Ksp question from TBR

[yellowjellybean]

When 0.1 gram ZnCO3(s) is added to 100mL of water at 25 degrees Celsius, which of the following statements best describes what is observed? (taken from passage X in TBR but I don't think you need the passage for this one)

A. The entire 0.10 g sample dissociates completely
B. Most of the 0.10 g sample dissociates into solution, with only a small portion not dissolving.
C. Only a small portion of the 0.10 g sample dissociates into solution, with most of the ZnCO3 remaining as a precipitate in the flask.
D. None of the 0.10 g sample dissociates into solution.

I picked C which is the right answer but their explanation completely confuses me. I thought C was the right answer because the given Ksp for ZnCO3 is 1.4*10^-5 means there are more reactants so it would dissociate very little. That was how I reasoned it but the explanation in the back of the book is really elaborate and talks about comparing Q with Ksp and makes my reasoning too simple... so what is wrong here? Thanks very much.

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[BerkReviewTeach]

When 0.1 gram ZnCO3(s) is added to 100mL of water at 25 degrees Celsius, which of the following statements best describes what is observed? (taken from passage X in TBR but I don't think you need the passage for this one)

A. The entire 0.10 g sample dissociates completely
B. Most of the 0.10 g sample dissociates into solution, with only a small portion not dissolving.
C. Only a small portion of the 0.10 g sample dissociates into solution, with most of the ZnCO3 remaining as a precipitate in the flask.
D. None of the 0.10 g sample dissociates into solution.

I picked C which is the right answer but their explanation completely confuses me. I thought C was the right answer because the given Ksp for ZnCO3 is 1.4*10^-5 means there are more reactants so it would dissociate very little. That was how I reasoned it but the explanation in the back of the book is really elaborate and talks about comparing Q with Ksp and makes my reasoning too simple... so what is wrong here? Thanks very much.

You are correct that with such a small Ksp that very little can dissociate into solution, but that is only half of what you need to consider. You also have to consider whether 0.10 g is very little. It ends up that 0.10 g is about 0.008 moles, which is 8.0*10^-3. That's a lot compared to 1.4*10^-5, so only a small amount of the 0.10 grams will dissociate into the solution.

It would be like telling you that the melting point of a compound is 31.5 degrees C and asking if it is a liquid and then saying "that's a high MP, so it must be a solid." You can't answer that question without also knowing the temperature.

 

[Meredith92]

Just a quick question- why are we comparing the moles we start with to the Ksp? Shouldnt we be comparing it to the molar solubility?

 

[yellowjellybean]

OK I see, if we were to apply the same problem but instead of using Ksp we used say Kp, could I then just look at the Kp and determine qualitatively the relative amounts of products vs. reactants. Since the reactants are now in the equilibrium expression whereas Ksp the reactant was not in the equilibrium expression. Does my question make any sense?

 

[yellowjellybean]

My bad, the 1.4*10^-5 IS the molar solubility and not the Ksp.

 

[Sarahka74]

You are correct that with such a small Ksp that very little can dissociate into solution, but that is only half of what you need to consider. You also have to consider whether 0.10 g is very little. It ends up that 0.10 g is about 0.008 moles, which is 8.0*10^-3. That's a lot compared to 1.4*10^-5, so only a small amount of the 0.10 grams will dissociate into the solution.

It would be like telling you that the melting point of a compound is 31.5 degrees C and asking if it is a liquid and then saying "that's a high MP, so it must be a solid." You can't answer that question without also knowing the temperature.

How can you tell between C and D? How do we know a little of it will dissociate, rather than none of it dissociating?