The solubility of CaF 2 K sp - 5-3- 10-9 in 0-1 M solution of NaF would be - Assume no reaction of cation-anion -

The correct option is C 5.3 × 10^ 7 M (C) 5.3× 10^ 7m CaF_2⇌ Ca^2++2F^ K_sp=[Ca^2+][F^ ]^2=S(S+0.1)^2=S× 0.1^2=5.3× 10^ 9 Note: S l ...

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Standard XII

Chemistry

Common Ion Effect

The solubilit...

Question

The solubility of CaF $_{2}$ (K $_{s p} = 5.3 \times 10^{- 9}$ ) in $0.1$ M solution of NaF would be : (Assume no reaction of cation/anion) .

5.3 \times 10^{- 10}

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5.3 \times 10^{- 8}

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5.3 \times 10^{- 7}

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5.3 \times 10^{- 11}

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Solution

The correct option is C $5.3 \times 10^{- 7}$ M
$(C) 5.3 \times 10^{- 7} m$

$C a F_{2} ⇌ C a^{2 +} + 2 F^{-}$

$K_{s p} = [C a 2 +] [F^{-}]^{2} = S (S + 0.1)^{2} = S \times {0.1}^{2} = 5.3 \times 10^{- 9}$

Note: $S << 0.1$ so, $S + 0.1 \approx 0.1$

$\Rightarrow S = 5.3 \times 10^{- 7} M$

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The solubility of CaF2 (Ksp=5.3×10−9) in 0.1 M solution of NaF would be : (Assume no reaction of cation/anion) .

The correct option is C 5.3×10−7 M(C) 5.3×10−7mCaF2⇌Ca2++2F−Ksp=[Ca2+][F−]2=S(S+0.1)2=S×0.12=5.3×10−9Note: S<<0.1 so, S+0.1≈0.1 ⇒S=5.3×10−7 M

The solubility of CaF $_{2}$ (K $_{s p} = 5.3 \times 10^{- 9}$ ) in $0.1$ M solution of NaF would be : (Assume no reaction of cation/anion) .

The correct option is C $5.3 \times 10^{- 7}$ M
$(C) 5.3 \times 10^{- 7} m$

$C a F_{2} ⇌ C a^{2 +} + 2 F^{-}$

$K_{s p} = [C a 2 +] [F^{-}]^{2} = S (S + 0.1)^{2} = S \times {0.1}^{2} = 5.3 \times 10^{- 9}$

Note: $S << 0.1$ so, $S + 0.1 \approx 0.1$

$\Rightarrow S = 5.3 \times 10^{- 7} M$