gchem problem ksp

[yakuza]

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Can someone please explain kaplan's solution





41.

What is the minimum amount of CrO42- that must be added to a liter of a saturated solution of AgCl in order to precipitate Ag2CrO4?
Ksp AgCl = 2.8 x 10^-10
Ksp Ag2CrO4 = 1.4 x 10^-22
ans: 5E-13 moles
​

Kaplan's solution:

This question about Ksp and the precipitation of a slightly soluble salt can be answered using the ion product expressions for Ksp: Ksp = [M][X] and Ksp = [M]2[X]. The first of these expressions is valid for any MX salt, including the silver chloride in this question, while the second expression applies to any M2X salt, such as the silver chromate which will precipitate upon sufficient addition of chromate ion. To solve for the necessary chromate ion concentration, we can rearrange the expression of Ag2CrO4 as follows:
Ksp = [Ag+]2
[CrO42-][CrO42-] =

--------wtf happened here?

We were given the Ksp, but we need the [Ag+]2 to complete the set up. Because the solution is saturated, we can get [Ag+]2 directly from the Ksp expression for AgCl, then substitute:
Ksp = [Ag+][Cl-] = x2
[Ag+]2 = 2.8 x 10-10
[CrO42-] = Ksp/[Ag+]2 = (1.4 x 10-22)/(2.8 x 10-10) = 0.5 x 10-12 = 5.0 x 10-13​

​

 

[Danny289]

Can someone please explain kaplan's solution




41.

What is the minimum amount of CrO42- that must be added to a liter of a saturated solution of AgCl in order to precipitate Ag2CrO4?
ans: 5E-13 moles
​

Kaplan's solution:

This question about Ksp and the precipitation of a slightly soluble salt can be answered using the ion product expressions for Ksp: Ksp = [M][X] and Ksp = [M]2[X]. The first of these expressions is valid for any MX salt, including the silver chloride in this question, while the second expression applies to any M2X salt, such as the silver chromate which will precipitate upon sufficient addition of chromate ion. To solve for the necessary chromate ion concentration, we can rearrange the expression of Ag2CrO4 as follows:
Ksp = [Ag+]2
[CrO42-][CrO42-] =

--------wtf happened here?

We were given the Ksp, but we need the [Ag+]2 to complete the set up. Because the solution is saturated, we can get [Ag+]2 directly from the Ksp expression for AgCl, then substitute:
Ksp = [Ag+][Cl-] = x2
[Ag+]2 = 2.8 x 10-10
[CrO42-] = Ksp/[Ag+]2 = (1.4 x 10-22)/(2.8 x 10-10) = 0.5 x 10-12 = 5.0 x 10-13​

​

(1)you know Ksp for Ag2CrO4 ======> ksp= [Ag+]2 [CrO4 2-]
(2)ksp for AgCl =======> ksp= [Ag+][Cl-] ====>[Ag+]=[Cl-] ksp=[Ag+]^2
now plug ksp in (2) for your [Ag+]2 in (1). clear?

 

[yakuza]

Isn't it suppose to be

Ksp Ag2CrO4 = 4[2.8 x 10^-10] ^2 [X]

 

[Danny289]

Isn't it suppose to be

Ksp Ag2CrO4 = 4[2.8 x 10^-10] ^2 [X]

No, we had a long disscusion before about that.

 

[yakuza]

The ksp formula is: [Ag+]^2 [CrO--] 100% sure

but when you do the math, you add in 2x into [Ag+]^2. I really hope kaplan's solution is wrong..lol
this isn't making sense

 

[doc3232]

Ksp of AgCl - is Ag x Cl
so
take square root of it and you get 1.67 x 10^-5

Now Ksp of the Ag CrO4 = 1.4 x 10^-22 = (1.67 x 10^-5)^2 x (How much Chromate)

Solve for how much Chromate and you get 5 x 10^-13

I think this is right, not sure.

 

[yakuza]

but its 2Ag CrO4