Ksp and Reaction Quotient *AAMC MATERIAL SPOILERS*

[GomerPyle]

Another question for ya'll! It is good practice

In aamc chem self assessment #91, "The addition of excess Ca2+ to a solution containing Ca2+ and CO3-2 ions causes CaCO3 to precipitate because:

A: Ksp for CaCO3 would increase due to the increase of Ca2+.
B: Ksp for CaCO3 would decrease due to the increase of Ca2+.
C: The reactivity quotient Q would be lower than the Ksp.
D: The reactivity quotient Q would be higher than the Ksp.

Immediately, A and B could be eliminated since Ksp cannot be changed (it is a constant).

I selected D because the reaction quotient (products over reactants) would decrease because of the increase of Ca2+ ions (reactants). This would shift the reaction towards more product (CaCO3), causing it to precipitate more.

The answer is C because "When CaCO3 is in solution, the following takes place: CaCO3 --> Ca2+ + CO3-2 and Ksp=[Ca][CO3]. When excess Ca is added, CaCO3 precipitates due to le chatliers principle. [CO3] is therefore decreased and Ksp is maintained.

I dont understand the reasoning. I understand that adding Ca increases CaCo3 due to le chatliers, and also decreases CO3 because more of it precipitates, but how does the reaction quotient become lower than the Ksp?

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[SSerenity]

When you add more of the reactants the [REACTANTS] portion increases. Q = [PRODUCTS]/[REACTANTS]

So Q is decreasing. When Q is less than K the reaction will go to the right.

You are correct to think that more Products will be formed. But the question asks what will happen BEFORE equilibrium is reached, "...causes CaCO3 to precipitate because...".

 

[GomerPyle]

When you add more of the reactants the [REACTANTS] portion increases. Q = [PRODUCTS]/[REACTANTS]

So Q is decreasing. When Q is less than K the reaction will go to the right.

You are correct to think that more Products will be formed. But the question asks what will happen BEFORE equilibrium is reached, "...causes CaCO3 to precipitate because...".

Okay, so because you have a precipitating product, that indicates that it is before equilibrium?

Also, I still don't understand how Q is greater than the Ksp, causing the CaCO3 to precipitate.

 

[SSerenity]

Okay, so because you have a precipitating product, that indicates that it is before equilibrium?

Also, I still don't understand how Q is greater than the Ksp, causing the CaCO3 to precipitate.

Yes. If a precipitate is about to form, equilibrium hasn't been reached.
If a reaction is about to explode, equilibrium hasn't been reached.

Q is not greater than KSP, I said its less than KSP. The moment you add Ca2+ to the reaction, you suddenly have more reactants than before. This will drive the reaction to the right. At this moment (before any precipitate has formed), your Q value is smaller than KSP because your [REACTANTS] just increased.

Now, as precipitate occurs, your [REACTANTS] value steadily decreases, hence INCREASING Q until Q=KSP.

Make sense?

 

[nOchemallday]

Yes. If a precipitate is about to form, equilibrium hasn't been reached.
If a reaction is about to explode, equilibrium hasn't been reached.

Q is not greater than KSP, I said its less than KSP. The moment you add Ca2+ to the reaction, you suddenly have more reactants than before. This will drive the reaction to the right. At this moment (before any precipitate has formed), your Q value is smaller than KSP because your [REACTANTS] just increased.

Now, as precipitate occurs, your [REACTANTS] value steadily decreases, hence INCREASING Q until Q=KSP.

Make sense?

I'm actually inclined to disagree (only about what is products vs reactants) LeChatlier's and all the rest is fine.

For all Ksp (solubility product) reactions, the reaction is ALWAYS listed as MX(s) <-> M+(aq) + X-(aq); Where Ksp = [M][X]. Because you are dissolving a solid (aka solubility). Exactly like how the explanation reads: CaCO3 <-> Ca2+ + (CO3)2-; Ksp=[Ca2+][(CO3)2-].

PRODUCTS = Ca2+ and (CO3)2-, and Solids are never included in equilibria expressions. An increase in Ca2+ is an increase in the products, not the reactants. Therefore if Q = [products]/[reactants], it would seem to me that Q > Ksp and thereby decreases as it forms CaCO3. (Decrease in products increase in reactants to reach equilibrium).

It would seem to me that D should be the correct answer. Obviously the answer is entirely dependent on which direction you consider the forward reaction; I'm just suggesting that solubility product reactions are always written as the dissolution of solids and not the formation of precipitates. Therefore Ca2+ is a product, not a reactant.

 

[eradicator]

Yes D is the correct answer. The eqn they are asking about is CaCO3 --> Ca2+ + CO32-.....where CaCO3 is the reactant....so the product Ca2+ is added....increasing Q....

Since K<Q....reaction proceeds to the left (favoring the reactants)....thereby producing CaCO3

 

[gettheleadout]

Thread closed as this is AAMC material outside the designated threads.