Kaplan Acid Base Test Question

[SaintJude]

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Which of the following will occur if ammonia were added to a saturated solution of sodium sulfate?

A. The Ksp will decrease and an additional sodium sulfate will dissolve.
B. The Ksp will be unchanged and additional sodium sulfate will dissolve.
C. The Ksp will increase and additional sodium sulfate will precipitate.
D. The Ksp will remain unchanged and additional sodium sulfate will precipitate.

Ok, I understand what happens to the Ksp but how can I figure out what happens to sodium sulfate? Read Kaplan's explanation and still don't get it. 🙁

 

[MedPR]

Which of the following will occur if ammonia were added to a saturated solution of sodium sulfate?

A. The Ksp will decrease and an additional sodium sulfate will dissolve.
B. The Ksp will be unchanged and additional sodium sulfate will dissolve.
C. The Ksp will increase and additional sodium sulfate will precipitate.
D. The Ksp will remain unchanged and additional sodium sulfate will precipitate.

Ok, I understand what happens to the Ksp but how can I figure out what happens to sodium sulfate? Read Kaplan's explanation and still don't get it. 🙁

Is it C?

 

[milski]

Ksp=[Na]^2[SO4] for the concentrations of the saturated solution. If Ksp goes down, the product on the right will become larger than Ksp, meaning that you have excess of Na/SO4 and you'll get some precipitation. If it goes down, that will allow you to increase the concentrations further.

The volume of the solution will increase with adding NH3, which will drop the concentrations, but I'm not sure what the effect of NH3 on Ksp is going to be. Na should not form complex ions so Ksp should stay the same. All that leads to answer B.

But I'm not sure about the Ksp changes.

 

[SaintJude]

Umm, I don't think you're right. Well, it's D.

Ksp does not change at all! This is a constant that is only dependent on the salt structure & temperature of the solution.

Sooo....if it helps anyone I can post the Kaplan explanation--maybe someone can translate that into something coherent?

 

[milski]

Umm, I don't think you're right. Well, it's D.

Which part? 🙂 I'm not saying that I'm right but I'd like to know what is wrong.

Ksp - does not change, good, that was the one I was least sure about.
Concentrations going down - they're moles/volume, moles stay the same, volume increases, so they should go down.
Ksp definition - that's from the textbook...

Ksp does not change at all! This is a constant that is only dependent on the salt structure & temperature of the solution.

Sooo....if it helps anyone I can post the Kaplan explanation--maybe someone can translate that into something coherent?

Sure. Let's see what they think.

 

[MedPR]

Umm, I don't think you're right. Well, it's D.

Ksp does not change at all! This is a constant that is only dependent on the salt structure & temperature of the solution.

Sooo....if it helps anyone I can post the Kaplan explanation--maybe someone can translate that into something coherent?

Oh crap. I was thinking of Keq. I assumed the balanced equation was something like 2Na + SO4 -> Na2SO4, so if you add NH3, you shift the equilibrium to the right, which would increase Keq, I think?

Yes, Ksp as you said is dependent on the salt and temperature.

I think my thought process for Keq helps explains why the salt will precipitate. Here's why. You add in NH3, which shifts the equilibrium to the right, thus forming more Na2SO4(solid). Since your solution is already saturated, you can't dissolve anymore salt into it, so the newly formed salt must precipitate.

The only way to supersaturate is to heat it up to dissolve more, then to slowly cool it right?

 

[SaintJude]

Why would adding NH3 shift the equilibrium to the right?

 

[milski]

Oh crap. I was thinking of Keq. I assumed the balanced equation was something like 2Na + SO4 -> Na2SO4, so if you add NH3, you shift the equilibrium to the right, which would increase Keq, I think?

Yes, Ksp as you said is dependent on the salt and temperature.

I think my thought process for Keq helps explains why the salt will precipitate. Here's why. You add in NH3, which shifts the equilibrium to the right, thus forming more Na2SO4(solid). Since your solution is already saturated, you can't dissolve anymore salt into it, so the newly formed salt must precipitate.

The only way to supersaturate is to heat it up to dissolve more, then to slowly cool it right?

But NH3 does not participate in the equation above, why would it shift the equilibrium?

 

[SaintJude]

And here is the explanation:

A saturated solution of sodium sulfate will be slightly basic. Because sulfate is the conjugate base of sulfuric acid, it is a weak base. But still as a weak base it will react with water molecules to pluck of protons, forming an equilibrium with the bisulfide ion (HSO4-) and creating hydroxide ions, thus creating a slightly basic solution. [May I add: WHAT?!?]

