Solubility of Zirconium phosphate `Zr_3(PO_4)_4` is 's' moles per litre. Solubility product of `K_(sp)` may be given as

Options

A. `6912s^7`
B. `108s^7`
C. `27s^4`
D. `6812s^7`

Answer

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

### Step 1: Write the dissociation equation
Zirconium phosphate dissociates in water according to the following equation:
\[ Zr_3(PO_4)_4 (s) \rightleftharpoons 3 Zr^{4+} (aq) + 4 PO_4^{3-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( Zr_3(PO_4)_4 \) be \( s \) moles per liter. This means that in a saturated solution, the concentration of \( Zr_3(PO_4)_4 \) is \( s \).

### Step 3: Determine the concentrations of ions
From the dissociation equation:
- For every 1 mole of \( Zr_3(PO_4)_4 \) that dissolves, 3 moles of \( Zr^{4+} \) ions are produced.
- For every 1 mole of \( Zr_3(PO_4)_4 \) that dissolves, 4 moles of \( PO_4^{3-} \) ions are produced.

Thus, the concentrations of the ions in terms of \( s \) are:
- Concentration of \( Zr^{4+} \) ions = \( 3s \)
- Concentration of \( PO_4^{3-} \) ions = \( 4s \)

### Step 4: Write the expression for the solubility product \( K_{sp} \)
The solubility product \( K_{sp} \) is given by the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation:
\[ K_{sp} = [Zr^{4+}]^3 \times [PO_4^{3-}]^4 \]

Substituting the concentrations we found:
\[ K_{sp} = (3s)^3 \times (4s)^4 \]

### Step 5: Simplify the expression
Calculating the powers:
\[ K_{sp} = (27s^3) \times (256s^4) \]
\[ K_{sp} = 6912 s^{7} \]

### Conclusion
Thus, the solubility product \( K_{sp} \) of zirconium phosphate \( Zr_3(PO_4)_4 \) is:
\[ K_{sp} = 6912 s^{7} \]