Qualitative analysis of Barium, Strontium, Calcium

[chound]

During qualitative analysis of Barium, Strontium, Calcium, why do we use NH4CO3, NH4OH, NH4Cl. Even if we were to apply common ion effect, two NH4+ ions are enough right? What is the need for third one?

 

[chemisttree]

I assume that you are referring to the ammonium chloride?

It might be there to completely describe what you might expect to find in your test tube after adding NH4OH or NH4CO3 [(NH4)2CO3? or (NH4)HCO3?] to your unknown which is probably a soluble chloride salt.

eg.

Sr(Cl)2 + 2NH4OH -----> Sr(OH)2(s) + 2 NH4Cl

Ca(Cl)2 + (NH4)2CO3 -----> CaCO3(s) + 2 NH4Cl

 

[ravenprp]

During qualitative analysis of Barium, Strontium, and Calcium, we use NH4CO3, NH4OH, and NH4Cl because they act as precipitating agents for these ions. This means that they help to form insoluble salts with these ions, allowing us to identify their presence in a solution.

The use of NH4CO3, NH4OH, and NH4Cl is necessary because each of these ions has different solubility properties. For example, Barium forms an insoluble carbonate salt with NH4CO3, while Strontium forms an insoluble hydroxide salt with NH4OH. Calcium, on the other hand, forms an insoluble chloride salt with NH4Cl.

The common ion effect is not the only consideration when choosing these precipitating agents. The solubility of the resulting salt also plays a crucial role. In some cases, the common ion effect may not be strong enough to precipitate all of the ions present. Therefore, the use of a third precipitating agent is necessary to ensure that all of the ions are effectively removed from the solution.

In conclusion, the use of NH4CO3, NH4OH, and NH4Cl in qualitative analysis of Barium, Strontium, and Calcium is essential for their ability to form insoluble salts with these ions and to ensure complete precipitation. The common ion effect is just one factor to consider, and the solubility properties of the resulting salts also play a crucial role in the choice of these precipitating agents.