Common Ion Effect and Solubility

[Teleologist]

Of the following solutions, the solubility of calcium hydroxide (solid) is least in:

1) 0.4 M calcium chloride
2) 0.2 M barium hydroxide
3) 0.3 M potassium hydroxide
4) 0.45 M sodium nitrate

I've narrowed it down to 1 or 2 and I'm thinking it's 1 because that solution has 0.4 moles of calcium ion while the second solution has 0.4 moles of hydroxide ion. According to the solubility equilibrium of calcium hydroxide, there is one mole of calcium ion for every two moles of hydroxide ion. It seems to me that adding calcium ions then to the right hand side of the equilibrium will have more of an effect on the system than adding hydroxide ions, correct?

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[jayoh]

It should be (2), and I calculated the molar solubilities just to make sure. But since that involves looking up the Ksp on a table and taking square roots (things you won't do on the MCAT), I think the best way to answer this would be to imagine trying to dissolve 0.4 moles of Ca(OH)2 into a liter of solution (1) and solution (2) separately. Solution (1) would have 0.6 moles of Ca2+ after mixing, and solution (2) would have 0.8 moles of OH- after mixing. Since there is more common OH- you can reason that it would be less soluble in this solution.

 

[Teleologist]

So stoichiometric ratios play no factor in this calculation? I just want to make sure, because for the solubility equilibrium there are two hydroxide ions on the right side and one Ca(OH)2 molecule on the left hand side.

 

[jayoh]

So stoichiometric ratios play no factor in this calculation? I just want to make sure, because for the solubility equilibrium there are two hydroxide ions on the right side and one Ca(OH)2 molecule on the left hand side.

If you wanted to calculate the solubility you would take the stoichiometry into account in the Ksp expression:

Ksp = [Ca][OH]^2

So the exact solubility would be equal to [Ca] since it's a one-to-one ratio there. So you would use

Ksp = [.s][2s]^2

and solve for "s" to find the solubility. But like I said, I didn't learn that until quantitative chem, and you don't have enough information (or a calculator) to do that on the MCAT.

 

[Teleologist]

If you wanted to calculate the solubility you would take the stoichiometry into account in the Ksp expression:

Ksp = [Ca][OH]^2

So the exact solubility would be equal to [Ca] since it's a one-to-one ratio there. So you would use

Ksp = [2s]^2

and solve for "s" to find the solubility. But like I said, I didn't learn that until quantitative chem, and you don't have enough information (or a calculator) to do that on the MCAT.

That makes more sense. Thanks for explaining! Good thing I won't be seeing this around either 🙂.

 

[getsome111]

so I came here to propose that Jayoh was incorrect in his reasoning to arrive at answer 2, which I still thought was correct, but after working through the problem I think I've found that answer 1 is actually correct and solubility is actually effected less negatively when the species producing the effect is doubly present in the precipitate… go figure(maybe it the OH's propensity to shift equilibrium is reduced when it encounters 2 sites to do so). If anyone could explain this it would make my day.. anyway, I guess the answers A


ksp= 5x10^-6 (literature value, for illustration purposes)

first .4M CaCl2
Ksp Ca(OH)2 = [Ca][OH]^2
Ksp = [.4M + X][2X]^2
Wolfram alpha - just to prove a point, leaving out the ksp since its basically zero
1.6*x^2+4x^3-(5x10^-6)
x=-.0018 .0024

now .2M Barium hydroxide
Ksp Ba(OH)2 = [Ba][OH]^2
Ksp = [X][.4M+2X]^2
WA
4*x^3+1.6x^2+.16x-(5x10^-6)
x=3x10^-5

 

[jayoh]

so I came here to propose that Jayoh was incorrect in his reasoning to arrive at answer 2, which I still thought was correct, but after working through the problem I think I've found that answer 1 is actually correct and solubility is actually effected less negatively when the species producing the effect is doubly present in the precipitate… go figure(maybe it the OH's propensity to shift equilibrium is reduced when it encounters 2 sites to do so). If anyone could explain this it would make my day.. anyway, I guess the answers A

ksp= 5x10^-6 (literature value, for illustration purposes)

first .4M CaCl2
Ksp Ca(OH)2 = [Ca][OH]^2
Ksp = [.4M + X][2X]^2
Wolfram alpha - just to prove a point, leaving out the ksp since its basically zero
1.6*x^2+4x^3-(5x10^-6)
x=-.0018

now .2M Barium hydroxide
Ksp Ba(OH)2 = [Ba][OH]^2
Ksp = [X][.4M+2X]^2
WA
4*x^3+1.6x^2+.16x-(5x10^-6)
x=3x10^-5

I agree that my reasoning may not have been the best, but I was looking for a way to solve this without doing the math (since that's what we would have to do on the MCAT). But I think that the right answer is still (2), so let me know what you think:

the equations I used were like this- (I assume the entire ion concentration was from the solution, something we always did in class, but it works either way)

using the Ksp = [Ca][OH]^2 and Ksp = 4.7x10^-6

Solubility is equal to Ca solubility since it is one to one, so solve for "s":

Ksp = [.s][2s]^2

so plugging in 0.4M for CaCl2 I got:

4.7x10^-6=[0.4M][2s]^2

=> s= 0.0024

plugging in for Ba(OH)2 i got:

4.7x10^-6= [.s][0.4M]^2

=> s= 0.000029

So our answers were pretty much the same, except you got a negative value for your first one. So I plugged what you got into wolfram alpha as well and it says that the solution could be -0.0017 or +0.0017, so I think in this case we would use the positive one since you can't have a negative solubility. (try plugging in "1.6*x^2+4x^3=4.7*10^-6 from -.1 to .1" into wolframapha)

So I think the answer is still (2), but I would love to hear a better way to reason through this one on the MCAT. Hopefully someone else can chime in.

