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

Concept: The solubility product constant ($K_{sp}$) is the equilibrium constant for a solid substance dissolving in an aqueous solution. For a salt that dissociates into multiple ions, the $K_{sp}$ is calculated by raising the concentration of each ion to the power of its stoichiometric coefficient.

Step 1: Write the dissociation equation.
Magnesium phosphate ($Mg_3(PO_4)_2$) dissociates in water as follows: \[ Mg_3(PO_4)_2(s) \xrightleftharpoons{} 3Mg^{2+}(aq) + 2PO_4^{3-}(aq) \]

Step 2: Express ion concentrations in terms of solubility ($S$).
If the solubility of the salt is $S$ mol/L, then according to the stoichiometry :
• $[Mg^{2+}] = 3S$
• $[PO_4^{3-}] = 2S$

Step 3: Calculate $K_{sp}$.
The expression for the solubility product is: \[ K_{sp} = [Mg^{2+}]^3 [PO_4^{3-}]^2 \] Substituting the concentration values: \[ K_{sp} = (3S)^3 (2S)^2 \] \[ K_{sp} = (27S^3) (4S^2) \] \[ K_{sp} = 108S^5 \]