Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: blightcutter on October 28, 2008, 06:22:37 PM
-
Hello all,
I have been reading these forums for a while and I must say that they help me understand a lot of these concepts. So, thanks!
Anyway, I am attepmting to find the solubility of AgIO3 at constant AND variable pH. The only given here is AgIO3 Ksp=3.1 x 10-8
My approach looks like this...
AgIO3 :larrow: :rarrow: Ag++IO3-
IO3-+H2O :larrow: :rarrow: HIO3+OH-
2H2O :larrow: :rarrow: H3O++OH-
Kb=[OH-][HIO3]
[IO3]
Kw=[H3O][+OH-]=1 x 10-14
For the mass balance I have:
[Ag+]=[IO3-]+[HIO3]
Now, I know I need to make some assumptions here. Can anyone lead me in the right direction? Is what I wrote correct?
-
The only given here is AgIO3 Ksp=3.1 x 10-8
If you don't have pKa of HIO3, you can't calculate pH dependence.
Besides, that's a strong acid, so its protonation will not play any significant role unless pH will be below 1.
Generally speaking what you wrote so far is OK, although Kw and Kb expressions are probably not needed. Ka will be enough.
-
Thank you for the reply last night. Just wanted to finish the problem on here.
Kb=[OH-][HIO3] = 5.9x10-14
[IO3]
@constant pH
if pH=7
Since the pH is fixed, [H3O+] and [OH-] are known.
From Kb,
[HIO3]=5.9x10-14[IO3-]/[OH-]
pH=7 so
[HIO3]=5.9x10-14[IO3-]/1x10-7=5.9x10-7[IO3-]
from the mass balance equation, now,
[Ag+]=[IO3-]+5.9x10-7[IO3-]=[IO3-]
This makes the Ksp value = [Ag+][Ag+]=3.1x10-8
and [Ag+]=1.8x10-4
:D
--Now, for variable pH I think its necessary to make a few assmptions. However, like Borek said, the pH needs to be below 1 in order to make significant amounts of HIO3. So could I say that any assumptions I make will be invalid unless the pH is below 1? Again, thanks for your help.
-
As Borek wrote you can safely neglect Kw and Ka. Then:
[Ag+]=SQRT(Ksp) = 1.8x10-4
-
It's starting to make more sense to me now =D thank you. (Chem minor here :-\ this stuff is a little confusing at first...Ooo mole snack!)