Molar Solubility

[MDwannabe7]

How do you figure out which compound has the highest molar solubility when given a list of compounds and their associated Ksps?

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[russellfx]

I wanna say the higher the Ksp the the higher the solubility bc it means that the rxn favors the dissociated products... if thats what your asking?

 

[MDwannabe7]

I wanna say the higher the Ksp the the higher the solubility bc it means that the rxn favors the dissociated products... if thats what your asking?

Yes and no - How does the number of dissociated ions play into this - b/c the answer that I have for the particular problem I am referring to actually has the lowest KSP but the greatest number of moles of dissociated ions. - Is that basically how you figure it out - the number of moles of dissociated ions times the ksp and the highest number is the highest molar solubility?

 

[russellfx]

Yes and no - How does the number of dissociated ions play into this - b/c the answer that I have for the particular problem I am referring to actually has the lowest KSP but the greatest number of moles of dissociated ions. - Is that basically how you figure it out - the number of moles of dissociated ions times the ksp and the highest number is the highest molar solubility?

Hmm well molar solubility I'm assuming refers to # of moles dissolved in solution.. so H2SO4 would have a higher molar solubility than HCl because if you had 1 mole of each the H2SO4 would dissociate into 3 moles and HCL only 2 (this assumes they both dissociate competely). So if you have some constraints on how much will dissolve you would have less moles in solution than the fully dissociated product so since the factor affecting the dissociation that varies based on substance is Ksp it would make sense to multiply it by the # of moles.. i think? lol im not exactly 100% on this but I think thats what i'd do.. whats the question anyway im curious?

 

[MDwannabe7]

whats the question anyway im curious?

Which of the following has the highest molar solubility?

BaCrO4 (Ksp = 2.1 x 10^-10)
AgCl (Ksp = 1.6 x 10^-10)
Al(OH)3 (Ksp = 3.7 x 10^-15)
PbCO3 (Ksp = 3.3 x 10^-14)

I'll provide the answer that I have if you indicate how you would figure this out and which one you would pick.

 

[ThePandaFactor]

BaCrO4 since its Ksp is the highest?

 

[russellfx]

Which of the following has the highest molar solubility?

BaCrO4 (Ksp = 2.1 x 10^-10)
AgCl (Ksp = 1.6 x 10^-10)
Al(OH)3 (Ksp = 3.7 x 10^-15)
PbCO3 (Ksp = 3.3 x 10^-14)

I'll provide the answer that I have if you indicate how you would figure this out and which one you would pick.

Okay molar solubility is the solubility of the compound in mol/L.

So solve for x:

I'm kinda lazy but if I had to guess you said the answer was the one w the smallest Ksp so ill do it for Al(OH)3.

Al(OH)3 -> Al3+ + 3OH-

Set up the equation:

Ksp = [Al][OH]^3

(solids arent included so thats why theres no denom. w AlOH3)

Now:

Ksp = [x][3x]^3 = 27x^4

3.7e-15 = 27x^4

Solve for x which is equal to the molar solubility. Repeat for all. Whichever one is the highest has the highest molar solubility.

 

[MDwannabe7]

Okay molar solubility is the solubility of the compound in mol/L.

So solve for x:

I'm kinda lazy but if I had to guess you said the answer was the one w the smallest Ksp so ill do it for Al(OH)3.

Al(OH)3 -> Al3+ + 3OH-

Set up the equation:

Ksp = [Al][OH]^3

(solids arent included so thats why theres no denom. w AlOH3)

Now:

Ksp = [x][3x]^3 = 27x^4

3.7e-15 = 27x^4

Solve for x which is equal to the molar solubility. Repeat for all. Whichever one is the highest has the highest molar solubility.

Okay, so

For Al(OH)3
((3.7 * (10^(-15)))^(1 / 4)) / 27 = 9.13453968 × 10-6

For BaCrO4
(2.1 * (10^(-10)))^(1 / 2) = 1.44913767 × 10-5

For AgCl
(1.6 * (10^(-10)))^(1 / 2) = 1.26491106 × 10-5

For PbCO3
(3.3 * (10^(-14)))^(1 / 2) = 1.81659021 × 10-7


Of course, if this were a real MCAT, I wouldn't have used a calculator, but I did this for speed - According to this method (if I did it all correctly) the answer should be BaCrO4, but it is Al(OH)3, at least according to my Kaplan instructor 2 years ago...

