Question

Distinguish between the following pairs of compounds using the test given within brackets.

1. Dilute hydrochloric acid and dilute Sulphuric acid,(using lead nitrate solution).

Answer 1

Dilute hydrochloric acid and dilute Sulphuric acid can be distinguished as:

Part 1: Dilute sulphuric acid

• Sulphuric acid reacts with lead nitrate to give lead Sulphate and nitric acid as a product.
• The white Lead sulphate formed is insoluble in both cold and warm water.
• Balanced chemical equations:

$\underset{\mathrm{Lead}\mathrm{nitrate}}{\underbrace{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)}}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{\underbrace{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}}\to \underset{\mathrm{Lead}\mathrm{sulphate}}{\underbrace{{\mathrm{PbSO}}_4\left(\mathrm{s}\right)\downarrow }}+\underset{\mathrm{Nitric}\mathrm{acid}}{\underbrace{2{\mathrm{HNO}}_3\left(\mathrm{aq}\right)}}$

Part 2: Dilute Hydrochloric acid

• While lead nitrate reacts with dilute hydrochloric acid to give white ppt. of lead chloride and nitric acid as a product.
• The white ppt.(Lead chloride) formed is soluble in warm water but remains insoluble in cold water.
• Balanced chemical equation:

$\underset{\mathrm{Lead}\mathrm{nitrate}}{\underbrace{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{aq}\right)}}+\underset{\mathrm{Hydrochloric}\mathrm{acid}}{\underbrace{2\mathrm{HCl}\left(\mathrm{aq}\right)}}\to \underset{\mathrm{Lead}\mathrm{chloride}}{\underbrace{{\mathrm{PbCl}}_2\left(\mathrm{s}\right)\downarrow }}+\underset{\mathrm{Nitric}\mathrm{acid}}{\underbrace{2{\mathrm{HNO}}_3\left(\mathrm{aq}\right)}}$