caf2---->ca2+ + 2F-
at aout pH 3.18 water, F- exist HF and F- in the ratio 1:1 coz pH=pka
Ksp= [ca2+][F-]^2
[Ca2+]=x
Can the equation be Ksp=(x)(x/2)^2 right??
i can't understand why [F-] is x/2 in that part.
2006-07-26 06:34:06 · 5 answers · asked by ym 1 in Science & Mathematics ➔ Chemistry
F- takes part in two different reactions.
So the concentration at equiliibrium should be satisfying both Ksp and Ka and thus the total amount of CaF2 that will dissolve will be a bit more than in simple water (part of the F is trapped as HF)
That's why if [Ca+2]=x then [F-] is not 2x, but [F-]total=2x
and [F-]total=[F-]+[HF]
but you said that pH=pKa,
thus [F-]=[HF]
thus [F-]total=[F-]+[HF]= [F-]+ [F-]= 2[F-]
and we said [F-]total=2x=>
Thus 2[F-]=2x => [F-]=x
thus Ksp=x*x^2=x^3
2006-07-26 07:12:57 · answer #1 · answered by bellerophon 6 · 1⤊ 0⤋
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2016-12-14 14:20:28 · answer #2 · answered by Anonymous · 0⤊ 0⤋
No, its Ksp=(x)(2x)^2
2006-07-26 06:48:56 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋
you're right...it should be 2x squared....
that would represent this solubility product constant corretly. not x/2.
2006-07-26 07:12:41 · answer #4 · answered by ucenigma 3 · 0⤊ 0⤋
ask jeeves or your chemistry teacher
2006-07-26 06:39:42 · answer #5 · answered by erohk2000 1 · 0⤊ 0⤋