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NEET Hard

The solubility of BaSO4BaSO_4 in water is 2.42×103 gL12.42 \times 10^{-3} \text{ gL}^{-1} at 298 K. The value of its solubility product (KspK_{sp}) will be (Given molar mass of BaSO4=233 g mol1BaSO_4 = 233 \text{ g mol}^{-1})

A

1.08×1014 mol2L21.08 \times 10^{-14} \text{ mol}^2 \text{L}^{-2}

B

1.08×1012 mol2L21.08 \times 10^{-12} \text{ mol}^2 \text{L}^{-2}

C

1.08×1010 mol2L21.08 \times 10^{-10} \text{ mol}^2 \text{L}^{-2}

D

1.08×108 mol2L21.08 \times 10^{-8} \text{ mol}^2 \text{L}^{-2}

Step-by-Step Solution

Solubility of BaSO4BaSO_4, s=2.42×103233 (mol L1)=1.04×105 (mol L1)s = \frac{2.42 \times 10^{-3}}{233} \text{ (mol L}^{-1}) = 1.04 \times 10^{-5} \text{ (mol L}^{-1}). BaSO4(s)Ba2+(aq)+SO42(aq)BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq). Ksp=[Ba2+][SO42]=s2=(1.04×105)2=1.08×1010 mol2L2K_{sp} = [Ba^{2+}][SO_4^{2-}] = s^2 = (1.04 \times 10^{-5})^2 = 1.08 \times 10^{-10} \text{ mol}^2 \text{L}^{-2}.

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