Selective Precipitation — Overview & Examples

Way back, we learned about separating mixtures. We know that solutions are a type of mixture. But, it's challenging to remove solutes from solutions. We can vaporize the solvent. But, what if we had a solution with multiple compounds? If we boil off the solvent, we still have a mixture. So, how do we do it?

What if we could take advantage of different solubilities? We can. The method is called selective precipitation. It's the separation of ions by causing a precipitation reaction. For example, imagine a solution with silver and copper ions. We know that silver chloride precipitates. It's one of the exceptions to our solubility rules. Typically, halogen-metal compounds are soluble. Not silver! Yet, copper chloride is not an exception. It's soluble! So, what happens if we add hydrochloric acid? Silver and chlorine react to form solid silver chloride. We can precipitate the silver!

Selective Percipitation Calculations

We can do some unique calculations with selective precipitation. We'll look at two types. First, we'll calculate the concentration we need to start the precipitation. Then we'll compare the solubilities of two ions. We can see if we can separate them.

Problem 1. A solution contains silver ions. It's molarity is $3.5\times10^{-3}$ molar. Silver's solubility product constant is $1.8\times10^{-10}$. Let's say our total volume is 1000.0 milliliters. What mass of hydrochloric acid should we add to start the precipitation?

• First, we need to write a balanced chemical equation. It's a single replacement reaction.

• $Ag^{+} + HCl \rightleftharpoons AgCl_{(s)} + H^{+}$.

• We can write it as a net ionic equation. $Ag^{+} + Cl^{-}\rightleftharpoons AgCl_{(s)}$.

• Precipitation starts when we are at equilibrium. So, we can use the $K_{sp}$ to calculate our chloride ion concentration.

• $K_{sp}=[Ag^{+}][Cl^{-}]\rightarrow 1.8\times10^{-10}=[3.5\times10^{-3}][Cl^{-}]\rightarrow [Cl^{-}]=5.14\times10^{-8}\frac{mols}{L}$.

• Now, we can use dimensional analysis. First, we'll convert concentration to moles. Then we can convert moles to mass. The molar mass of hydrochloric acid is $36.45\frac{g}{mol}$.

• This problem uses stoichiometry. If you need a refresher, look at the stoichiometry roadmap.

• $5.14\times10^{-8}\frac{mols}{L}\times 1.0 L\times 36.45\frac{g}{mol}=1.87\times10^{-6} grams$

Problem 2. Let's say we have a solution with $4.0\times10^{-5}$ molar silver and $0.01$ molar mercury ions. Silver's $K_{sp}= 1.8\times 10^{-10}$. Mercury's is $K_{sp}= 2.6\times 10^{-15}$. Both mercury and silver chlorides are insoluble. What concentration of chloride is needed to start each precipitation reaction? Which will precipitate first?

• For silver chloride, $K_{sp}=[Ag^{+}][Cl^{-}]\rightarrow 1.8\times 10^{-10}=[4.0\times10^{-5}][Cl^{-}]\rightarrow [Cl^{-}]=4.5\times10^{-7}$

• For Mercury, $K_{sp}=[Hg^{2+}][Cl^{-}]^{2} \rightarrow [Cl^{-}]=\sqrt{\frac{K_{sp}}{[Hg^{2+}]}}=\sqrt{\frac{2.6\times 10^{-15}}{0.01}}=5.1\times10^{-7}$

• Less chlorine is required to precipitate silver compared to mercury. Yet, they are relatively close. From the scientific notation, we see that they have the same order of magnitude. So, we would expect silver to precipitate first. But, it would be challenging to use chlorine as a separation tool. Why did this happen? Silver chloride has a much larger $K_{sp}$. So we expect it to precipitate well before mercury chloride. The reason is the immense difference in the initial concentrations. The initial concentration of mercury is three orders of magnitude larger.

(Video) Selective Precipitation

by Professor Dave Explains

In this video, Professor Dave explains selective precipitation. You'll learn how to compare the solubility product constants. Why? So you can select what ions precipitate from a solution. It is a great separation technique!