Question

What is the solubility of \(\text{BaSO}_4\) in g /\(\text{dm}^3\) if its solubility product is \(1.0\times 10^{-10}\) at \(25^{\circ}\)C. [Molar mass of \(\text{BaSO}_4 = 233\) g/mol]

• A. \(2.33\times 10^{-3}\) g/\(\text{dm}^3\)
• B. \(4.66\times 10^{-3}\) g/\(\text{dm}^3\)
• C. \(3.48\) g/\(\text{dm}^3\)
• D. \(1.16\times 10^{-3}\) g/\(\text{dm}^3\)

Hint:

For a 1:1 salt, Ksp = s squared. Then multiply s (mol/dm3) by the molar mass.

Answer 1

The Correct Option is A

Step 1: Square-root route:
For an $\text{AB}$ type salt, solubility is $\sqrt{K_{sp}}$. So $s = \sqrt{10^{-10}} = 10^{-5}$ mol per dm$^3$.

Step 2: Change the unit:
Mass per dm$^3$ = moles per dm$^3$ times molar mass = $10^{-5} \times 233$ g.

Step 3: Simplify:
$233 \times 10^{-5} = 2.33 \times 10^{-3}$ g/dm$^3$. This matches option A.

Final Answer:
The molar solubility is 1e-5 mol/dm3, which is 2.33e-3 g/dm3. \[ \boxed{\text{(A) }2.33 \times 10^{-3}\ \text{g/dm}^3} \]