The molar solubility of CaF_2 ( K_sp = 5.3 × 10^-11 ) in 0.1 M solution of NaF will be: - CHEMISTRY MCQ NEET 2019 2026 | ExamTest.live

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The molar solubility of CaF_{2} (K_{s p} = 5.3 \times 1 0^{- 11}) in 0.1 M solution of NaF will be:

Solution & Step-by-step Explanation

In the presence of NaF (0.1 M), the concentration of fluoride ion [F^{-}] is dominated by NaF due to the common ion effect. [F^{-}] \approx 0.1 M .For CaF_{2} ⇌ Ca^{2 +} + 2 F^{-} : K_{s p} = [Ca^{2 +}] [F^{-}]^{2} 5.3 \times 1 0^{- 11} = (s) \times (0.1)^{2} s = \frac{5.3 \times 1 0 ^{- 11}}{1 0 ^{- 2}} = 5.3 \times 1 0^{- 9} mol L^{- 1} .

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The molar solubility of CaF_{2} (K_{s p} = 5.3 \times 1 0^{- 11}) in 0.1 M solution of NaF will be:

5.3 \times 1 0^{- 11} mol L^{- 1}

5.3 \times 1 0^{- 8} mol L^{- 1}

5.3 \times 1 0^{- 9} mol L^{- 1}

5.3 \times 1 0^{- 10} mol L^{- 1}

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