Question

The solubility of \(Mg_3(PO_4)_2\) is \(S\) mol L\(^{-1}\). The solubility product is given by the relation:

• A. $S^5$
• B. $36S^6$
• C. $6S^5$
• D. $108S^5$

Hint:

For a general salt of type $A_x B_y$, the shortcut formula for the solubility product in terms of solubility $S$ is: \[ K_{sp} = x^x y^y S^{(x+y)} \] For $Mg_3(PO_4)_2$, $x=3$ and $y=2$. Thus, $3^3 \cdot 2^2 \cdot S^{(3+2)} = 27 \cdot 4 \cdot S^5 = 108S^5$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
\(K_{sp}\) is derived from the ion product of the dissolved salt components at saturation.
Step 2: Key Formula or Approach:
For \(A_{x}B_{y} \rightleftharpoons xA + yB\), \(K_{sp} = x^{x} y^{y} S^{x+y}\).
Step 3: Detailed Explanation:
\(Mg_{3}(PO_{4})_{2} \to 3Mg^{2+} + 2PO_{4}^{3-}\).
\([Mg^{2+}] = 3S\), \([PO_{4}^{3-}] = 2S\).
\(K_{sp} = (3S)^{3} (2S)^{2} = 27S^{3} \times 4S^{2} = 108S^{5}\).
Step 4: Final Answer:
The relation is \(108S^{5}\).