The conductivity of a saturated solution of CaF2 after deducting the contribution from water is 4.0×10−5Scm−1 If molar conductivity of CaF2 at infinite dilution is 200 S cm2mol−1 then the solubility product of CaF2, is

The solubility of CaF2 is c=κΛ=4.0×10−5Scm−1200Scm2mol−1=2.0×10−7molcm−3 =2.0×10−7mol10−1dm−3=2.0×10−4moldm−3From CaF2(s)⇌Ca2+c(aq)+2F−2c(aq) we find Ksp=Ca2+F−2=4c3=42.0×10−4M3=3.2×10−11M3

The conductivity of a saturated solution of CaF2 after deducting the contribution from water is 4.0×10−5Scm−1 If molar conductivity of CaF2 at infinite dilution is 200 S cm2mol−1 then the solubility product of CaF2, is

The solubility of CaF2 is c=κΛ=4.0×10−5Scm−1200Scm2mol−1=2.0×10−7molcm−3 =2.0×10−7mol10−1dm−3=2.0×10−4moldm−3From CaF2(s)⇌Ca2+c(aq)+2F−2c(aq) we find Ksp=Ca2+F−2=4c3=42.0×10−4M3=3.2×10−11M3

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The conductivity of a saturated solution of ${CaF}_{2}$ after deducting the contribution from water is $4.0 \times 10^{- 5} S {cm}^{- 1}$ If molar conductivity of ${CaF}_{2}$ at infinite dilution is $200 S {cm}^{2} {mol}^{- 1}$ then the solubility product of ${CaF}_{2}$ , is

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$3.2 \times 10^{- 20} M^{3}$

$8.0 \times 10^{- 12} M^{3}$

$3.2 \times 10^{- 11} M^{3}$

$1.6 \times 10^{- 8} M^{3}$

answer is C.

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Detailed Solution

The solubility of ${CaF}_{2}$ is $c = \frac{κ}{Λ} = \frac{4.0 \times 10^{- 5} S {cm}^{- 1}}{200 S {cm}^{2} {mol}^{- 1}} = 2.0 \times 10^{- 7} mol {cm}^{- 3}$ $= 2.0 \times 10^{- 7} mol {(10^{- 1} dm)}^{- 3} = 2.0 \times 10^{- 4} mol {dm}^{- 3}$

From ${CaF}_{2} (s) ⇌ \underset{c}{{Ca}^{2 +}} (aq) + \underset{2 c}{2 F^{-}} (aq)$ we find $K_{sp} = [{Ca}^{2 +}] {[F^{-}]}^{2} = 4 c^{3} = 4 {(2.0 \times 10^{- 4} M)}^{3} = 3.2 \times 10^{- 11} M^{3}$