I'll bet good money that you didn't square the F^-
1 mg/L F = 0.001/19 = 5.26E-5 mols/L
200 mg/L Ca = 0.200/40.078 = 4.99E-3 M
Qsp = (Ca^2+)(F^-)^2 =
(4.99E-3)(5.26E-5)^2 = 1.38E-11. This is smaller than 3.45E-11; therefore, no ppt; i.e., Ksp is not exceeded. Ppts occur when Qsp exceeds Ksp.
Will a fluoride concentration of 1.0 mg/L be soluble in a water containing 200 mg/L of calcium?
I got Qsp 1.38 * 10^-8 and Ksp = 3.45*10^-11 does that seem right? and does this mean fluride concentration will be soluble in water containing calcium?
3 answers
By the way, 1.38E-8 is larger than Ksp by a factor of about 1000; using your figures Ksp is exceeded and you get a ppt.
oh alright thank you i found out my mistake. thank you sir.