What am I missing?

 

[MDwannabe7]

BaCrO4 since its Ksp is the highest?

That's what I thought too, but that's not correct according to my Kaplan instructor from 2 years ago...

 

[russellfx]

hmm idk if you did those numbers right i rlly have no idea why its not working out right.. im pretty sure thats how to do it, unless im forgetting something stupid. anyone else ?

 

[MDwannabe7]

lol im trying to find a calculator but cant .. but what if you now took that number since x is mol/L and multiply it by the # of dissociated ions .. so Al(OH)3 ud do 9e-6 * 4 .. and then after that is Aloh3 the highest?

That seemed to work, but it doesn't make sense why you should have to do that... seems like we're pulling for straws or something...

For Al(OH)3

(9.13453968 × (10^(-6))) * 4 = 3.65381587 × 10-5



For BaCrO4

(1.44913767 × (10^(-5))) * 2 = 2.89827534 × 10-5



For AgCl

(1.26491106 × (10^(-5))) * 2 = 2.52982212 × 10-5



For PbCO3

(1.81659021 × (10^(-7))) * 2 = 3.63318042 × 10-7


Now Al(OH)3 is the largest, but idk if this is the right way to do it - any other opinions or facts?

 

[ThePandaFactor]

I think your instructor might just be confused

 

[MDwannabe7]

I think your instructor might just be confused

You know - that's quite possible - I didn't really like him and didn't think he explained certain things very well - but now that I'm taking the MCAT again and reviewing all my notes, I'm finding more things that were probably incorrect - you'd think that the rest of my classmates would have figured it out tho - most of them are pretty smart - oh well.

 

[RPedigo]

Okay, let's settle this.

BaCrO4 (Ksp = 2.1 x 10^-10)
AgCl (Ksp = 1.6 x 10^-10)
Al(OH)3 (Ksp = 3.7 x 10^-15)
PbCO3 (Ksp = 3.3 x 10^-14)


BaCrO4 (Ksp = 2.1 x 10^-10)
Ksp = [Ba++][C2O4--]
Ksp =
Ksp = s^2
(Ksp)^(1/2) = s
s = 1.45 x 10^-5


AgCl (Ksp = 1.6 x 10^-10)
Ksp = [Ag+][Cl-]
Ksp =
Ksp = s^2
(Ksp)^(1/2) = s
s = 1.26 x 10^-5


Al(OH)3 (Ksp = 3.7 x 10^-15)
Ksp = [Al+++][OH-]^3
Now remember for every 1 Al+++ we have we will have three OH-, so the [OH-] is three times that of the [Al+++]
Ksp = [3s]^3
Ksp = 27s^4
(Ksp/24)^(1/4) = s
s = 1.11 x 10^-4


PbCO3 (Ksp = 3.3 x 10^-14)
Ksp = [Pb++][CO3--]
Ksp =
Ksp = s^2
(Ksp)^(1/2) = s
s = 1.82 x 10^-7



So let's compare:
BaCrO4, s = 1.45 x 10^-5
AgCl, s = 1.26 x 10^-5
Al(OH)3, s = 1.11 x 10^-4
PbCO3, s = 1.82 x 10^-7

By far (by a factor of ten), aluminum hydroxide has the highest molar solubility.



The way to solve this without a calculator would be seeing that everything breaks up into two ions except Al(OH)3. So everything that breaks up into two ions will have the Ksp square rooted to get the molar solubility. The Al(OH)3 would need the Ksp taken to the (1/4) power because it breaks up into four ions (and you'd have to divide by 27). So just roughly estimate it.

 

[ThePandaFactor]

Wow yeah I didn't even think about that, good call

 

[riceman04]

Ksp is not always a good indicator of a salts relative solubility b/c the units of solubility vary with the # ions.

 

[Craig McDermott]

Ksp is not always a good indicator of a salts relative solubility b/c the units of solubility vary with the # ions.

It is key to recognize how many ions a salt will break into. Thanks RPedigo for the excellent breakdown!