Question:
The solubility of \(Mg_3(PO_4)_2\) is \(S\) mol L\(^{-1}\). The solubility product is given by the relation
The solubility of \(Mg_3(PO_4)_2\) is \(S\) mol L\(^{-1}\). The solubility product is given by the relation
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Keep these common electrolyte relations in mind to quickly solve solubility product questions:
- $1:1$ electrolyte (e.g., $\text{AgCl}$): $K_{sp} = S^2$
- $1:2$ or $2:1$ electrolyte (e.g., $\text{CaF}_2$): $K_{sp} = 4S^3$
- $1:3$ or $3:1$ electrolyte (e.g., $\text{AlCl}_3$): $K_{sp} = 27S^4$
- $3:2$ or $2:3$ electrolyte (e.g., $\text{Mg}_3(\text{PO}_4)_2$): $K_{sp} = 108S^5$
- $1:1$ electrolyte (e.g., $\text{AgCl}$): $K_{sp} = S^2$
- $1:2$ or $2:1$ electrolyte (e.g., $\text{CaF}_2$): $K_{sp} = 4S^3$
- $1:3$ or $3:1$ electrolyte (e.g., $\text{AlCl}_3$): $K_{sp} = 27S^4$
- $3:2$ or $2:3$ electrolyte (e.g., $\text{Mg}_3(\text{PO}_4)_2$): $K_{sp} = 108S^5$
Updated On: Oct 6, 2026
- $S^5$
- $36S^6$
- $6S^5$
- $108S^5$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
The question asks to find the mathematical relationship between the solubility product constant ($K_{sp}$) and the molar solubility ($S$) for the sparingly soluble salt magnesium phosphate, $\text{Mg}_3(\text{PO}_4)_2$.
Step 2: Key Formula or Approach:
For a sparingly soluble salt of type $A_x B_y$ that dissociates into ions:
\[ A_x B_y(s) \rightleftharpoons x A^{y+}(aq) + y B^{x-}(aq) \] The solubility product ($K_{sp}$) in terms of molar solubility ($S$) is given by:
\[ K_{sp} = x^x \cdot y^y \cdot S^{(x+y)} \]
Step 3: Detailed Explanation:
• Let us write the balanced chemical equation for the dissociation of magnesium phosphate in its saturated aqueous solution:
\[ \text{Mg}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Mg}^{2+}(aq) + 2\text{PO}_4^{3-}(aq) \]
• If $S$ is the molar solubility of $\text{Mg}_3(\text{PO}_4)_2$ in $\text{mol L}^{-1}$, then according to stoichiometry, the equilibrium concentrations of the dissociated ions will be:
- Concentration of magnesium ions:
\[ [\text{Mg}^{2+}] = 3S \] - Concentration of phosphate ions:
\[ [\text{PO}_4^{3-}] = 2S \]
• The expression for the solubility product constant ($K_{sp}$) is:
\[ K_{sp} = [\text{Mg}^{2+}]^3 \cdot [\text{PO}_4^{3-}]^2 \]
• Substituting the concentration terms in terms of $S$ into the $K_{sp}$ expression:
\[ K_{sp} = (3S)^3 \cdot (2S)^2 \]
• Simplifying the mathematical terms:
\[ K_{sp} = (27S^3) \cdot (4S^2) \] \[ K_{sp} = (27 \cdot 4) \cdot S^{3+2} \] \[ K_{sp} = 108S^5 \]
• This gives the final relationship between the solubility product and molar solubility for this specific salt stoichiometry ($3:2$ electrolyte).
Step 4: Final Answer:
The solubility product of $\text{Mg}_3(\text{PO}_4)_2$ is given by $108S^5$.
The question asks to find the mathematical relationship between the solubility product constant ($K_{sp}$) and the molar solubility ($S$) for the sparingly soluble salt magnesium phosphate, $\text{Mg}_3(\text{PO}_4)_2$.
Step 2: Key Formula or Approach:
For a sparingly soluble salt of type $A_x B_y$ that dissociates into ions:
\[ A_x B_y(s) \rightleftharpoons x A^{y+}(aq) + y B^{x-}(aq) \] The solubility product ($K_{sp}$) in terms of molar solubility ($S$) is given by:
\[ K_{sp} = x^x \cdot y^y \cdot S^{(x+y)} \]
Step 3: Detailed Explanation:
• Let us write the balanced chemical equation for the dissociation of magnesium phosphate in its saturated aqueous solution:
\[ \text{Mg}_3(\text{PO}_4)_2(s) \rightleftharpoons 3\text{Mg}^{2+}(aq) + 2\text{PO}_4^{3-}(aq) \]
• If $S$ is the molar solubility of $\text{Mg}_3(\text{PO}_4)_2$ in $\text{mol L}^{-1}$, then according to stoichiometry, the equilibrium concentrations of the dissociated ions will be:
- Concentration of magnesium ions:
\[ [\text{Mg}^{2+}] = 3S \] - Concentration of phosphate ions:
\[ [\text{PO}_4^{3-}] = 2S \]
• The expression for the solubility product constant ($K_{sp}$) is:
\[ K_{sp} = [\text{Mg}^{2+}]^3 \cdot [\text{PO}_4^{3-}]^2 \]
• Substituting the concentration terms in terms of $S$ into the $K_{sp}$ expression:
\[ K_{sp} = (3S)^3 \cdot (2S)^2 \]
• Simplifying the mathematical terms:
\[ K_{sp} = (27S^3) \cdot (4S^2) \] \[ K_{sp} = (27 \cdot 4) \cdot S^{3+2} \] \[ K_{sp} = 108S^5 \]
• This gives the final relationship between the solubility product and molar solubility for this specific salt stoichiometry ($3:2$ electrolyte).
Step 4: Final Answer:
The solubility product of $\text{Mg}_3(\text{PO}_4)_2$ is given by $108S^5$.
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