The solubility of an insoluble phosphate, `M_(3)(PO_(4))_(2)` of molecular weight w in water is x grams per litre, its solubility product is proportional to

Options

A. `(x//w)^(2//3)`
B. `(x//W)`
C. `(x//W)^(3//2)`
D. `(X//w)^(5)`

Answer

To solve the problem regarding the solubility product of the insoluble phosphate \( M_3(PO_4)_2 \), we will follow these steps:

### Step 1: Write the dissociation equation
When \( M_3(PO_4)_2 \) dissolves in water, it dissociates into its ions:
\[ M_3(PO_4)_2 \rightleftharpoons 3M^{2+} + 2PO_4^{3-} \]

### Step 2: Define solubility
Let the solubility of \( M_3(PO_4)_2 \) be \( s \) moles per liter. Thus, if \( s \) is the solubility:
- The concentration of \( M^{2+} \) ions will be \( 3s \) (since there are 3 moles of \( M^{2+} \) for every mole of \( M_3(PO_4)_2 \) that dissolves).
- The concentration of \( PO_4^{3-} \) ions will be \( 2s \) (since there are 2 moles of \( PO_4^{3-} \) for every mole of \( M_3(PO_4)_2 \) that dissolves).

### Step 3: Write the expression for the solubility product (Ksp)
The solubility product \( K_{sp} \) is given by the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation:
\[ K_{sp} = [M^{2+}]^3 [PO_4^{3-}]^2 \]
Substituting the concentrations:
\[ K_{sp} = (3s)^3 (2s)^2 \]

### Step 4: Simplify the expression
Calculating the powers:
\[ K_{sp} = 27s^3 \cdot 4s^2 = 108s^5 \]

### Step 5: Relate solubility to grams per liter
Given that the solubility in grams per liter is \( x \) and the molecular weight is \( W \), we can relate the solubility in moles to the solubility in grams:
\[ s = \frac{x}{W} \]

### Step 6: Substitute for \( s \) in the Ksp expression
Now substitute \( s \) in the \( K_{sp} \) expression:
\[ K_{sp} = 108 \left( \frac{x}{W} \right)^5 \]

### Final Expression
Thus, the solubility product \( K_{sp} \) is proportional to:
\[ K_{sp} \propto \frac{108x^5}{W^5} \]

### Conclusion
The final answer is:
\[ K_{sp} \propto \frac{x^5}{W^5} \]