The solubility of BaSO_4 in water is 2.42 × 10^-3 g L^-1 at 298 K . The value of its solubility product ( K_sp ) will be (Given molar mass = 233 g mol - Chemistry MCQ NEET (UG) 2018 2026 | ExamTest.live

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The solubility of BaSO_{4} in water is 2.42 \times 1 0^{- 3} g L^{- 1} at 298 K . The value of its solubility product (K_{s p}) will be (Given molar mass = 233 g mol^{- 1}):

Solution & Step-by-step Explanation

1. Convert solubility from g L^{- 1} to mol L^{- 1} (s): s = \frac{2.42 \times 1 0 ^{- 3}}{233} \approx 1.038 \times 1 0^{- 5} mol L^{- 1} .2. Dissociation: BaSO_{4} (s) ⇌ Ba^{2 +} (a q) + SO_{4}^{2 -} (a q) .3. K_{s p} = [Ba^{2 +}] [SO_{4}^{2 -}] = s^{2} .4. K_{s p} = (1.038 \times 1 0^{- 5})^{2} \approx 1.08 \times 1 0^{- 10} mol^{2} L^{- 2} .

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The solubility of BaSO_{4} in water is 2.42 \times 1 0^{- 3} g L^{- 1} at 298 K . The value of its solubility product (K_{s p}) will be (Given molar mass = 233 g mol^{- 1}):

1.08 \times 1 0^{- 14} mol^{2} L^{- 2}

1.08 \times 1 0^{- 12} mol^{2} L^{- 2}

1.08 \times 1 0^{- 10} mol^{2} L^{- 2}

1.08 \times 1 0^{- 8} mol^{2} L^{- 2}

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