4 = 256 \]
Thus,
\[ K_{sp} = 27S^3 \cdot 256S^4 \]
Combine the terms:
\[ K_{sp} = 27 \cdot 256 \cdot S^{3+4} = 6912S^7 \]

### Final Result
The solubility product constant \( K_{sp} \) of \( Zr_3(PO_4)_4 \) in terms of solubility \( S \) is:
\[ K_{sp} = 6912S^7 \]