Solubility Problem

[The Law]

A saturated solution of CaF2 is created by adding excess CaF2 to water.

The excess CaF2 precipitate is filtered out of solution. 0.1 M HCl is added to the solution. Does a precipitate form?

 

A) Yes, because chloride ions from the dissociation of HCl act as a common ion causing precipitation

Yes b/c HCl is a strong acid

C) No because acid increases the solubility of CaF2

D) No because protons push the equilibrium of the overall reaction to the right balancing the leftward shift caused by HCl

 

 

 

Answer: D

 

 

Can someone explain the concepts being tested here please. There's no common ion present, so I'm a bit confused as to what is happening...

 

Merci in advance.

[war485]

They're testing you on equilibrium here, nothing more.

 

One way to think is to use equilibrium constant K.

 

HCl + CaF2 --> Ca2+ + 2F- + H+ + Cl-

 

compare K1 = [H+][Cl-][Ca2+]([F-]^2) / [HCl]

 

to the K in the original saturated solution

 

CaF2 --> Ca2+ + 2F-

K2 = [Ca2+]*[F-]^2

 

in K1, if you had [HCl] in less than 1 M, then you are dividing by a small number, effectively increasing K (in this case, 10x), and thus is much bigger than K2, so no solid should be present since it will favor the products much more (not the solid CaF2, which is a reactant).

 

Another way to think of this is look at the overall equation:

HCl + CaF2 --> Ca2+ + 2F- + H+ + Cl-

If you added the HCl, you are effectively adding more reactants, which will push it towards the products.

[Guest Juicy.Fruit]

You approach doesn't make sense to me. It is an equilibrium question but your approach has HCl and CaF2 in one equilibrium equation. I don't think you can include HCl + CaF2 into a single equilibrium constant since they dissociate independently.

 

Here is my approach::

 

CaF2 <--> Ca2+ + 2F-

Adding HCl will shift the equilibrium to the left since the solution is saturated to begin with. HCl is a strong acid so it will dissociate nearly 100%, pushing Ca2+ + F- out of solution. Remember, when saturated, the number of particles the water can hold has been maxed out. So, if HCl dissociates into ions, Ca + F will have to exit the solution.

HCl --> H+ + Cl-

 

But,

HF <--> H+ + F-

 

HF is a weak acid. So its equilibrium lies far to the left. F- will be trapped as HF, pulling the CaF2 <--> Ca2+ + 2F- back to the right.

 

So, the equilibrium: CaF2 <--> Ca2+ + 2F- will not be affected by the addition of HCl. (Net effect of HCl pushing to the left and formation of HF pushing to the right.)

 

Hence, the choice D.

 

On the real thing, I think it would be best to use process of elimination for this Q. A,B,C are all obviously wrong answers.

 

If anyone sees a flaw in my reasoning feel free to point it out..

&& Good luck with the MCAT as well~

[The Law]

Good explanation. That was my reasoning juicy.fruit... but I couldn't quite figure out what'd push equilibrium to the left for HF... and now it's obvious, HF is a weak acid!

 

Thanks for your help!