Zirconium phosphate `[Zr_3(PO_4)_4]` dissociates into three zirconium cations of charge +4 and four phosphate anions of charge -3 . If molar solubility of zirconium phosphate is denoted by S and its solubility product by `K_(sp)` then which of the following relationship between `S and K_(sp)` is correct ?

Options

A. `S={K_(sp)//6912}^7`
B. `S={K_(sp)//144}^(1//7)`
C. `S={K_(sp)//(6912)^(1//7)}`
D. `S=(K_(sp)//6912)^(1//7)`

Answer

To solve the problem, we need to analyze the dissociation of zirconium phosphate and establish the relationship between its molar solubility (S) and its solubility product (Ksp).

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
Zirconium phosphate, \( Zr_3(PO_4)_4 \), dissociates in water as follows:
\[ Zr_3(PO_4)_4 (s) \rightleftharpoons 3 Zr^{4+} (aq) + 4 PO_4^{3-} (aq) \]

2. **Define Molar Solubility (S)**:
Let the molar solubility of zirconium phosphate be denoted by \( S \). This means that when \( Zr_3(PO_4)_4 \) dissolves, it produces:
- 3 moles of \( Zr^{4+} \) for every mole of \( Zr_3(PO_4)_4 \), leading to a concentration of \( 3S \) for \( Zr^{4+} \).
- 4 moles of \( PO_4^{3-} \) for every mole of \( Zr_3(PO_4)_4 \), leading to a concentration of \( 4S \) for \( PO_4^{3-} \).

3. **Write the Expression for Ksp**:
The solubility product \( K_{sp} \) is given by the product of the concentrations of the ions raised to the power of their coefficients in the balanced equation:
\[ K_{sp} = [Zr^{4+}]^3 \times [PO_4^{3-}]^4 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (3S)^3 \times (4S)^4 \]

4. **Calculate Ksp**:
Expanding the expression:
\[ K_{sp} = 27S^3 \times 256S^4 \]
\[ K_{sp} = 6912 S^{7} \]

5. **Express S in Terms of Ksp**:
To find \( S \) in terms of \( K_{sp} \), we rearrange the equation:
\[ S^{7} = \frac{K_{sp}}{6912} \]
Taking the seventh root of both sides:
\[ S = \left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}} \]

### Conclusion:
The relationship between the molar solubility \( S \) and the solubility product \( K_{sp} \) for zirconium phosphate is:
\[ S = \left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}} \]