# 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

### 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.