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 molar solubility (S) of zirconium phosphate \(Zr_3(PO_4)_4\) and its solubility product constant (\(K_{sp}\)), we will follow these steps:

### Step-by-Step Solution:

1. **Dissociation of the Compound**:
The first step is to write the dissociation equation for zirconium phosphate in water:
\[ Zr_3(PO_4)_4 (s) \rightleftharpoons 3 Zr^{4+} (aq) + 4 PO_4^{3-} (aq) \]

2. **Identify the Molar Solubility (S)**:
Let the molar solubility of \(Zr_3(PO_4)_4\) be \(S\). From the dissociation equation:
- For every 1 mole of \(Zr_3(PO_4)_4\) that dissolves, it produces 3 moles of \(Zr^{4+}\) ions and 4 moles of \(PO_4^{3-}\) ions.
- Therefore, the concentrations of the ions at equilibrium will be:
- \([Zr^{4+}] = 3S\)
- \([PO_4^{3-}] = 4S\)

3. **Write the Expression for \(K_{sp}\)**:
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 values from the previous step:
\[ K_{sp} = (3S)^3 \times (4S)^4 \]

4. **Calculate \(K_{sp}\)**:
Now, we will calculate \(K_{sp}\):
\[ K_{sp} = 27S^3 \times 256S^4 \]
\[ K_{sp} = 6912S^{7} \]

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

### Final Relation:
Thus, the relationship between the molar solubility \(S\) and the solubility product \(K_{sp}\) is given by:
\[ S = \left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}} \]