molar solubility question

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stevvo111

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question molar solubility....

I know it's the max number of moles of substance that can be dissolved before it starts precipitating out, but I'm bugging out over what the exact value of it is.

for instance given M(OH)2 the Ksp=4x^3

ksp= 36 then I plug in 36 and get 36/4=x^3 ...

so x=3.... is 3 the molar solubility for M(OH)2, as a molecule, or is it the molar solubility for only M (since x=[M])?? I watched some random videos online and they never specified, help.

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I think Ksp of M(OH)2 would be 4x^3 since you have Ksp = [x][2x]^2 = 4x^3, not 4x^2. so you would equate 4x^3 = 16. this means that x^3 = 4 so x = square root of 2 which is like 1.41
 
oh right right...

But the more important question I have is does x equal molar solubility for M (with the molar solubility for OH being 2x) or is x the molar solubility for the whole compound M(OH)2?
 
I had trouble with this as well, but I think this is the solution:

so for M(OH)2 Ksp = (x)(2x)^2

The x corresponds to M, and the 2x corresponds to OH

You're solving for x, so in reality you're solving for the molar solubility of M. If you wanted to solve for OH, you would have to divide the solubility by two (x/2).

This was my source: http://mcat-review.org/solution-chemistry.php#ksp

I'm not entirely sure of this explanation so if someone could confirm that would be great
 
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I had trouble with this as well, but I think this is the solution:

so for M(OH)2 Ksp = (x)(2x)^2

The x corresponds to M, and the 2x corresponds to OH

You're solving for x, so in reality you're solving for the molar solubility of M. If you wanted to solve for OH, you would have to divide the solubility by two (x/2).

This was my source: http://mcat-review.org/solution-chemistry.php#ksp

I'm not entirely sure of this explanation so if someone could confirm that would be great

There is not much to explain since you are 100% right....
 
There is not much to explain since you are 100% right....

😀 haha that made me lol

I did some more reading and as a point of interest when you solve for x, you're solving for the compound M(OH)2, which is the same molar solubility of M. This is for when the question just asks for the molar solubility of the compound.

But if the question asked for the amount of ion generated when the salt dissociates for let's say OH, then it would be double (i.e. 2x). And of course if it asked for the amount of M ion dissociated it's the same as solubility, x.

So look out for if the question is asking you for molar solubility or for the amount of ion yielded upon dissociation.
 
😀 haha that made me lol

I did some more reading and as a point of interest when you solve for x, you're solving for the compound M(OH)2, which is the same molar solubility of M. This is for when the question just asks for the molar solubility of the compound.

But if the question asked for the amount of ion generated when the salt dissociates for let's say OH, then it would be double (i.e. 2x). And of course if it asked for the amount of M ion dissociated it's the same as solubility, x.

So look out for if the question is asking you for molar solubility or for the amount of ion yielded upon dissociation.

Sweet thanks, I did some research on my own and found just that. x is the molar solubility, to find other stuff use it and plug it in etc.
 
One thing to note here is that (x) is defined to be the molar solubility, i.e. the number of moles solid that can be dissolved to make 1L of solution. This means that even for a solid M2Y3, where you have no lone (x) in the solubility product ( Ksp = (2x)^2 x (3x)^3 ), the molar solubility of M2Y3 is still just (x).
 
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