kaplan error? Ksp Q

[ashlander]

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I need help on this one.
I thought I had it right but Kaplan got me very very confused.
Q: What is the minimum amount of CrO4 ^2- that must be added to a liter of a saturated solution of AgCl in order to precipitate Ag2CrO4?
(Ksp of AgCl = 2.8 x 10^-10 Ksp of Ag2CrO4 = 1.4 x 10^-22)

I set up like this.
2.8 x 10^-10 = [Ag][Cl] , so [Ag] is 1.6 x 10^-5
Then,
1.4 x 10^-22 = [(2 x 1.6) x 10^-5]^2 x [CrO4] ...since (2x)^2 and X
solved for [CrO4]
Calculation came out close to 1 x 10^-13.

Kaplan solution says
Ksp = [Ag][Cl] = X^2 ...........i got that part
[Ag]^2 = 2.8 x 10^-10 ......i get this as well
[CrO4] = Ksp / [Ag]^2 = (1.4 x 10^-22) / (2.8 x 10^-10) = 5 x 10^-13
....I don't get this last part and was thinking if this is a mistake.
I think they left out 2 from (2x)^2
I am stuck on this Q and this really sux.
Thanx in advance.

 

[Xenon]

I need help on this one.
I thought I had it right but Kaplan got me very very confused.
Q: What is the minimum amount of CrO4 ^2- that must be added to a liter of a saturated solution of AgCl in order to precipitate Ag2CrO4?
(Ksp of AgCl = 2.8 x 10^-10 Ksp of Ag2CrO4 = 1.4 x 10^-22)

I set up like this.
2.8 x 10^-10 = [Ag][Cl] , so [Ag] is 1.6 x 10^-5
Then,
1.4 x 10^-22 = [(2 x 1.6) x 10^-5]^2 x [CrO4] ...since (2x)^2 and X
solved for [CrO4]
Calculation came out close to 1 x 10^-13.

Kaplan solution says
Ksp = [Ag][Cl] = X^2 ...........i got that part
[Ag]^2 = 2.8 x 10^-10 ......i get this as well
[CrO4] = Ksp / [Ag]^2 = (1.4 x 10^-22) / (2.8 x 10^-10) = 5 x 10^-13
....I don't get this last part and was thinking if this is a mistake.
I think they left out 2 from (2x)^2
I am stuck on this Q and this really sux.
Thanx in advance.

I think this is associated with the common ion effect thing. Since [Ag] has already been pre-determined, you just plug in the value of [Ag]^2 = 2.8 x 10^-10 in the second calculation. x * (2x)^2 = Ksp is used only when you're given a liter of a saturated solution of Ag2CrO4. I hope this helps clear up somewhat.

 

[nze82]

I need help on this one.
I thought I had it right but Kaplan got me very very confused.
Q: What is the minimum amount of CrO4 ^2- that must be added to a liter of a saturated solution of AgCl in order to precipitate Ag2CrO4?
(Ksp of AgCl = 2.8 x 10^-10 Ksp of Ag2CrO4 = 1.4 x 10^-22)

I set up like this.
2.8 x 10^-10 = [Ag][Cl] , so [Ag] is 1.6 x 10^-5
Then,
1.4 x 10^-22 = [(2 x 1.6) x 10^-5]^2 x [CrO4] ...since (2x)^2 and X
solved for [CrO4]
Calculation came out close to 1 x 10^-13.

Kaplan solution says
Ksp = [Ag][Cl] = X^2 ...........i got that part
[Ag]^2 = 2.8 x 10^-10 ......i get this as well
[CrO4] = Ksp / [Ag]^2 = (1.4 x 10^-22) / (2.8 x 10^-10) = 5 x 10^-13
....I don't get this last part and was thinking if this is a mistake.
I think they left out 2 from (2x)^2
I am stuck on this Q and this really sux.
Thanx in advance.

The answer provided by KAPLAN is correct. Here's how you should approach this problem:

Ag2CrO4 <--> 2Ag+ + CrO4 2-, Ksp = 1.4E-22 = [Ag+]^2x[CrO42-]

AgCl <--> Ag+ + Cl- , Ksp = 2.8E-10 = [Ag+]x[Cl-]

So, as you can see, Ag+ is the common ion. But what does this mean?

It means that when we pour in our Ag2CrO4 into our saturated AgCl solution, even before Ag2CrO4 starts to dissociate, Ag+ ions are available in the mixture (These Ag+ ions come from dissociation of AgCl). So, let's see how much Ag+ ions are available in the mixture right at the begining, even before Ag2CrO4 begins dissociating:

Ksp = 2.8E-10 = [Ag+]x[Cl-] = X^2 -->X = [Ag+] = sqr (2.8E-10) = 1.67E-5

Now, let's see what happens, when the added Ag2CrO4 begins to dissociate:
As dissociation begins, the compound breaks into 2 moles of Ag+ for every 1 mole of CrO42-. However, dissociation of Ag2CrO4 has only a minor contribution to the total amount of Ag+ ions. Why?

Ksp = 1.4E-22 = [Ag+]^2x[CrO42-]= 4X^3 = 1.4E-22

--> X = [Ag+] = 3.3E-8 = 0.000000033 ~ 0


So, you can ignore this amount. Subsequently, the total concentration of Ag+ ions in the mixture is:


[Ag+] = (1.67E-5 + 0.000000033) = 1.67E-5

Now that you have the total amount of Ag+ ions, you can replace it in the following equation to solve for the maximum solubility of CrO4 2-, which is also indicative of the maximum solubility of Ag2CrO4 (Because one mole of Ag2CrO4 releases one mole of CrO4 2-), and that's what the question is asking for:


Ksp = 1.4E-22 = [Ag+]^2x[CrO42-]

1.4E-22/[Ag+]^2 = [CrO4 2-] = 1.4E-22 / (1.67E-5)^2 = 5.0E-13



Hope this helps!

 

[UCB05]

I need help on this one.
I thought I had it right but Kaplan got me very very confused.
Q: What is the minimum amount of CrO4 ^2- that must be added to a liter of a saturated solution of AgCl in order to precipitate Ag2CrO4?
(Ksp of AgCl = 2.8 x 10^-10 Ksp of Ag2CrO4 = 1.4 x 10^-22)

I set up like this.
2.8 x 10^-10 = [Ag][Cl] , so [Ag] is 1.6 x 10^-5
Then,
1.4 x 10^-22 = [(2 x 1.6) x 10^-5]^2 x [CrO4] ...since (2x)^2 and X
solved for [CrO4]
Calculation came out close to 1 x 10^-13.

Kaplan solution says
Ksp = [Ag][Cl] = X^2 ...........i got that part
[Ag]^2 = 2.8 x 10^-10 ......i get this as well
[CrO4] = Ksp / [Ag]^2 = (1.4 x 10^-22) / (2.8 x 10^-10) = 5 x 10^-13
....I don't get this last part and was thinking if this is a mistake.
I think they left out 2 from (2x)^2
I am stuck on this Q and this really sux.
Thanx in advance.

Don't multiply by two. That's a technique used when the concentration is an unknown and you're working from a solid A2B starting point. In this case, Ksp = [Ag]^2[CrO4] and you've already calculated the value of [Ag]. Multiplying it by 2 means youre pulling another 1.6x10^-5 M out of thin air.