Solubility Product
The solubility of BaSO_4 in water at 25°C is 1.05 \times 10^{-5} mol L^{-1}. What is the value of its solubility product K_{sp}?
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About This Question
- Subject
- chemistry
- Chapter
- equilibrium
- Topic
- solubility product
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
1.10×10−10
For a sparingly soluble salt that dissociates as BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq), the solubility product K_{sp} = [Ba^{2+}][SO_4^{2-}]. If the molar solubility is s mol/L, then [Ba^{2+}] = s and [SO_4^{2-}] = s, giving K_{sp} = s^2. With s = 1.05 \times 10^{-5}: K_{sp} = (1.05 \times 10^{-5})^2 = 1.1025 \times 10^{-10} \approx 1.10 \times 10^{-10}. Option 2.10 \times 10^{-5} is wrong because it is 2s rather than s^2, confusing a simple multiplication with a squaring operation. Option 1.05 \times 10^{-5} is the solubility s itself, not K_{sp}. Option 4.41 \times 10^{-25} would come from cubing s (appropriate for salts dissociating into 3 ions), not squaring, which is the correct operation for a 1:1 electrolyte. This K_{sp} calculation is a fundamental NCERT ionic equilibrium problem and appears regularly in JEE Main. Plausibility check: the known K_{sp} of BaSO_4 is approximately 1.1 \times 10^{-10}, confirming our calculated value.
This easy difficulty chemistry question is from the chapter equilibrium, covering the topic of solubility product. It appeared in the 2025 exam.
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