[vickyvoo2]

1. The problem statement, all variables and given/known data

An aqueous effluent contains 12 M Cd²⁺ and 10 M Mg²⁺ as solution of nitrates. Current practice to clean up the effluent is to selectively precipitate the metals with sodium hydroxide for recovery (this is a two-step process in whichone ofthe metals is precipitated first, removed and then the second metal isprecipitated). One of your process chemists suggested using sodium fluoride as a selective precipitating agent claiming that one could reduce the cadmium impurities in the magnesium precipitate to zero.

(a) Is what your process chemist proposes feasible (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 fluoride as precipitating agent instead of hydroxide?

(c)If you consider switching the precipitating agent during theprocess (from hydroxide to fluoride or vice versa) which sequence should you choose?

2. Relevant 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

3. 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^-8/0.023² = 6.99*10^-8

Purity of CdF₂: [Cd²⁺]/[Cd²⁺]+[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