hard (?) solubility Qs

[cloak25]

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So given in the passage is a table of different generic salts, their Ksp and molar solubility: http://i.imgur.com/SM3g7.jpg

1) To preferentially precipitate Z2+(aq) from a solution with Q+ (aq), it would be best to add:

A. X- (aq)
B. Y- (aq)
C. either X-(aq) or Y-(aq), as both would work equally well.
D. neither X-(aq) or Y-(aq), because neither would work.

2) To dissolve ZX2(s) into solution so that the greatest amount of Z2+(aq) is present in solution, what salt can be added?

A. LX(s)
B. QX(s)
C. MY(s)
D. QY(s)

Answers: A, C respectively.

Imagine you have 1 minute, 2 minutes max to do each question. What would be your approach to answering each question? Solubility always gets me, but I feel as if the second question is tougher..

thanks!

 

[Night Hawk]

So given in the passage is a table of different generic salts, their Ksp and molar solubility: http://i.imgur.com/SM3g7.jpg

1) To preferentially precipitate Z2+(aq) from a solution with Q+ (aq), it would be best to add:

A. X- (aq)
B. Y- (aq)
C. either X-(aq) or Y-(aq), as both would work equally well.
D. neither X-(aq) or Y-(aq), because neither would work.

2) To dissolve ZX2(s) into solution so that the greatest amount of Z2+(aq) is present in solution, what salt can be added?

A. LX(s)
B. QX(s)
C. MY(s)
D. QY(s)

Answers: A, C respectively.

Imagine you have 1 minute, 2 minutes max to do each question. What would be your approach to answering each question? Solubility always gets me, but I feel as if the second question is tougher..

thanks!

1) With X- you end up with a max [Z2+] of 1.0 x 10^-4 M and with Y- you end up with a max [Z2+] of 1.0 x 10^-3 M, most mcat prep books suggest that any salt with a concentration less than 0.01M is insoluble, so I would say "C."

2) This is a common ion effect question. You can immediately eliminate LX and QX because they're more soluble than ZX2. Out of MY and QY, I would add MY because MX is less soluble than ZX2 and as such it will steal the X- ions in solution, shifting the equilibrium of ZX2 and more Z2+ will be liberated.

These are very doable in the alotted time but you really need to understand Ksp's in and out.

 

[cloak25]

Thanks for your response. For #1 the explanation says that the molar solubility of ZX2 is less than the molar solubility of QX by a factor of 100 and by a factor of 2 between QY and ZY2 so addition of Y- will work but just not as well as X- because adding Y- will not result in pure separation (molar solubility values are too close to one another). So C is true but A is the "better" answer I guess. thanks again.

 

[cloak25]

Hmm. I still don't really get #2. Isn't MY more soluble than ZX2 too? so if MX was a choice, MY would still be correct because although MX is less soluble than ZX2, adding MX will cause less dissociation of ZX2(s) due to the common effect..?

 

[Night Hawk]

Thanks for your response. For #1 the explanation says that the molar solubility of ZX2 is less than the molar solubility of QX by a factor of 100 and by a factor of 2 between QY and ZY2 so addition of Y- will work but just not as well as X- because adding Y- will not result in pure separation (molar solubility values are too close to one another). So C is true but A is the "better" answer I guess. thanks again.

Fair enough, didn't see that you put the answers lol. I don't think I've ever encountered a question like that, what prep are you using?

 

[Night Hawk]

Hmm. I still don't really get #2. Isn't MY more soluble than ZX2 too?

Yes it is more soluble, for MY you have Ksp = x^2, resulting in a molar solubility of 3.0 x 10^-3, while ZX2 is Ksp = 4x^3 and has a molar solubility of 1.0 x 10^-4

so if MX was a choice, MY would still be correct because although MX is less soluble than ZX2, adding MX will cause less dissociation of ZX2(s) due to the common effect..?

You got it.

MX is less soluble than ZX2 and you aren't putting any more X- into the solution. Lets look at this as two seperate problems. First, we have [Z2+] and we can assume that it's concentration in solution is dependant on [X-] so if we remove more X-, we will get more Zn2+ in solution. Second, which option allow up to remove X- from solution (which salt is less soluble than ZnX2)? The only choice that fits is MY.

 

[sotto voce]

So given in the passage is a table of different generic salts, their Ksp and molar solubility: http://i.imgur.com/SM3g7.jpg

1) To preferentially precipitate Z2+(aq) from a solution with Q+ (aq), it would be best to add:

A. X- (aq)
B. Y- (aq)
C. either X-(aq) or Y-(aq), as both would work equally well.
D. neither X-(aq) or Y-(aq), because neither would work.

2) To dissolve ZX2(s) into solution so that the greatest amount of Z2+(aq) is present in solution, what salt can be added?

A. LX(s)
B. QX(s)
C. MY(s)
D. QY(s)

Answers: A, C respectively.

Imagine you have 1 minute, 2 minutes max to do each question. What would be your approach to answering each question? Solubility always gets me, but I feel as if the second question is tougher..

thanks!

Hi, maybe I can make things clearer.

1. -The first thing is to calculate molar solubility from Ksp and vice versa (takes a few seconds and it will save you a headache later).

-We're trying to separate Z2+ and Q+. For X- and then Y-, think about the two molecules that will form. We're looking for the two with the biggest difference in solubility.

-ZX2 vs QX ==> 1E-4 vs 1E-2. We see that there is a difference of 1000 between these two numbers (and the less soluble one has Z2+, which we want).

-ZY2 vs QY ==> 1E-3 vs 2E-3. We see that there is a difference of 2 between these two numbers (not as good as if you add X). So X- is better to add (answer A), because you obtain a larger solubility gap.

2. -Yes, this one's tricky. 🙂

-Start with the basics. What's the objective? The objective is to get the most Z2+ in solution. So, we need to form the compound, that contains Z2+, that is the most soluble.

-Look at the chart (and at your paper, on which you've written the missing molar solubilities). Immediately notice that ZY2 is the most soluble. So that's the compound we need to form. Now we know the answer is C or D, as those two contain Y-.

-But is it C or D? It's C, MY, because MY is more soluble. There's another way to do this problem, which would focus on removing as much X- from the solution. However, notice that if you use my (our) way, ZY2 forms (and subsequently dissociates) and you're left with the remaining two elements, which form MX. MX has the absolute lowest solubility of all the compounds in the entire table. So either way, you want to add MY.

-If the answers are not A and C, let me know and I'll see if I made an error somewhere or just don't know what I'm talking about lol. Good luck. 🙂

 

[cloak25]

The answers are A and C. Thanks for the responses!