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NEET 2019
Exam
NEET
Year
2019
Subject
Chemistry
Chapter
Ionic Equilibrium

NEET 2019 · Chemistry · Ionic Equilibrium

The molar solubility of CaF_2 (K_ sp =5.3 × 10^ -11 ) in 0.1 M...

The molar solubility of CaF2\mathrm{CaF}_2 (Ksp=5.3×10−11)\left(\mathrm{K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right) in 0.1M0.1 \mathrm{M} solution of NaF\mathrm{NaF} will beThe molar solubility of CaF_2 (K_ sp =5.3 × 10^ -11 ) in 0.1 M solution of NaF will be

Options

  1. A5.3×1011 mol L−15.3 \times 10^{11} \mathrm{~mol} \mathrm{~L}^{-1}
  2. B5.3×10−8 mol L−15.3 \times 10^{-8} \mathrm{~mol} \mathrm{~L}^{-1}
  3. C5.3×10−9 mol L−15.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}
  4. D5.3×10−10 mol L−15.3 \times 10^{-10} \mathrm{~mol} \mathrm{~L}^{-1}

Correct answer

C. 5.3×10−9 mol L−15.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}

Step-by-step solution

Let the solubility of CaF2\mathrm{CaF}_2 in 0.1MNaF0.1 \mathrm{M} \mathrm{NaF} is ' S\mathrm{S} ' molL−1\mathrm{mol} \mathrm{L}^{-1} The molar solubility of CaF_2 (K_ sp =5.3 × 10^ -11 ) in 0.1 M solution of NaF will be
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