DAT Destroyer GC question

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The Ksp of Fe(OH)2 at 25 degrees celsius is 1.6E-14. What is the solubility of Fe(OH)2 in 0.025 M FeCl2?

What confuses me about this question is the way Destroyer sets up the problem--

Ksp= [Fe+2][OH-]^2
Ksp=[x][2x]^2
1.6E-14=(.025)(4x^2)

Why the 2x? I understand its differentiated between the other "x" variable-- but we are given that x?

I tried the problem without the "2x" and I get the wrong answer...

I've watched Chad's videos, and thats not the way he does those problem...

so why the 2x?? help?!

answer comes out to 4E-7 M

 

[Stacker]

The Ksp of Fe(OH)2 at 25 degrees celsius is 1.6E-14. What is the solubility of Fe(OH)2 in 0.025 M FeCl2?

What confuses me about this question is the way Destroyer sets up the problem--

Ksp= [Fe+2][OH-]^2
Ksp=[x][2x]^2
1.6E-14=(.025)(4x^2)

Why the 2x? I understand its differentiated between the other "x" variable-- but we are given that x?

I tried the problem without the "2x" and I get the wrong answer...

I've watched Chad's videos, and thats not the way he does those problem...

so why the 2x?? help?!

answer comes out to 4E-7 M

1 Fe(OH)2 > 1 Fe + 2 OH, there are more OH particles than there are Fe particles, in fact there are 2 times more OH than Fe so if x =Fe, then 2x=OH
------------------------------------------------------------------------------
Total solution moles = 3 moles
1 mole Fe = 1/3 solution moles...3 moles Fe = 1 solution mole
1 mole OH = 2/3 solution moles...3/2 moles OH = 1 solution mole
3 moles Fe = 3/2 moles OH...2 moles Fe =1 mole OH
[Fe] = x, 2x = [OH]
Ksp = [Fe]*[OH]^2, Ksp = [x]*[2x]^2, Ksp = 4x^3

 

[bittersweet008x]

ooooh okay got it! thanks!

 

[Stacker]

ooooh okay got it! thanks!

But now I don't get it! Please tell me the mechanism because the question is now confusing.

Plugging in 0.025 for x in Ksp = 4x^3 gives you a Ksp of 6.25 x 10^-5, not 1.6E-14, so what is going on?!!?!? It's 2:26 am and I can't go to sleep now.

How could the question give you both the question and the answer with Ksp = 1.6E-14 and the x concentration = 0.025 M in the equation Ksp = 4x^3 because 1.6E-14 does not equal 6.25 x 10^-5.
AHHHHHHH!

 

[Potatochips]

But now I don't get it! Please tell me the mechanism because the question is now confusing.

Plugging in 0.025 for x in Ksp = 4x^3 gives you a Ksp of 6.25 x 10^-5, not 1.6E-14, so what is going on?!!?!? It's 2:26 am and I can't go to sleep now.

How could the question give you both the question and the answer with Ksp = 1.6E-14 and the x concentration = 0.025 M in the equation Ksp = 4x^3 because 1.6E-14 does not equal 6.25 x 10^-5.
AHHHHHHH!

you only plug in 0.025 for x that represents IRON not the whole thing

so it would be like 0.025 * 4x^2 = 1.6E^-14

x would be a solubility not Ksp too

 

[BYU4you]

Sorry to resurrect an old thread, but I thought I'd bring up the same question.
I understand why it is the way it is, but I am wondering if these types of problems will be on the real DAT?

Chad's videos don't really cover it like this and Bootcamp doesn't seem to have these problems that require the coefficient to be added in front of the concentration as well as being the power it is raised to.

 

[FeralisExtremum]

I am wondering if these types of problems will be on the real DAT?

I would expect it to - this is pretty fundamental equilibrium stuff. It's a piece of cake once you learn the logic behind setting up ICE tables, which is definitely worth learning.

