Step 1: Understanding the Concept:
Solubility (\( S \)) is defined as the maximum concentration of a solute that can dissolve in a given amount of solvent at equilibrium.
When a sparingly soluble salt dissolves, it dissociates into its constituent ions.
The Solubility Product Constant (\( K_{sp} \)) is the equilibrium constant for this dissociation process.
It is calculated by multiplying the concentrations of the ions, with each concentration raised to the power of its stoichiometric coefficient in the balanced equation.
Step 2: Key Formula or Approach:
For a salt with the general formula \( A_x B_y \), the dissociation equilibrium is:
\[ A_x B_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq) \]
If the solubility is \( S \), then:
\[ [A^{y+}] = xS \text{ and } [B^{x-}] = yS \]
The \( K_{sp} \) expression is:
\[ K_{sp} = [A^{y+}]^x \cdot [B^{x-}]^y = (xS)^x \cdot (yS)^y \]
Simplifying gives: \( K_{sp} = x^x \cdot y^y \cdot S^{(x+y)} \).
Step 3: Detailed Explanation:
Let's apply this to Magnesium phosphate, \( Mg_3(PO_4)_2 \).
1. **Dissociation Equation**:
\[ Mg_3(PO_4)_2(s) \rightleftharpoons 3Mg^{2+}(aq) + 2PO_4^{3-}(aq) \]
From this equation, we see that for every \( 1 \) mole of solid that dissolves, \( 3 \) moles of Magnesium ions and \( 2 \) moles of Phosphate ions are produced.
2. **Equilibrium Concentrations**:
If the molar solubility is \( S \):
- Concentration of Magnesium ions, \( [Mg^{2+}] = 3S \).
- Concentration of Phosphate ions, \( [PO_4^{3-}] = 2S \).
3. **Ksp Calculation**:
Substitute these into the solubility product expression:
\[ K_{sp} = [Mg^{2+}]^3 \cdot [PO_4^{3-}]^2 \]
\[ K_{sp} = (3S)^3 \cdot (2S)^2 \]
Calculate the individual terms:
- \( (3S)^3 = 3^3 \cdot S^3 = 27S^3 \)
- \( (2S)^2 = 2^2 \cdot S^2 = 4S^2 \)
4. **Final Multiplication**:
\[ K_{sp} = 27S^3 \times 4S^2 \]
\[ K_{sp} = (27 \times 4) \cdot S^{(3+2)} \]
\[ K_{sp} = 108S^5 \]
This specific mathematical relationship corresponds to a \( 3:2 \) type electrolyte. Comparing this result with the options, we find it matches option (D).
Step 4: Final Answer:
The solubility product of \( Mg_3(PO_4)_2 \) is \( 108S^5 \).