Asked by John
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?
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?
Answers
Answered by
DrBob222
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.
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.
Answered by
DrBob222
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.
Answered by
John
oh alright thank you i found out my mistake. thank you sir.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.