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 \([Zr_3(PO_4)_4]\) and establish the relationship between its molar solubility \(S\) and its solubility product \(K_{sp}\).

### Step 1: Write the dissociation equation
The dissociation of zirconium phosphate can be expressed as follows:

\[ Zr_3(PO_4)_4 (s) \rightleftharpoons 3 Zr^{4+} (aq) + 4 PO_4^{3-} (aq) \]

### Step 2: Determine the stoichiometry
From the dissociation equation, we can see that:
- 3 moles of \(Zr^{4+}\) ions are produced for every mole of \(Zr_3(PO_4)_4\) that dissolves.
- 4 moles of \(PO_4^{3-}\) ions are produced for every mole of \(Zr_3(PO_4)_4\) that dissolves.

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

### Step 4: Write the expression for \(K_{sp}\)
The solubility product \(K_{sp}\) is defined as the product of the concentrations of the ions raised to the power of their coefficients in the balanced equation:

\[ K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4 \]

Substituting the expressions for the concentrations in terms of \(S\):

\[ K_{sp} = (3S)^3 (4S)^4 \]

### Step 5: Simplify the expression
Now we can simplify the expression for \(K_{sp}\):

\[ K_{sp} = (27S^3)(256S^4) = 6912 S^{7} \]

### Step 6: Solve for \(S\)
To find the relationship between \(S\) and \(K_{sp}\), we can rearrange the equation:

\[ S^7 = \frac{K_{sp}}{6912} \]

Taking the seventh root of both sides gives us:

\[ S = \left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}} \]

### Conclusion
Thus, the relationship between the molar solubility \(S\) and the solubility product \(K_{sp}\) is:

\[ S = \left(\frac{K_{sp}}{6912}\right)^{\frac{1}{7}} \]