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: Understanding the Concept:
\(\text{BaSO}_4\) is a sparingly soluble salt. Its solubility product \(K_{sp}\) is the product of the ion concentrations at saturation.

Step 2: Write the dissociation:
\[ \text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]
If the solubility is \(s\) mol/dm\(^3\), then \([\text{Ba}^{2+}] = [\text{SO}_4^{2-}] = s\).

Step 3: Find the solubility in mol/dm3:
\[ K_{sp} = s \times s = s^2 = 1.0 \times 10^{-10} \]
\[ s = 1.0 \times 10^{-5} \text{ mol/dm}^3 \]

Step 4: Convert to g/dm3:
\[ s = 1.0 \times 10^{-5} \times 233 = 2.33 \times 10^{-3} \text{ g/dm}^3 \]

Step 5: Why the other options are wrong.
4.66e-3 comes from taking \(s = 2 \times 10^{-5}\), as if there were two ions of each. 1.16e-3 comes from halving the right value. 3.48 g/dm\(^3\) does not follow from any correct step.

Final Answer:
The solubility of BaSO4 is \(2.33 \times 10^{-3}\) g/dm3. \[ \boxed{\text{(A) }2.33 \times 10^{-3}\ \text{g/dm}^3} \]