 

[BYU4you]

It seems like most practice problems I do the common ion being added would be something like the original

The Ksp of Fe(OH)2 at 25 degrees celsius is 1.6E-14. What is the solubility of Fe(OH)2 in 0.025 M FeCl2?

but instead of adding FeCl2 they would be adding something like KOH which would make the original 2x of OH negligible and you can skip to the end instead of the way this original was set up.

if that makes sense.

 

[FeralisExtremum]

That makes sense, in that given situation the original concentration of [OH] would make differentiating between X and 2X for [OH] unnecessary towards actually solving the problem. But if they give you a question, even a simpler one (that could reasonable show up on the DAT) like:

The Ksp of Fe(OH)2 at 25 degrees celsius is 1.8E-12. What is the molar solubility of OH-?

Then that knowledge would definitely be necessary towards getting the right answer, without it being a particularly tricky question.

 

[PocketRocket]

Ok now Im confused.

The Ksp of Fe(OH)2 at 25 degrees celsius is 1.8E-12. What is the molar solubility of OH in a 0.025 M solution of Fe(OH)2?

What is the answer to that?

 

[FeralisExtremum]

Ksp of Fe(OH)2 = [Fe 2+][OH-]^2
1.8E-12 = [Fe 2+][OH-]^2

Now we look at how Fe(OH)2 dissolves:
Fe(OH)2 --> Fe (2+) + 2 OH-
- X --> +X + 2X (the spacing on this is messed up. -X should be underneath Fe(OH)2, +X should be underneath Fe 2+, and +2X should be underneath 2 OH-). In case you're wondering why it's 2X here: every time a molecule of Fe(OH)2 breaks apart, we get one Fe 2+ atom and two OH-.

We plug this in to our Ksp expression:
1.8E-12 = [Fe 2+][OH-]^2
1.8E-12 = [X][2X]^2
1.8E-12 = [X][4X^2]
1.8E-12 = 4X^3
.45E-12 = X^3

Next we solve for X, since I made up the numbers randomly they don't actually work out cleanly without a calculator but it should be:
x=0.000077

But the molar solubility for [OH-] isn't just X, it's 2X. So the answer is:
2 * 0.000077
0.000154

 

[BYU4you]

I think he was confused by the .025 M solution of Fe(OH)2 part

 

[FeralisExtremum]

Wow. Why did I even put that there. Now I am also confused.

EDIT: Ok, concentration of solution shouldn't have been there, chopped that out. My bad.

 

[Artdent97]

Ok. Destroyer has the right answer here. First of all Fe(OH)2 dissociates and produces 2OH's. The compound is not in water, it is in FeCl2. So, there will be a common ion effect because FeCl2 produces Fe2+. So, Fe is X, Oh is 2x. If you do the equilibrium expression, Ksp=(Fe)*(Oh)^2. However, you are adding Fe, so you will add + 0.025 to the x. Since the Ksp is 1.6*10^-14, it is very small, so you can neglect the x. It will be a very small value. You will end up with Ksp=[0.025][2x]^2.

Hope this helps.

 

[BYU4you]

Ok. Destroyer has the right answer here. First of all Fe(OH)2 dissociates and produces 2OH's. The compound is not in water, it is in FeCl2. So, there will be a common ion effect because FeCl2 produces Fe2+. So, Fe is X, Oh is 2x. If you do the equilibrium expression, Ksp=(Fe)*(Oh)^2. However, you are adding Fe, so you will add + 0.025 to the x. Since the Ksp is 1.6*10^-14, it is very small, so you can neglect the x. It will be a very small value. You will end up with Ksp=[0.025][2x]^2.

Hope this helps.

It really helps to read the whole thread through first haha the answer was answered a couple of years ago.

 

[le dentist94]

I'm a little confused too now... is this the right setup:

Fe(OH)2 ---> Fe + 2OH
I .025 .025 0
C -x +x +x
E .025-x .025 + x +x