# How would you separate sodium chloride and copper sulfate?

I've tried dissolving both compounds in water and evaporating the mixture, but this leaves a combination of brown and green crystals. Is there was any way to cleanly separate both compounds? Any advice would be much appreciated. :)

• They are not there anymore. You can't separate them without chemically transforming at least some of the ions involved. Typically, when you separate a mixture, you need just one of the components. Which one is it? – Ivan Neretin Apr 3 '18 at 6:17

Suppose you have a solution of sodium chloride $\ce{(NaCl)}$ and copper sulfate ($\ce{CuSO4}$), in a beaker/conical flask. First, you treat this mixture with excess potassium ferrocyanide $(\ce{K_4[Fe(CN)_6]})$. Due to this addition, a choclate brown precipitate of $\ce{Cu_2[Fe(CN)_6]}$ (cupric ferrocyanide) is formed while the $\ce{NaCl}$ remain unreacted.
$$\ce{CuSO_4 + K_4[Fe(CN)_6] -> Cu_2[Fe(CN)_6](ppt) + K_2SO_4}$$
Then you filter the solution, and you will have the precipitate separated from the solution, which now only contains $\ce{NaCl}$. Now take the cupric ferrocyanide and treat with excess $\ce{NH_4OH}$, and it will form a precipitate of cupric hydroxide $(\ce{Cu(OH)_2})$.
Again, filter the precipitate and treat the compound with a concentrated solution of sulfuric acid ($\ce{H_2SO_4}$). With this, you will get back original copper sulfate, and now, it is seperated from $\ce{NaCl}$.