Question

Using dilute sulphuric acid only how will you distinguish between:

Magnesium metal and lead metal.

Answer 1

The use of dilute Sulphuric acid $(\mathrm{dil}.{\mathrm{H}}_2{\mathrm{SO}}_4)$ helps to do the qualitative analysis of the given sample. This is known as dil. Sulphuric acid test.

• Qualitative analysis: The analysis which aims at detecting the presence of any substance in a given sample.
• To perform this test, a sample of the given substance is taken in a test-tube and a few drops of dilute Sulphuric acid is added and then warmed gently. The observations are then recorded.

S.No	Magnesium $\left(\mathrm{Mg}\right)$	Lead $\left(\mathrm{Pb}\right)$
1	Magnesium reacts with dilute Sulphuric acid and a colourless and odourless gas is realeased. When a burning match stick is brought closer, it burns with a a pop sound which is a confirmatory test for Hydrogen $\left({\mathrm{H}}_2\right)$ gas. $\underset{\left(\mathrm{Magnesium}\right)}{\mathrm{Mg}\left(\mathrm{s}\right)}+\underset{(\mathrm{Dil}.\mathrm{Sulphuric}\mathrm{acid})}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\stackrel{\mathrm{Warm}}{\to }\underset{(\mathrm{Magnesium}\mathrm{sulphate})}{{\mathrm{MgSO}}_4\left(\mathrm{aq}\right)}+\underset{\left(\mathrm{Hydrogen}\right)}{{\mathrm{H}}_2\left(\mathrm{g}\right)}$	Lead does not evolve Hydrogen gas on reaction with dilute Sulphuric acid because Lead sulphate $\left({\mathrm{PbSO}}_4\right)$ initially formed is insoluble and thus prevents further reaction due to which Hydrogen gas is not evolved.

Therefore, it can be concluded that if:

1	Hydrogen gas is liberated	Magnesium $\left(\mathrm{Mg}\right)$
2	Hydrogen gas is not liberated	Lead $\left(\mathrm{Pb}\right)$