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NEET 2018
Exam
NEET
Year
2018
Subject
Chemistry
Chapter
Ionic Equilibrium

NEET 2018 · Chemistry · Ionic Equilibrium

The solubility of BaSO _ 4 in water is 2 .42 × 10 ^ - 3 gL ^ - 1 at...

The solubility of BaSO4{BaSO}_{4} in water is 2.42×10−3gL−12 .42 \times {10}^{- 3}{gL}^{- 1} at 298K.298K . The value of its solubility product (Ksp)\left({K}_{sp}\right) will be

(Given- the molar mass of BaSO4=233gmol−1{BaSO}_{4} = 233g{mol}^{- 1})

Options

  1. A1.08×10−14mol2L−21 .08 \times {10}^{- 14}{mol}^{2}{L}^{- 2}
  2. B1.08×10−12mol2L−21 .08 \times {10}^{- 12}{mol}^{2}{L}^{- 2}
  3. C1.08×10−10mol2L−21 .08 \times {10}^{- 10}{mol}^{2}{L}^{- 2}
  4. D1.08×10−8mol2L−21 .08 \times {10}^{- 8}{mol}^{2}{L}^{- 2}

Correct answer

C. 1.08×10−10mol2L−21 .08 \times {10}^{- 10}{mol}^{2}{L}^{- 2}

Step-by-step solution

Solubility of BaSO4=2.42×10−3gL−1{BaSO}_{4} = 2 .42 \times {10}^{- 3}{gL}^{- 1}

∴s=2.42×10−3233=1.038×10−5molL−1∴ s = \dfrac{2 .42 \times {10}^{- 3}}{233} = 1 .038 \times {10}^{- 5}mol{L}^{- 1}

Ksp=s2=(1.038×10−5)2{K}_{sp} = {s}^{2} = {\left(1 .038 \times {10}^{- 5}\right)}^{2}

=1.08×10−10mol2L−2= 1 .08 \times {10}^{- 10}{mol}^{2}{L}^{- 2}
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