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## 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.