If ammonia is added to the solution, it will react in the same manner, forming ammonium and a hydroxide ion. This common effect will then allow less sodium sulfate to dissolve, precipitating some out, as ammonia sulfate has a higher Ksp than sodium sulfate [How was I supposed to know Ksp values?!] The MCAT expects you to know that Ksp is a function of the structure of the salt and the temperature of the solution. Adding a stress to the system will not affect the Ksp if the temperature remains the same, so the answer is D.

 

[milski]

So they're saying that you actually have (NH4)2SO4 and Na2SO4 dissolved, you have to consider the common ion effect for the two and know that the former has a lower Ksp. I can agree with the logic of the having both solutes but expecting to know the Ksp order seems a bit too much.

 

[MedPR]

Why would adding NH3 shift the equilibrium to the right?

But NH3 does not participate in the equation above, why would it shift the equilibrium?

Doesn't adding more particles into a solution result in shifting the equilibrium to the side with less particles? Just like increasing the pressure would shift to the side with less particles and increasing temperature should shift to the product side in an endothermic reaction?

 

[mcloaf]

Doesn't adding more particles into a solution result in shifting the equilibrium to the side with less particles? Just like increasing the pressure would shift to the side with less particles and increasing temperature should shift to the product side in an endothermic reaction?

As far as I know the two are not comparable. In the Le Chatelier's example about gases the gases are (in theory) the only things occupying your reaction vessel, so shifting the equilibrium can help account for added pressure by reducing moles of gas, PV=nRT, etc. In an aqueous reaction almost the entire volume is occupied by solvent, and your reaction vessel (unless otherwise indicated) is probably not closed and is at atmospheric pressure. Adding additional ions to solution is not going to appreciably change the volume of the reaction in the same way that going from 1 mol to 2 moles of gases will (i.e. doubling the pressure x volume, other things being equal).

As for the original question, it's probably harder than what the MCAT would ask, but you should be able to figure it out. The Ksp, as others have said, won't change. Draw out the three equilibria (leaving out charges for clarity):

Na2SO4 --> 2Na + SO4

H20 + SO4 --> OH + HSO4

NH3 + H20 --> NH4 + OH

You don't need to know the relative Ksps to see that adding ammonia will push the sulfate/bisulfate equilibrium left, and in so doing precipitate some sodium sulfate by pushing the top equilibrium left.

 

[milski]

As for the original question, it's probably harder than what the MCAT would ask, but you should be able to figure it out. The Ksp, as others have said, won't change. Draw out the three equilibria (leaving out charges for clarity):

Na2SO4 --> 2Na + SO4

H20 + SO4 --> OH + HSO4

NH3 + H20 --> NH4 + OH

You don't need to know the relative Ksps to see that adding ammonia will push the sulfate/bisulfate equilibrium right, and in so doing precipitate some sodium sulfate by pushing the top equilibrium right.

Aha, nice! That makes sense now, thank you!


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I am here: http://maps.google.com/maps?ll=47.678468,-122.327500

 

[mcloaf]

Aha, nice! That makes sense now, thank you!


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I am here: http://maps.google.com/maps?ll=47.678468,-122.327500

🙂

I think most of these equilibria questions get a lot easier when you actually draw out what's going on.


p.s. Greenlake, eh?! Nice. I live on Queen Anne at the moment.

 

[SaintJude]

Mmhhm, you're right. Those 3 equations were the key. Thank you!!

 

[MT Headed]

I got the right answer with a slightly different approach.

I knew Na2SO4 was in equilibrium with Na+ and SO4(2-)
I also knew that SO4(2-) is a weak base conjugate of the weak acid HSO4-
So all these are in equilibrium. Yay.

Suddenly somebody adds NH3, a well known base. What happens?
first, the base will eat available protons, shifting the
H+ SO4(2-) ---> HSO4- equilibrium to the left.
well geez, that means that suddenly we have a lot more SO4(2-) in solution.
And that will cause more precipitate.

 

[milski]

🙂

I think most of these equilibria questions get a lot easier when you actually draw out what's going on.


p.s. Greenlake, eh?! Nice. I live on Queen Anne at the moment.

Yes, writing down all that was going on would have helped. I was trying to reason through it.

PS: I can see most of the North side of Queen Anne from my window.. Look wet right now. 😛