Question

The solubility of sapringly soluble salt \(\text{AB}_2\) is \(18\cdot 78\times 10^{-4}\text{ g/dm}^3\) What is its solubility product ? (Molar mass of \(\text{AB}_2 = 187\cdot 8\text{ g mol}^{-1}\))

• A. \(2\times 10^{-15}\)
• B. \(4\times 10^{-15}\)
• C. \(6\times 10^{-15}\)
• D. \(8\times 10^{-15}\)

Hint:

Convert g per dm3 to mol per dm3 and use Ksp = 4S cubed.

Answer 1

The Correct Option is B

Step 1: Moles dissolved per litre:
Molar mass is 187.8 g/mol, and 18.78 $\times 10^{-4}$ g dissolves per dm$^3$, which is $\frac{18.78\times10^{-4}}{187.8} = 10^{-5}$ mol.

Step 2: Ion concentrations:
Each formula unit gives one $\text{A}^{2+}$ and two $\text{B}^-$ ions, so $[\text{A}^{2+}] = 10^{-5}$ and $[\text{B}^-] = 2\times 10^{-5}$.
\[ K_{sp} = 10^{-5}\times(2\times 10^{-5})^2 = 4\times 10^{-15} \]
This is option (B).

Final Answer:
$K_{sp} = 4\times 10^{-15}$. \[ \boxed{4\times 10^{-15}} \]