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  4. Solubility of CaF2 is 0.5 × 10-4 mol L-1. The value of Ksp for the salt is

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Q. Solubility of $CaF_2$ is $0.5 \times 10^{-4}\, mol\, L^{-1}$. The value of $K_{sp}$ for the salt is

Equilibrium

Solution:

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$K_{sp}=0.5\times10^{-4}\times1\times10^{-8}$
$=0.5\times10^{-12}$ or $5\times10^{-13}$