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 will follow these steps:

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

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

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [Zr^{4+}]^3 \times [PO_4^{3-}]^4 \]
Substituting the values from the dissociation:
\[ K_{sp} = (3s)^3 \times (4s)^4 \]

### Step 4: Calculate \( K_{sp} \)
Now we calculate each part:
- \( (3s)^3 = 27s^3 \)
- \( (4s)^4 = 256s^4 \)

Now, substituting these into the \( K_{sp} \) expression:
\[ K_{sp} = 27s^3 \times 256s^4 \]
\[ K_{sp} = 6912s^{7} \]

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