TBR 3.27 -- Common Ion Effect Q

[PremedicalS]

---

Members don't see this ad.

3.27. Lithium carbonate is most solube in which of the following solutions?
a. 0.01 M Li+
b. 0.01 M CO3 2-
c. 0.10 M Li+
d. 0.10 M CO3 2-

Answer: B.

"Lithium carbonate is most soluble in the solution with the least common ion effect. Because two Li+ cations are formed upon the dissociation of Li2CO3, Li+ has a greater impact on the equilibrium than CO32-."

I understand what the explanation is saying, but am wondering if someone could please explain this a little better? I am still somewhat confused. Since lithium carbonate dissociates into 2Li+ ions for every 1 carbonate ion, wouldn't that mean that the presence of carbonate ions (versus the presence of the same amount of lithium ions) would affect the equilibrium the most...and thus, choice A would be correct?

Thank you in advance for your help 🙂

 

[Postictal Raiden]

3.27. Lithium carbonate is most solube in which of the following solutions?
a. 0.01 M Li+
b. 0.01 M CO3 2-
c. 0.10 M Li+
d. 0.10 M CO3 2-

Answer: B.

"Lithium carbonate is most soluble in the solution with the least common ion effect. Because two Li+ cations are formed upon the dissociation of Li2CO3, Li+ has a greater impact on the equilibrium than CO32-."

I understand what the explanation is saying, but am wondering if someone could please explain this a little better? I am still somewhat confused. Since lithium carbonate dissociates into 2Li+ ions for every 1 carbonate ion, wouldn't that mean that the presence of carbonate ions (versus the presence of the same amount of lithium ions) would affect the equilibrium the most...and thus, choice A would be correct?

Thank you in advance for your help 🙂

For lack of better word, think of it as if lithium saturation in the solution is twice as much as carbonate is. Therefore, the presence of more lithium ions would hinder the solubility of the compound twice as much as the presence of carbonate ion.

I hope this resolves for you.

 

[milski]

Another way to think about it: the equilibrium is Li2CO3 <--> 2Li+ CO3(2-) which leads to Ksp=[Li]^2[CO3]/[Li2CO3]. Since the concentration of Li is squared, changes in it will have larger effect on where the equilibrium is.

 

[PremedicalS]

ok thank you!

 

[LoLCareerGoals]

Another way to think about it: the equilibrium is Li2CO3 <--> 2Li+ CO3(2-) which leads to Ksp=[Li]^2[CO3]/[Li2CO3]. Since the concentration of Li is squared, changes in it will have larger effect on where the equilibrium is.

I like this framework. So let's plug #s in:

A: Ksp = [0.01]^2*[CO3^2-]/[LiCO3(aq??? or S???) = 0.0001*[CO3^2-]
B: Ksp = [Li+]^2*[0.01]/[LiCO3(aq??? or S???) = 0.01*[Li+]

Still a few questions remain for me:
1) Should we use solid Li2CO3 and thus not include in the Ksp formula?
2) Even 2 Ksp values come out as they do above, what do you put in for [CO3^2-] and [Li+]? Zeros? Isn't this the case where we have to do they whole initial, delta, final computation?

Don't really feel the intuition for this yet, so a bulletproof algebraic proof would be deeply appreciated.
-----------------------
Edit:

Ok so I am trying to work this one:
A:
Li CO3
0.01 0
+2x +x
-----------
0.01+2x x
Ksp_A = (0.01+2x)x = 2x^2+ 0.01x

B:
Li CO3
0 0.01
+2x +x
---------------
2x 0.01+x
Ksp_B = 2x(0.01+x) = 2x^2 + 0.02x


Clearly Ksp_B > Ksp_A but ONLY IF x from A is the same as x from B which I frankly don't see why it would be. Did I do this right and if so, can you explain the justification for x from A is same as x from B?
Thanks.

 

[milski]

You are right, [Li2CO3] should be 1, since it is solid. That does not really matter since it changes in the same way in all cases. If you want to be formal about it, you can consider Qsp1/Qsp2 where [Li2CO3] cancels and arrive to the same conclusions.

The reaction quotients that you want to compare are (0.01+&#916;x)*(2&#916;x)^2 and &#916;x*(0.01+2&#916;x)^2. If you expand both expressions is should be easy to see that the second one grows faster with &#916;x and will allow you to add less Li2CO3.

In the first part of your work you're using 0.01 - that's incorrect. If you are adding Li2CO3, you are adding both Li and CO3, the concentrations should be 2&#916;x and 0.01+&#916;x or &#916;x and 0.01+2&#916;x, depending on weather the 0.01 is lithium or carbonate.

In the second part you have that correct, but have missed the fact that [Li] needs to be squared.