Ksp question

[myair]

How many moles of AgIO3 (KSP=3.1x 10^-8) will dissolve in one liter of a 10^-5 M solution of NaIo3?

answer: (radical 3.1x 10^-8)- 10^-5

I don't think it is correct. will show me the steps plz

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[Contach]

How many moles of AgIO3 (KSP=3.1x 10^-8) will dissolve in one liter of a 10^-5 M solution of NaIo3?

answer: (radical 3.1x 10^-8)- 10^-5

I don't think it is correct. will show me the steps plz

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-

 

[myair]

I do not understand what you are doing

is their answer correct?

I know how they did it but I do not think it is a correct way. I believe it is from Kaplan but not sure.

 

[myair]

How many moles of AgIO3 (KSP=3.1x 10^-8) will dissolve in one liter of a 10^-5 M solution of NaIo3?

answer: (radical 3.1x 10^-8)- 10^-5

I don't think it is correct. will show me the steps plz

any idea? if the answer is correct
answer: (radical 3.1x 10^-8)- 10^-5????

 

[adizzle87]

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

 

[adizzle87]

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

 

[myair]

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

hummmm, I am still have doubts

 

[myair]

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

 

[adizzle87]

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

I don't see why they would do that, do you?

 

[harrygt]

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-

IO3- is the iodate ion. You can't say IO3- => 3IO-

 

[harrygt]

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

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]