Question

The solubility product of a sparingly soluble salt BA is \(6.4\times 10^{-13}\). Calculate it's solubility in \(\text{g dm}^{-3}\) .
Molar mass of salt is \(190 \text{g mol}^{-1}\) .

• A. \(1.52\times 10^{-4}\)
• B. \(1.25\times 10^{-4}\)
• C. \(2.1\times 10^{-4}\)
• D. \(1.9\times 10^{-4}\)

Hint:

For a 1:1 salt, S = sqrt(Ksp); multiply by molar mass for g per dm3.

Answer 1

The Correct Option is A

Step 1: Setup:
Let the solubility be $s$ mol per dm$^3$. Then $[\text{B}^+] = [\text{A}^-] = s$.

Step 2: Solve for s:
$s^2 = 6.4\times10^{-13} = 64\times10^{-14}$, so $s = 8\times10^{-7}$ M.

Step 3: Convert to mass:
Mass per dm$^3$ = $8\times10^{-7}\times190 = 1520\times10^{-7} = 1.52\times10^{-4}$ g. This is option (A).

Final Answer:
The salt dissolves to 1.52e-4 g per dm3. \[ \boxed{\text{(A) }1.52\times10^{-4}\ \text{g dm}^{-3}} \]