Ksp question
Started by myair
i think this is how you should do it:
you do ice chart (common ion effect)
agio3 <--> ag3+ 3 IO-
I__A___________ 10^-5
C -x ______+x __10^-5 + 3x
E A-x ______x___10^-5 + 3x
plug this into the ksp formula and solve for x, you can probably ignore the x though because the ksp is about 3 magnitures smaller than the initial concentration of IO-
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Common way to mistakenly do it:
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, given [IO3-] solve for [Ag+]
Proper way:
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, [IO3-] = 10^-5
let x = amount of AgIO3 that needs to be added
[x][(10^-5)+x] = 3.1 x 10^-8
(10^-5)x + x^2 = 3.1 x 10^-8
x^2 = 3.1 x 10^-8, since x^2 is much bigger than .00001x
x = (3.1 x 10^-8)^0.5, don't know how they added the 10^-5 part though
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, given [IO3-] solve for [Ag+]
Proper way:
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, [IO3-] = 10^-5
let x = amount of AgIO3 that needs to be added
[x][(10^-5)+x] = 3.1 x 10^-8
(10^-5)x + x^2 = 3.1 x 10^-8
x^2 = 3.1 x 10^-8, since x^2 is much bigger than .00001x
x = (3.1 x 10^-8)^0.5, don't know how they added the 10^-5 part though
Common way to mistakenly do it:
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, given [IO3-] solve for [Ag+]
Proper way:
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, [IO3-] = 10^-5
let x = amount of AgIO3 that needs to be added
[x][(10^-5)+x] = 3.1 x 10^-8
(10^-5)x + x^2 = 3.1 x 10^-8
x^2 = 3.1 x 10^-8, since x^2 is much bigger than .00001x
x = (3.1 x 10^-8)^0.5, don't know how they added the 10^-5 part though
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, given [IO3-] solve for [Ag+]
Proper way:
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, [IO3-] = 10^-5
let x = amount of AgIO3 that needs to be added
[x][(10^-5)+x] = 3.1 x 10^-8
(10^-5)x + x^2 = 3.1 x 10^-8
x^2 = 3.1 x 10^-8, since x^2 is much bigger than .00001x
x = (3.1 x 10^-8)^0.5, don't know how they added the 10^-5 part though
Common way to mistakenly do it:
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, given [IO3-] solve for [Ag+]
this is the way if we are looking for solubility
Ksp = [Ag+][IO3-] => x2 = 3.1 x 10^-8 => x= radical 3.1 x 10^-8
substract the ions of NaIo3. =>> (radical 3.1 x 10^-8) - 10^-5
that's how they solved it
Ksp = [Ag+][IO3-] = 3.1 x 10^-8, given [IO3-] solve for [Ag+]
this is the way if we are looking for solubility
Ksp = [Ag+][IO3-] => x2 = 3.1 x 10^-8 => x= radical 3.1 x 10^-8
substract the ions of NaIo3. =>> (radical 3.1 x 10^-8) - 10^-5
that's how they solved it
I don't see why they would do that, do you?
This is the right method of doing it. THe answer is pretty close to the one you got from the source. There is no logic for subtracting 10^-5 as far as I can see [since we already considered the 10^-5 by the common ion effect]
