Question

Solubility of $\text{Ca}_3(\text{PO}_4)_2$ is ' S ' $\text{moldm}^{-3}$. Find solubility product.

• A. $\text{S}^5$
• B. $108 \text{ S}^5$
• C. $54 \text{ S}^5$
• D. $12 \text{ S}^5$

Hint:

For any salt $A_xB_y$, use the shortcut formula $K_{sp} = x^x y^y S^{x+y}$.

Answer 1

The Correct Option is B

Step 1: Concept
The solubility product ($K_{sp}$) is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution.

Step 2: Meaning
For a salt $A_xB_y$, the dissociation is $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$. If solubility is $S$, then $K_{sp} = [A^{y+}]^x [B^{x-}]^y = (xS)^x (yS)^y = x^x y^y S^{(x+y)}$.

Step 3: Analysis
$\text{Ca}_3(\text{PO}_4)_2$ dissociates as: $\text{Ca}_3(\text{PO}_4)_2 \rightleftharpoons 3\text{Ca}^{2+} + 2\text{PO}_4^{3-}$. Here, $x = 3$ and $y = 2$. Substituting the values: $K_{sp} = 3^3 \cdot 2^2 \cdot S^{(3+2)}$.

Step 4: Conclusion
Calculation: $K_{sp} = 27 \cdot 4 \cdot S^5 = 108 S^5$. Final Answer: (B)