The Ksp of calcium fluoride is 3.2×10−11 Calculate the ΔG∘ in kJ⋅mol−1 for the dissolving of solid calcium fluoride at 25° C.

ΔG∘=−2.303×8.314×2981000log10⁡3.2×10−11=+59.9

ΔG∘=−2.303×8.314×2981000log10⁡3.2×10−11=+59.9

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The $K_{s p}$ of calcium fluoride is $3.2 \times 10^{- 11}$ Calculate the ${ΔG}^{\circ} (in kJ \cdot {mol}^{- 1})$ for the dissolving of solid calcium fluoride at 25° C.

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2.18

5.02

26.0

59.9

answer is D.

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$Δ G^{\circ} = \frac{- 2.303 \times 8.314 \times 298}{1000} \log_{10} (3.2 \times 10^{- 11})$

$= + 59.9$

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