(O level Chem) How to prepare salts

Given a salt, how do you decide what method to use? First of all, ask yourself whether it is soluble or insoluble.

• All Group I (sodium, potassium) compounds are soluble.
  
• All ammonium salts are soluble.
  
• All nitrates are soluble.
  
• Examples of insoluble salts: AgCl, PbCl₂, AgI, PbI₂, BaSO₄, CaSO₄.
  

Precipitation method: If it's insoluble, then the steps are simple. Mix two solutions in a beaker and the desired salt precipitates out. Then filter out the precipitate and dry it.

Worked Example: Describe, with the aid of balanced chemical and ionic equations, how you would prepare lead iodide.
[Solution]
Lead iodide PbI₂ is an insoluble salt. So, use the precipitation method.
Choose two solutions of soluble compounds that are readily available in the laboratory: lead(II) nitrate and potassium iodide (Remember: All nitrates are soluble. All potassium salts are soluble.)
Steps:

• Add colorless solution of lead(II) nitrate to colorless solution of potassium iodide in a beaker. A yellow precipitate of lead(II) iodide is formed.
  

Chemical equation: Pb(NO₃)₂(aq) + 2KI(aq) --> PbI₂(s) + 2KNO₃(aq)
Ionic equation: Pb²⁺(aq) + 2I^-(aq) --> PbI₂(s)

• Filter the suspension to obtain the residue of lead(II) iodide.
  
• Dry the lead(II) iodide salt under an infrared lamp.
  

Titration method: If it's a soluble salt of Group I metals or ammonium, use the titration method. The description of this method is pretty long. Basically, a titration between an acid and an alkali is performed using an indicator. This is to determine the volume of the acid or alkali needed to completely neutralize the other. The mixture is discarded and the step is repeated without the indicator. Now, you get the solution of the required salt. The salt is finally obtained from its solution by crystallization.

Worked Example: Describe how you would prepare rubidium sulfate.
[Solution]
Check the Periodic Table and you find that rubidium is in Group I. So Rb₂SO₄ is soluble. The acid part of the salt clearly comes from H₂SO₄ while the base part of the salt can come from rubidium hydroxide RbOH (similar to NaOH and KOH). The titration method should be used.
Steps:

• Place RbOH in a burette.
  
• Pipette 25.0 cm³ of H₂SO₄ into a conical flask.
  
• Add an indicator into the H₂SO₄.
  
• Run RbOH from the burette into the acid until the indicator just changes color.
  

2RbOH(aq) + H₂SO₄(aq) --> Rb₂SO₄(aq) + 2H₂O(l)

• Repeat the above steps without the indicator so that you obtain a colorless Rb₂SO₄ solution.
  
• Evaporate some of the water to obtain a saturated solution of Rb₂SO₄. Let the saturated solution to cool so that Rb₂SO₄ crystallizes out. Filter the crystals and dry them under an infrared lamp.
  

Now, salts of Group I metals and ammonium are not the only soluble salts. What if you are asked to prepare copper(II) sulfate or zinc chloride? You cannot simply suggest titrating Cu(OH)₂ or Zn(OH)₂ with H₂SO₄ or HCl because Cu(OH)₂ and Zn(OH)₂ are insoluble solids (precipitates) if you remember your Qualitative Tests for Metals. Here are other methods you can use.

• acid + insoluble base --> salt + water
  
• acid + insoluble carbonate --> salt + water + carbon dioxide
  
• acid + metal --> salt + hydrogen gas
  

All these three methods are similar in that the insoluble base / insoluble carbonate / metal are added in excess to the acid so as to ensure that all the acid has reacted. You then filter away the unreacted solid reagent so as to obtain the filtrate only. Finally, the salt is then obtained from the filtrate solution by crystallization.

• Most bases (e.g. CuO) are insoluble. Only oxides of Group I metals and some Group II metals are soluble.
  
• Most carbonates (e.g. CaCO₃, CuCO₃) are insoluble. Only ammonium carbonate, Group I carbonates are soluble.
  
• Copper does not react with dilute HCl and H₂SO₄. Do not suggest preparing CuSO₄ or CuCl₂ by reacting Cu with dilute acids.
  

Worked Example: Describe how you would prepare copper(II) sulfate.
[Solution]
The acid part of copper(II) sulfate is clearly from sulfuric acid. The base part can come from CuO or CuCO₃ but not from Cu because Cu does not react with dilute H₂SO₄.
Steps:

• Add excess CuO (a black solid powder) or CuCO₃ (a green solid powder) to a beaker of dilute H₂SO₄ and stir. The solid dissolves and the solution turns blue.
  

Either: CuO(s) + H₂SO₄(aq) --> CuSO₄(aq) + H₂O(l)
Or: CuCO₃(s) + H₂SO₄(aq) --> CuSO₄(aq) + H₂O(l) + CO₂(g)

• Filter the unreacted CuO or CuCO₃ to obtain the blue filtrate solution of CuSO₄.
  
• Evaporate some of the water to obtain a saturated solution of CuSO₄. Let the saturated solution to cool so that CuSO₄ crystallizes out. Filter the crystals and dry them under an infrared lamp.