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Ionic Equilibrum question

2018 · Q106
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Ionic Equilibrum question

2018 · Q106

NEETChemistryIonic EquilibrumMCQ+4 / −1
The solubility of BaSO4 in water is 2.42 × 10–3 g L–1 at 298 K. The value of its solubility product (Ksp) will be (Given molar mass of BaSO4 = 233 g mol–1)
  1. A
    1.08 × 10–10 mol2 L–2
  2. B
    1.08 × 10–12 mol2 L–2
  3. C
    1.08 × 10–14 mol2 L–2
  4. D
    1.08 × 10–8 mol2 L–2
View written solutionFree

Correct answer: A

Given, Solubility of BaSO4 = 2.42 × 10–3 g L–1

Convert solubility in mol/lit.

s = 2.42×10−3233=1.04×10−5{{2.42 \times {{10}^{ - 3}}} \over {233}} = 1.04 \times {10^{ - 5}}2332.42×10−3​=1.04×10−5 mol L-1

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BaSO4(s)⇌Ba2+(aq)+SO42-(aq)
ss

Ksp = [Ba2+][SO42-] = s2

= $${\left( {1.04 \times {{10}^{ - 5}}} \right)^2}$$

= 1.08 $ \times $ 10-10 mol2 L–2
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