Selective precipitation, purity and yield of precipitate

[vickyvoo2]

Homework Statement

An aqueous effluent contains 12 M Cd2+ and 10 M Mg2+ as solution of nitrates. Current practice is use NaOH to selectively precipitate the metals.

(a) Is it feasible to get 0 Cd impurities in Mg ppt by using NaF instead of NaOH (back up your answer with adequate calculations)

(b) What trade-off with respect to yield and purity do you have to accept if you use NaF instead of NaOH?

(c) If consider switching the precipitating agent during the process which sequence should be used

Homework Equations

Ksp MgF2 = 3.7*10^-8
Ksp CdF2 = 6.44*10^-3

KspMgF2 = [Mg2+][F-]^2
KspCdF2 = [Cd2+][F-]^2

Ksp Mg(OH)2 = 1.8*10^-11
Ksp Cd(OH)2 = 2.5*10^-14

KspMg(OH)2 = [Mg2+][OH-]^2
KspCd(oh)2 = [Cd2+][OH-]^2

The Attempt at a Solution

Attempt at answer:
(a) Using NaF:
As the Ksp of MgF2 is smaller it is less soluble and will precipitate first. You need to add F- to the point where CdF2 will just not start precipitating. The limiting concentration is when the ion product=Ksp

Limiting concentration for MgF2:
[F-]=sqrt(KspMgF2/[Mg2+]) = sqrt(3.7*10^-8/10) = 6.08*10^-5 M

Limiting concentration for CdF2:
[F-]=sqrt(KspCdF2/[Cd2+]) = sqrt(6.44*10^-3/12) = 0.023 M

The limiting concentration of Mg2+ is lower so this will precipitate first. By adding F- up to the limiting concentration of Cd2+ no CdF2 will precipitate so MgF2 precipitate will be pure.

(b) I assume you need to find the yield and purity for both methods (using NaF or NaOH).
To find the purity of CdF2 need to know how much Md2+ is left in solution at the point when CdF2 starts precipitating.
[Mg2+]sol = KspMgF2/[F-]^2 = 3.7*10^-8/0.023^2 = 6.99*10^-8

Purity of CdF2: [Cd]/[Cd]+[Mg2+]sol *100 = 99.99%
I assume the yield of CdF2 would be 100 as you can keep adding F- until it's all precipitated out.

I'm not sure if to calculate the yield of Magnesium I need to do the original concentration - impurities in CdF2/original concentration. If I did this then:
10-6.99*10^-8/10 *100 = 99.99%

To compare with using NaOH as the precipitating agent:

As the Ksp of Cd(OH)2 is smaller it is less soluble and will precipitate first.

Limiting concentration for Mg(OH)2:
[F-]=sqrt(KspMg(OH)2/[Mg2+]) = sqrt(1.8*10^-11/10) = 1.43*10^-6 M

To find the purity of Mg(OH)2 need to know how much Cd2+ is left in solution at the point when Mg(OH)2 starts precipitating.
[Cd2+]sol = KspCd(OH)2/[F-]^2 = 2.5*10^-14/(1.43*10^-6 M )^2 = 0.0138 M

Purity of Mg: 10/10+0.0138 *100 = 99.86%
Yield of Cd: 12-0.0138/12 *100 = 99.86%

For part B the way the question is phrased I assume the purity or yield using NaF shouldn't be as great as using NaOH but I don't know if I have done my calculations right as from my results it would seem theyre both accurate methods.

For part C i assume I must've done something wrong in my previous calculations, as what I've calculated it would seem just using NaF is good enough.

 

[Borek]

At a quick glance I see nothing wrong with your approach - logic looks OK.

Sadly, this is a very bad question. First, it is poorly worded and it is not clear what it asks. Second, it calls for an impossible solution, I am not aware of any salts of Mg and Cd that could be used to produce solution that is 10 M in Mg and 12 M in Cd. 1 M and 1.2 M would be much more likely. Third, it wants you to assume purity of the precipitate is driven only by the Ksp, which is simply not true. Some traces of other ions will be always built into a precipitate, this is called coprecipitation and is especially prominent in the case of amorphous precipitates (like hydroxides).

 

[vickyvoo2]

At a quick glance I see nothing wrong with your approach - logic looks OK.

Sadly, this is a very bad question. First, it is poorly worded and it is not clear what it asks. Second, it calls for an impossible solution, I am not aware of any salts of Mg and Cd that could be used to produce solution that is 10 M in Mg and 12 M in Cd. 1 M and 1.2 M would be much more likely. Third, it wants you to assume purity of the precipitate is driven only by the Ksp, which is simply not true. Some traces of other ions will be always built into a precipitate, this is called coprecipitation and is especially prominent in the case of amorphous precipitates (like hydroxides).

Is there another way to calculate the purity of the precipitate not only using Ksp? I just think the results I'm getting don't fit what part b wants.

 

[Borek]

Is there another way to calculate the purity of the precipitate not only using Ksp?

None that I am aware of. This is rather something that has to be determined experimentally.