MgF2 has Ksp = 7.4 x 10-11. What is the solubility of MgF2 in mol L-1?
At equilibrium:
MgF2
<----> Mg2+
+ 2
F-
s
2s
Ksp = [Mg2+][F-]^2
7.4*10^-11=(s)*(2s)^2
7.4*10^-11= 4(s)^3
s = 2.6*10^-4 mol/L
Answer: 2.6*10^-4 mol/L
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