Question

The solubility of AgBr is $7.1 \times 10^{-7} \text{ moldm}^{-3}$. Calculate its solubility product at the same temperature.

• A. $7.08 \times 10^{-13}$
• B. $3.67 \times 10^{-13}$
• C. $5.89 \times 10^{-13}$
• D. $5.04 \times 10^{-13}$

Hint:

For $AB$ type salts (like $AgCl, AgBr, AgI$), $K_{sp} = S^2$. For $AB_2$ type salts (like $PbCl_2$), $K_{sp} = 4S^3$.

Answer 1

The Correct Option is D

Step 1: Concept
For a 1:1 salt like $AgBr$, the solubility product ($K_{sp}$) is related to solubility ($S$) by the formula $K_{sp} = S^2$.

Step 2: Meaning
$AgBr$ dissociates as $AgBr \rightleftharpoons Ag^+ + Br^-$. If solubility is $S$, then $[Ag^+] = S$ and $[Br^-] = S$.

Step 3: Analysis
- $S = 7.1 \times 10^{-7} \text{ moldm}^{-3}$. - $K_{sp} = S^2 = (7.1 \times 10^{-7})^2$. - $K_{sp} = 50.41 \times 10^{-14} = 5.041 \times 10^{-13}$.

Step 4: Conclusion
The solubility product is $5.04 \times 10^{-13}$. Final Answer: (D)