# Selective precipitation, purity and yield of precipitate

• vickyvoo2
In summary: Note that for amorphous precipitates, the composition is not well defined, as the "precipitate" is actually a mixture of individual particles, each with slightly different composition, and the composition of the "precipitate" may be different from the composition of the solution in equilibrium with it. Also, the purity may be affected by kinetics of the precipitation reaction (which of course is not considered in the thermodynamic treatment).
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

(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.

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).

Borek said:
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.

vickyvoo2 said:
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.

## 1. What is selective precipitation?

Selective precipitation is a method used in chemistry to separate and purify a specific substance from a mixture. It involves adding a reactant that will only react with the desired substance and form a precipitate, which can then be filtered out.

## 2. How does selective precipitation work?

Selective precipitation works by taking advantage of the differing solubility of substances in a solvent. The reactant added will react with the desired substance to form a solid precipitate, while leaving the other substances in solution. The precipitate can then be separated through filtration.

## 3. What factors affect the purity of the precipitate?

The purity of the precipitate can be affected by several factors, including the choice of reactant and solvent, the temperature and pH of the solution, and the rate of precipitation. It is important to carefully control these factors to ensure a pure precipitate is obtained.

## 4. How is the purity of a precipitate determined?

The purity of a precipitate can be determined through various analytical techniques, such as melting point determination, chromatography, and spectroscopy. These methods can help identify any impurities present in the precipitate and determine its purity level.

## 5. What is the relationship between yield and purity of a precipitate?

The yield of a precipitate refers to the amount of desired substance obtained through the selective precipitation process. The purity of the precipitate is directly related to its yield, as a higher purity will result in a higher yield. However, it is important to balance purity and yield, as some impurities may also be precipitated along with the desired substance.

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