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Answer to Question #8274 in Other Chemistry for raven

Question #8274
The solubility product constant KS for BaF2 is 1.70*10^–6. How much barium
difluoride can be dissolved in 2.00 liters of pure water?
Expert's answer
s=ks * Mw
s =1.70*10^–6 * 175=2.975*10^-4 g/L
m= 2 L *2.975*10^-4
g/L=0.000595 g.

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