 

[getsome111]

Ok so I went back and plugged everything in, I don't know how I only saw the negative number, must have just been tired. I agree that it works out that it is more soluble in 2. Intially the way that I saw it was that the solubility product = the decimal multiple of all the ions in solution from that specific compound when it dissociates. the .4M CaCl2 decreases 1 term in the equation, while the Ba(OH)^2 decreases 2, thus a decrease in 2 terms will have a greater effect in decreasing the solubility.

 

[Teleologist]

So the answer is 1 or 2? Key says 1.

And I gotta agree with the key since I just worked it out using solubility products and everything. You don't necessarily have to go through with the math and find the exact S value; you can see through inspection that S of the barium hydroxide solution is greater than the S of the calcium chloride solution. In other words, calcium hydroxide is least soluble in a 0.4 M solution of calcium chloride, and more soluble in a solution of 0.2 M barium hydroxide.

 

[Teleologist]

I agree that my reasoning may not have been the best, but I was looking for a way to solve this without doing the math (since that's what we would have to do on the MCAT). But I think that the right answer is still (2), so let me know what you think:

the equations I used were like this- (I assume the entire ion concentration was from the solution, something we always did in class, but it works either way)

using the Ksp = [Ca][OH]^2 and Ksp = 4.7x10^-6

The Ksp value I found for calcium hydroxide is 6.5 * 10^-6. Source: my chem textbook (Chemistry by Silberberg). The Ksp value, however, shouldn't really matter in the calculations - it's just, in essence, an arbitrary constant that remains the same for any of the answer choices.

Solubility is equal to Ca solubility since it is one to one, so solve for "s":

Ksp = [.s][2s]^2

so plugging in 0.4M for CaCl2 I got:

4.7x10^-6=[0.4M][2s]^2

=> s= 0.0024

plugging in for Ba(OH)2 i got:

4.7x10^-6= [.s][0.4M]^2

=> s= 0.000029

So our answers were pretty much the same, except you got a negative value for your first one. So I plugged what you got into wolfram alpha as well and it says that the solution could be -0.0017 or +0.0017, so I think in this case we would use the positive one since you can't have a negative solubility. (try plugging in "1.6*x^2+4x^3=4.7*10^-6 from -.1 to .1" into wolframapha)

So I think the answer is still (2), but I would love to hear a better way to reason through this one on the MCAT. Hopefully someone else can chime in.

I'm not sure if your approach is valid since I'm getting the opposite answer - that the answer is 1. I think the best way is to set up an ICE table. Your approach is leaving out the change in ion concentration. For example, in your setup for putting calcium hydroxide (solid) in a solution of calcium chloride, you plug 0.4 in the Ksp equation for the concentration of calcium ion. I think you have to account for both the 0.4 moles contributed by the 1 liter of 0.4 M calcium chloride solution and whatever arbitrary but constant amount (S) of calcium ion contributed by the solid chunk of calcium hydroxide chucked into solution.

So Ksp should look something like this:

Ksp = (0.4 + S)(2s)^2 = [Ca^2+][OH^-]^2.

 

[getsome111]

Well, now that you said the answer is 1 it seems like the only reasonable explanation is that barium hydroxide- as it also forms a precipitate, can be partially driven out of solution, making way for more calcium hydroxide to dissolve. I would be torn to choose between this explanation and the results of the calculations. The calculation I did above was an ice table-- just condensed, jayohs method of plugging in molarities also works but it isn't easy to follow

 

[jayoh]

The Ksp value I found for calcium hydroxide is 6.5 * 10^-6. Source: my chem textbook (Chemistry by Silberberg). The Ksp value, however, shouldn't really matter in the calculations - it's just, in essence, an arbitrary constant that remains the same for any of the answer choices.



I'm not sure if your approach is valid since I'm getting the opposite answer - that the answer is 1. I think the best way is to set up an ICE table. Your approach is leaving out the change in ion concentration. For example, in your setup for putting calcium hydroxide (solid) in a solution of calcium chloride, you plug 0.4 in the Ksp equation for the concentration of calcium ion. I think you have to account for both the 0.4 moles contributed by the 1 liter of 0.4 M calcium chloride solution and whatever arbitrary but constant amount (S) of calcium ion contributed by the solid chunk of calcium hydroxide chucked into solution.

So Ksp should look something like this:

Ksp = (0.4 + S)(2s)^2 = [Ca^2+][OH^-]^2.

Yeah, I intentionally left it out to simplify the math (which is fairly common in problems like these). And the equation above is also what getsome11 did, and both seem to work out to show that (2) is the correct answer. But if the key says its 1, then that must be right but I don't understand how or why. It must be some reason outside of straight calculations

 

[jayoh]

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