Answer to Question #40175 in Inorganic Chemistry for Farhan Mutlib
M(BaF2) = 175.34 g mol–1
Calculate the solubility of BaF2 in mol L–1.
s(BaF2) = 7.60427 × 10–3 mol L–1
Calculate the ion concentrations from the molar solubility.
[Ba2+] = 7.60427 × 10–3
[F–] = 1.520854 × 10–2
Ks(BaF2) = × 10–6 (3 sf)
How do i work out Ks?
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