If solubility product of `Zr_(3)(PO_(4))_(4)` is denoted by `S,` then which of the following relation between `S` and `K_(sp)` is correct ?

Options

A. `S=((K_(sp))/(144))^(1//6)`
B. `S=((K_(sp))/(6912))^(1//7)`
C. `S=(K_(sp)/(929))^(1//9)`
D. `S=((K_(sp))/(216))^(1//7)`

Answer

To solve the problem regarding the relationship between the solubility product (denoted as \( S \)) of \( Zr_3(PO_4)_4 \) and the solubility product constant (\( K_{sp} \)), we will follow these steps:

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

### Step 2: Define solubility
Let the solubility of \( Zr_3(PO_4)_4 \) be \( S \). This means:
- The concentration of \( Zr^{4+} \) ions will be \( 3S \) (since 3 moles of \( Zr^{4+} \) are produced for every mole of \( Zr_3(PO_4)_4 \)).
- The concentration of \( PO_4^{3-} \) ions will be \( 4S \) (since 4 moles of \( PO_4^{3-} \) are produced for every mole of \( Zr_3(PO_4)_4 \)).

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) can be expressed in terms of the concentrations of the ions at equilibrium:
\[ K_{sp} = [Zr^{4+}]^3 \cdot [PO_4^{3-}]^4 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (3S)^3 \cdot (4S)^4 \]

### Step 4: Simplify the expression
Calculating the powers:
\[ K_{sp} = 27S^3 \cdot 256S^4 = 6912S^{7} \]

### Step 5: Relate \( S \) to \( K_{sp} \)
From the equation \( K_{sp} = 6912S^{7} \), we can express \( S \) in terms of \( K_{sp} \):
\[ S^{7} = \frac{K_{sp}}{6912} \]
Taking the seventh root of both sides:
\[ S = \left( \frac{K_{sp}}{6912} \right)^{\frac{1}{7}} \]

### Conclusion
Thus, the relationship between \( S \) and \( K_{sp} \) can be summarized as:
\[ S = \left( \frac{K_{sp}}{6912} \right)^{\frac{1}{7}} \]