HSC Stuff > HSC Chemistry
Predicting Formation of Precipitate
avocadinq:
--- Quote from: myopic_owl22 on January 18, 2019, 01:53:34 pm ---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! :)
--- End quote ---
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:
--- Quote from: avocadinq on January 16, 2019, 11:14:01 pm ---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 :)
--- End quote ---
Thank you for the clarification!
janeaustin:
--- Quote from: myopic_owl22 on January 18, 2019, 01:53:34 pm ---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! :)
--- End quote ---
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!
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