Ksp and Solubility

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hypnosis3000

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This question was asked a few years back on sdn, but I don't get the solution.

The question asked: 1M NaOH is added to the following solutions (each is 1M) which will precipitate first?
AgOH (Ksp = 1.5 x 10 ^-8)
Al(OH)3 (ksp = 3.7 x 10^-15)
Mg(OH)2 (ksp = 1.2 x 10 ^-11)
Mn(OH)2 (ksp = 2 x 10^ -13)

The answer is AgOH.

I solved for x(where x=solubility) and I understand that the higher the x, the lower the precipitation. But with my calculations, I get Mn(OH)2 as the answer with the lowest solubility (x=3.6X10^-5)

For AgOH, I get x=1.22X10^-4
For Al(OH)3, I get x=1.08X10^-4
For Mg(OH)2, I get x=1.44X10^-4

In these calculations, I accounted for the stochiometric coefficients, but obviously must have messed up somewhere. So could someone please explain with detailed calculations what I messed up on.

Thanks

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This question was asked a few years back on sdn, but I don't get the solution.

The question asked: 1M NaOH is added to the following solutions (each is 1M) which will precipitate first?
AgOH (Ksp = 1.5 x 10 ^-8)
Al(OH)3 (ksp = 3.7 x 10^-15)
Mg(OH)2 (ksp = 1.2 x 10 ^-11)
Mn(OH)2 (ksp = 2 x 10^ -13)

The answer is AgOH.

I solved for x(where x=solubility) and I understand that the higher the x, the lower the precipitation. But with my calculations, I get Mn(OH)2 as the answer with the lowest solubility (x=3.6X10^-5)

For AgOH, I get x=1.22X10^-4
For Al(OH)3, I get x=1.08X10^-4
For Mg(OH)2, I get x=1.44X10^-4

In these calculations, I accounted for the stochiometric coefficients, but obviously must have messed up somewhere. So could someone please explain with detailed calculations what I messed up on.

Thanks

It would be nice if you could post a link to that thread, because it would be proper to quote the source of the question.

This a common ion effect question, so you can't use the method you have used; that's for when the salt was added to pure water. To solve a common ion question, you must consider the molarity of the ions in solution (in this case OH-). Also, the solutions could not start as 1M as the question states (given that the Ksp values are so much less than 1, they would wouldn't dissociate enough to make 1M solutions), so there is a problem with the wording of the question as you've written it. That aside, let's figure out which salt is least soluble under those conditions.

Choice A: AgOH (Ksp = 1.5 x 10^-8)
1.5 x 10^-8 = [Ag+][OH-] = (x)(1.0)
x = 1.5 x 10^-8 M

Choice B: Al(OH)3 (Ksp = 3.7 x 10^-15)
3.7 x 10^-15 = [Al3+][OH-]^3 = (x)(1.0)^3
x = 3.7 x 10^-15 M

Choice C: Mg(OH)2 (Ksp = 1.2 x 10^-11)
1.2 x 10^-11 = [Mg2+][OH-]^2 = (x)(1.0)^2
x = 1.2 x 10^-11 M

Choice D: Mn(OH)2 (Ksp = 2.0 x 10^-13)
2.0 x 10^-13 = [Mn2+][OH-]^2 = (x)(1.0)^2
x = 2.0 x 10^-13 M

The correct answer I believe should be Al(OH)3. AgOH would precipitate the least, so are you 100% of the wording of the original question? Where is that question originally from?
 
Would using NaOH be any different from using H2O? I would thinking since they are both sources of OH- and Na is not in any of your solutes.

If this was the case wouldn't you just calculate using the following:

AgOH
1.5x10^-8=(x)(x)
AlOH3
3.7x10^-15=(x)(3x)^3
MgOH2
1.2x10^-11=(x)(2x)^2
MnOH2
2x10^-13=(x)(2x)^2

Doing this gives AgOH the lowest solubility and thus would precipitate first, right?

On second thought, this is obviously the common ion effect, I think Berktech is correct. Everyone in that thread seems to think otherwise...
 
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I don't see why the problem couldn't be solved using hypnosis' method. When a common ion is added to the solution, wouldn't the compound which was least soluble in its original solution be the first to precipitate?

According to my calculations: MnOH2 had the lowest solubility before the common ion (OH-) was added.

The answer doesn't make sense to me either.
 
I think they got this on the other thread.

Yes, it's common ion, and yes, the one with the lowest solubility will precipitate first.

Just use:

Ksp(Al(OH)3) = [x][3x]^3 = 27x^4
...etc

I believe AgOH comes out to be smallest Ksp, and thus the first to precipitate.
 
When a common ion is added to the solution, wouldn't the compound which was least soluble in its original solution be the first to precipitate?

Yes, that's why you can compute the original solution and then just reason it out.
 
I think they got this on the other thread.

Yes, it's common ion, and yes, the one with the lowest solubility will precipitate first.

Just use:

Ksp(Al(OH)3) = [x][3x]^3 = 27x^4
...etc

I believe AgOH comes out to be smallest Ksp, and thus the first to precipitate.

The problem with this is if it is common ion, having 1M of NaOH, shouldn't it be:
Ksp(Al(OH)3)=[x][3x+1]^3, and as explained by Berk since KSP is <<<<1 we can ignore 3x making this [x][1]^3
 
The question is asking what happens when you ADD NaOH to a sol'n made of each of the given solutes. So the one that would precipitate out first would be the one with the HIGHEST concentration BEFORE the NaOH was added, since that sol'n would of course then be the most saturated...hence more likely to make a ppt?
 
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The question is asking what happens when you ADD NaOH to a sol'n made of each of the given solutes. So the one that would precipitate out first would be the one with the HIGHEST concentration BEFORE the NaOH was added, since that sol'n would of course then be the most saturated...hence more likely to make a ppt?

The question actually states that the solutions are assumed to all be 1M before adding 1.0 M NaOH, which is physically impossible (that is beyond the saturation point of all four salts). But let's assume they meant saturated solutions of the salts. If all of the salts are saturated, then the addition of NaOH will cause immediate precipitation in all four. So, the question again must be missing something as translated. But for the sake of getting an answer, let's consider adding NaOH to saturated salt solutions. The salt that has three OH ions will be most impacted by the addition of OH anion, resulting in the greatest magnitude of crashing out of solution.

No matter how you spin it though, this question is utter nonsense as it was translated by the original poster two years ago (see other thread). There must be something else given to actually have a question that can be answered.
 
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