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 determine the solubility product (Ksp) of the insoluble phosphate \( M_3(PO_4)_2 \) based on its solubility in water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Dissociation Reaction**:
The compound \( M_3(PO_4)_2 \) dissociates in water as follows:
\[ M_3(PO_4)_2 (s) \rightleftharpoons 3M^{2+} (aq) + 2PO_4^{3-} (aq) \]

2. **Define Solubility**:
Let the solubility of \( M_3(PO_4)_2 \) be \( s \) moles per liter. Therefore, when it dissolves:
- The concentration of \( M^{2+} \) ions will be \( 3s \) (since 3 moles of \( M^{2+} \) are produced for every mole of \( M_3(PO_4)_2 \)).
- The concentration of \( PO_4^{3-} \) ions will be \( 2s \) (since 2 moles of \( PO_4^{3-} \) are produced for every mole of \( M_3(PO_4)_2 \)).

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:
\[ K_{sp} = [M^{2+}]^3 \cdot [PO_4^{3-}]^2 \]
Substituting the expressions for the concentrations:
\[ K_{sp} = (3s)^3 \cdot (2s)^2 \]
Simplifying this gives:
\[ K_{sp} = 27s^3 \cdot 4s^2 = 108s^5 \]

4. **Relate Solubility to Given Mass and Molecular Weight**:
The solubility \( s \) can also be expressed in terms of the given mass \( x \) and the molecular weight \( w \):
\[ s = \frac{x}{w} \]
where \( x \) is the mass of the solute in grams and \( w \) is the molecular weight in grams per mole.

5. **Substitute \( s \) into the Ksp Expression**:
Now substituting \( s \) into the Ksp expression:
\[ K_{sp} = 108 \left(\frac{x}{w}\right)^5 \]
Thus, we can express \( K_{sp} \) as:
\[ K_{sp} = \frac{108x^5}{w^5} \]

6. **Determine Proportionality**:
From the final expression, we can see that:
\[ K_{sp} \propto \frac{x^5}{w^5} \]
Therefore, the solubility product \( K_{sp} \) is proportional to \( \frac{x}{w}^5 \).

### Conclusion:
The solubility product \( K_{sp} \) of the insoluble phosphate \( M_3(PO_4)_2 \) is proportional to \( \frac{x^5}{w^5} \).