Predicting Formation of Precipitate

[janeaustin]

Hey could you please do this question for me?

Predict whether a precipitate will form when a 1.0L solution contains 0.0040 molL^-1 (NH4)2CO3 and 3.0x10^-5 molL^-1 Mg(CH3COO)2

[avocadinq]

Hi janeaustin! here is the solution,

So essentially what I did was figure out the mass of both (NH4)2CO3 and Mg(CH3COO)2, using c=n/v and n=m/mm. I then determined that Mg2CO3 was insoluble because of the solubility rules (group two carbonates are insoluble) and found the ksp of Mg2CO3 from the formula sheet. After I used the V2= C1V1/V2 (rearranged from c1v1=c2v2), and plugged in respective values for Mg and CO3. Using those values, I calculated the ionic product (q), which was 4.301810372 x 10^-25. As Q is less than Ksp, a precipitate will not form.

Hopefully this helps, and let me know if you need further clarification

[janeaustin]

Hi janeaustin! here is the solution,

So essentially what I did was figure out the mass of both (NH4)2CO3 and Mg(CH3COO)2, using c=n/v and n=m/mm. I then determined that Mg2CO3 was insoluble because of the solubility rules (group two carbonates are insoluble) and found the ksp of Mg2CO3 from the formula sheet. After I used the V2= C1V1/V2 (rearranged from c1v1=c2v2), and plugged in respective values for Mg and CO3. Using those values, I calculated the ionic product (q), which was 4.301810372 x 10^-25. As Q is less than Ksp, a precipitate will not form.

Hopefully this helps, and let me know if you need further clarification

Hey, thank you for your reply! I have a question... why does the mass of each have to be found? Also, how did you find the values of V2 for each?

[avocadinq]

Hey, thank you for your reply! I have a question... why does the mass of each have to be found? Also, how did you find the values of V2 for each?

Hello! The reason why we find the mass of each is because we use that mass to find the volume for v1 and when we add the volumes of both, for v2. To find v2, we simply just add the v1 values of (NH4)2CO3 and Mg(CH3COO)2. Sorry that wasn't clear, i made a typo in my response - i was supposed to write C2=C1V1/V2 hahaha. Also, just realised I made a mistake of not converting from grams to litres, so i've uploaded the updated solutions

[myopic_owl22]

Hi there,
The way I've been doing these questions is a little different to what's been mentioned. Click here if you'd like some more info.


Essentially, we have two equations:
1. a precipitation reaction
2. a dissociation reaction.
We use the first to determine how many Mg2+ ions will be in the solution (which'll be the same as the amount of Magnesium Acetate), and likewise, the amount of carbonate.
The 2nd equation determines our Q expression. As everything is in 1:1 ratios, this is straightforward.
Compare Q to Ksp, given on the formula sheet. If it's larger, there's too many ions (of either kind) to be all dissolved, and the solid will precipitate. In this case, since Q of 1.2*10^-7 is less than the Ksp of 6.82*10^-6, then the precipitate hasn't formed.

Manipulating volumes and concentrations (with c₁v₁ = c₂v₂) would be used for questions like "Will there be a precipitate if I add (x)ml of (a)molL^-1 Substance 1 to (y)ml of (b)molL^-1 Substance 2, if that makes sense. Since the stuff in your question is all in one litre, just use the concentrations given and multiply by their subscripts if required. I'm also not too sure if you can just convert grams into litres as easily as that... to the best of my knowledge, water is the only compound that can be reliably converted, I don't think the same logic same can be applied to aqueous solutions.

Hopefully this clears up things a fraction!

[avocadinq]

Hi there,
The way I've been doing these questions is a little different to what's been mentioned. Click here if you'd like some more info.
(Image removed from quote.)

Essentially, we have two equations:
1. a precipitation reaction
2. a dissociation reaction.
We use the first to determine how many Mg2+ ions will be in the solution (which'll be the same as the amount of Magnesium Acetate), and likewise, the amount of carbonate.
The 2nd equation determines our Q expression. As everything is in 1:1 ratios, this is straightforward.
Compare Q to Ksp, given on the formula sheet. If it's larger, there's too many ions (of either kind) to be all dissolved, and the solid will precipitate. In this case, since Q of 1.2*10^-7 is less than the Ksp of 6.82*10^-6, then the precipitate hasn't formed.

Manipulating volumes and concentrations (with c₁v₁ = c₂v₂) would be used for questions like "Will there be a precipitate if I add (x)ml of (a)molL^-1 Substance 1 to (y)ml of (b)molL^-1 Substance 2, if that makes sense. Since the stuff in your question is all in one litre, just use the concentrations given and multiply by their subscripts if required. I'm also not too sure if you can just convert grams into litres as easily as that... to the best of my knowledge, water is the only compound that can be reliably converted, I don't think the same logic same can be applied to aqueous solutions.

Hopefully this clears up things a fraction!

Hi myopic_owl22! Thanks for submitting your solution (easier to understand imo) and so sorry to everyone who was very confused by my answer - haven't covered these types of questions in class

[janeaustin]

Hello! The reason why we find the mass of each is because we use that mass to find the volume for v1 and when we add the volumes of both, for v2. To find v2, we simply just add the v1 values of (NH4)2CO3 and Mg(CH3COO)2. Sorry that wasn't clear, i made a typo in my response - i was supposed to write C2=C1V1/V2 hahaha. Also, just realised I made a mistake of not converting from grams to litres, so i've uploaded the updated solutions

Thank you for the clarification!

[janeaustin]

Hi there,
The way I've been doing these questions is a little different to what's been mentioned. Click here if you'd like some more info.
(Image removed from quote.)

Essentially, we have two equations:
1. a precipitation reaction
2. a dissociation reaction.
We use the first to determine how many Mg2+ ions will be in the solution (which'll be the same as the amount of Magnesium Acetate), and likewise, the amount of carbonate.
The 2nd equation determines our Q expression. As everything is in 1:1 ratios, this is straightforward.
Compare Q to Ksp, given on the formula sheet. If it's larger, there's too many ions (of either kind) to be all dissolved, and the solid will precipitate. In this case, since Q of 1.2*10^-7 is less than the Ksp of 6.82*10^-6, then the precipitate hasn't formed.

Manipulating volumes and concentrations (with c₁v₁ = c₂v₂) would be used for questions like "Will there be a precipitate if I add (x)ml of (a)molL^-1 Substance 1 to (y)ml of (b)molL^-1 Substance 2, if that makes sense. Since the stuff in your question is all in one litre, just use the concentrations given and multiply by their subscripts if required. I'm also not too sure if you can just convert grams into litres as easily as that... to the best of my knowledge, water is the only compound that can be reliably converted, I don't think the same logic same can be applied to aqueous solutions.

Hopefully this clears up things a fraction!

Thank you so much! Your explanation is very clear and simple, I understand now

Question though... what difference would it make if the litre of the solution weren't 1.0L?

[myopic_owl22]

Hi janeaustin,
Since we're working in concentrations (units of molL^-1) for ksp values etc, anything that isn't in one litre should be converted by c=n/v or c1v1=c2v2 depending on what information is given. The new concentrations that you'll get can be used just as normal, no need to worry about volumes after that!
Alternatively, you could solve concentration problems intuitively. If you had a concentrations in molL^-1 for 2 litres, the concentration would halve, as you'd have half the number of moles per litre.
And thanks for your kind words! If I'm not making sense, let me know!