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In hard water, [Ca2+] is about 2.0×103 M. Water is fluoridated with 1.0 g F per 1.0×103 L of water. Will CaF2(s) precipitate from hard water upon fluoridation?
Find the reaction quotient for CaF2 and compare it to Ksp.
CaF2(s)Ca2+(aq) + 2 F(aq)
Q = [Ca2+]init[F]2init
[Ca2+]init = 2.0×103 M
[F]init = (1.0 g)/(19.0 g/mol)/( 1.0×103 L) = 5.3×105 M
so Q = [2.0×103][ 5.3×105]2 = 5.6×1012
Ksp = 5.3×109
Since Ksp > Q, there are not enough ions to satisfy the equilibrium condition so no precipitation occurs.