# Purity of precipitate in selective precipitation

An aqueous effluent contains $$12\ \mathrm{mol\ l^{-1}}$$ $$\ce{Cd^2+}$$ and $$10\ \mathrm{mol\ l^{-1}}$$ $$\ce{Mg^2+}$$ as solution of nitrates. Current practice is use $$\ce{NaOH}$$ to selectively precipitate the metals.

(a) Is it feasible to get 0 $$\ce{Cd}$$ impurities in $$\ce{Mg}$$ precipitate by using $$\ce{NaF}$$ instead of $$\ce{NaOH}$$ (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 $$\ce{NaF}$$ instead of $$\ce{NaOH}$$?

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

Attempt at answer:

### Part (a)

Using $$\ce{NaF}$$:

• $$K_\mathrm{sp}(\ce{MgF2}) = 3.7\times10^{-8}$$
• $$K_\mathrm{sp}(\ce{CdF2}) = 6.44\times10^{-3}$$
• $$K_\mathrm{sp}(\ce{MgF2}) = [\ce{Mg^2+}][\ce{F-}]^2$$
• $$K_\mathrm{sp}(\ce{CdF2}) = [\ce{Cd^2+}][\ce{F-}]^2$$

As the $$K_\mathrm{sp}$$ of $$\ce{MgF2}$$ is smaller it is less soluble and will precipitate first. You need to add $$\ce{F-}$$ to the point where $$\ce{CdF2}$$ will just not start precipitating. The limiting concentration is when the ion product = $$K_\mathrm{sp}$$.

Limiting concentration for $$\ce{MgF2}$$: $$[\ce{F-}]=\sqrt\frac{K_\mathrm{sp}(\ce{MgF2})}{[\ce{Mg^2+}]} = \sqrt\frac{3.7\times10^{-8}}{10} = 6.08\times10^{-5}\ \mathrm{mol\ l^{-1}}$$

Limiting concentration for $$\ce{CdF2}$$: $$[\ce{F-}]=\sqrt\frac{K_\mathrm{sp}(\ce{CdF2})}{[\ce{Cd^2+}]} = \sqrt\frac{6.44\times10^{-3}}{12} = 0.023\ \mathrm{mol\ l^{-1}}$$

The limiting concentration of $$\ce{Mg^2+}$$ is lower so this will precipitate first. By adding $$\ce{F-}$$ up to the limiting concentration of $$\ce{Cd^2+}$$ no $$\ce{Cd}$$ will precipitate so $$\ce{MgF2}$$ precipitate will be pure.

### Part (b)

I assume you need to find the yield and purity for both methods (using $$\ce{NaF}$$ or $$\ce{NaOH}$$). To find the purity of $$\ce{CdF2}$$ need to know how much $$\ce{Mg^2+}$$ is left in solution at the point when $$\ce{CdF2}$$ starts precipitating.
$$[\ce{Mg^2+}]_\text{sol} = \frac{K_\mathrm{sp}(\ce{MgF2})}{[\ce{F-}]^2} = \frac{3.7\times10^{-8}}{0.023^2} = 6.99\times10^{-8}$$

Purity of $$\ce{CdF2}$$: $$\frac{[\ce{Cd}]}{[\ce{Cd}]+[\ce{Mg2+}]_\text{sol}} \times 100\ \% = 99.99\ \%$$
I assume the yield of $$\ce{CdF2}$$ would be $$100$$ as you can keep adding $$\ce{F-}$$ 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 $$\ce{CdF2}$$/original concentration. If I did this then:
$$\frac{10-6.99\times10^{-8}}{10} \times 100\ \% = 99.99\ \%$$

To compare with using $$\ce{NaOH}$$ as the precipitating agent:

• $$K_\mathrm{sp}(\ce{Mg(OH)2}) = 1.8\times10^{-11}$$
• $$K_\mathrm{sp}(\ce{Cd(OH)2}) = 2.5\times10^{-14}$$
• $$K_\mathrm{sp}(\ce{Mg(OH)2}) = [\ce{Mg^2+}][\ce{OH-}]^2$$
• $$K_\mathrm{sp}(\ce{Cd(OH)2}) = [\ce{Cd^2+}][\ce{OH-}]^2$$

As the $$K_\mathrm{sp}$$ of $$\ce{Cd(OH)2}$$ is smaller it is less soluble and will precipitate first.

Limiting concentration for $$\ce{Mg(OH)2}$$:
$$[\ce{F-}]=\sqrt\frac{K_\mathrm{sp}(\ce{Mg(OH)2})}{[\ce{Mg^2+}]} = \sqrt\frac{1.8\times10^{-11}}{10} = 1.43\times10^{-6}\ \mathrm{mol\ l^{-1}}$$

To find the purity of $$\ce{Mg(OH)2}$$ need to know how much $$\ce{Cd^2+}$$ is left in solution at the point when $$\ce{Mg(OH)2}$$ starts precipitating.
$$[\ce{Cd^2+}]_\text{sol} = \sqrt\frac{K_\mathrm{sp}(\ce{Cd(OH)2})}{[\ce{F-}]^2} = \frac{2.5\times10^{-14}}{(1.43\times10^{-6})^2} = 0.0138\ \mathrm{mol\ l^{-1}}$$

Purity of $$\ce{Mg}$$: $$\frac{10}{10+0.0138}\times100\ \% = 99.86\ \%$$

Yield of $$\ce{Cd}$$: $$\frac{12-0.0138}{12}\times100\ \% = 99.86\ \%$$

For part B the way the question is phrased I assume the purity or yield using $$\ce{NaF}$$ shouldn’t be as great as using $$\ce{NaOH}$$ but I don’t know if I have done my calculations right as from my results it would seem they’re both accurate methods.