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NEET CHEMISTRYMedium

The solubility of BaSO4BaSO_4 in water is 2.42×10−3 g L−12.42 \times 10^{-3} \text{ g L}^{-1} at 298 K298 \text{ K}. The value of the solubility product will be: (Molar mass of BaSO4=233 g mol−1BaSO_4 = 233 \text{ g mol}^{-1})

A

1.08×10−10 mol2 L−21.08 \times 10^{-10} \text{ mol}^2 \text{ L}^{-2}

B

1.08×10−12 mol2 L−21.08 \times 10^{-12} \text{ mol}^2 \text{ L}^{-2}

C

1.08×10−14 mol2 L−21.08 \times 10^{-14} \text{ mol}^2 \text{ L}^{-2}

D

1.08×10−8 mol2 L−21.08 \times 10^{-8} \text{ mol}^2 \text{ L}^{-2}

Step-by-Step Solution

Solubility of BaSO4=2.42×10−3 g L−1BaSO_4 = 2.42 \times 10^{-3} \text{ g L}^{-1} Molar mass of BaSO4=233 g mol−1BaSO_4 = 233 \text{ g mol}^{-1} Molar solubility (SS) =2.42×10−3 g L−1233 g mol−1=1.038×10−5 mol L−1= \frac{2.42 \times 10^{-3} \text{ g L}^{-1}}{233 \text{ g mol}^{-1}} = 1.038 \times 10^{-5} \text{ mol L}^{-1} The dissociation of BaSO4BaSO_4 is given by: BaSO4(s)⇌Ba2+(aq)+SO42−(aq)BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq) Here, [Ba2+]=S[Ba^{2+}] = S and [SO42−]=S[SO_4^{2-}] = S The solubility product expression is: Ksp=[Ba2+][SO42−]=S×S=S2K_{sp} = [Ba^{2+}][SO_4^{2-}] = S \times S = S^2 Ksp=(1.038×10−5)2=1.077×10−10≈1.08×10−10 mol2 L−2K_{sp} = (1.038 \times 10^{-5})^2 = 1.077 \times 10^{-10} \approx 1.08 \times 10^{-10} \text{ mol}^2 \text{ L}^{-2}

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