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My search came up with nothing and I couldn't understand the solution in a textbook. Could someone please help me at least set up the equation?
1) The solubility of CaF2 (Ksp = 4 X 10 ^-11) in a 0.1 M solution of Ca(NO3)2 is approximately: Ans: 1 X 10 ^-5
For this I guess you separate the CaF2 into it equilibrium and assign the Ca an "s" and the F2 "2S", but then you square the 2s b/c there are 2 molecules of it? I guess I don't know where s and 2s and squaring the numbers came from. That's how the book said to do it.
Thanks!
1) The solubility of CaF2 (Ksp = 4 X 10 ^-11) in a 0.1 M solution of Ca(NO3)2 is approximately: Ans: 1 X 10 ^-5
For this I guess you separate the CaF2 into it equilibrium and assign the Ca an "s" and the F2 "2S", but then you square the 2s b/c there are 2 molecules of it? I guess I don't know where s and 2s and squaring the numbers came from. That's how the book said to do it.
Thanks!