Question

Using dilute sulphuric acid only how will you distinguish between:

Magnesium carbonate and magnesium sulphide.

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 carbonate $\left({\mathrm{MgCO}}_3\right)$	Magnesium sulphide $\left(\mathrm{MgS}\right)$
1	A colourless and odourless gas is observed with brisk effervescence.	A colourless gas which smells like rotten eggs is observed.
2	Magnesium carbonate reacts with dil. Sulphuric acid as shown below: $\underset{(\mathrm{Magnesium}\mathrm{carbonate})}{{\mathrm{MgCO}}_3\left(\mathrm{s}\right)}+\underset{(\mathrm{Dil}.\mathrm{Sulphuric}\mathrm{acid})}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\stackrel{\mathrm{Warm}}{\to }\underset{(\mathrm{Magnesium}\mathrm{sulphate})}{{\mathrm{MgSO}}_4\left(\mathrm{aq}\right)}+\underset{(\mathrm{Carbon}\mathrm{dioxide})}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\left(\mathrm{Water}\right)}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$	Magnesium sulphide reacts with dil. Sulphuric acid as shown below: $\underset{(\mathrm{Magnesium}\mathrm{sulphide})}{\mathrm{MgS}\left(\mathrm{s}\right)}+\underset{(\mathrm{Dil}.\mathrm{Sulphuric}\mathrm{acid})}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\stackrel{\mathrm{Warm}}{\to }\underset{(\mathrm{Magnesium}\mathrm{sulphate})}{{\mathrm{MgSO}}_4\left(\mathrm{aq}\right)}+\underset{(\mathrm{Hydrogen}\mathrm{sulphide})}{{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)}$
3	The gas evolved i.e.; Carbon dioxide $\left({\mathrm{CO}}_2\right)$ can be confirmed by passing it through limewater. It turns limewater milky. Carbon dioxide reacts with limewater to produce a white precipitate of Calcium carbonate (CaCO3) which makes the solution to look cloudy or milky in appearance. $\underset{(\mathrm{Carbon}\mathrm{dioxide})}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{(\mathrm{Calcium}\mathrm{hydroxide})}{\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)}\to \underset{(\mathrm{Calcium}\mathrm{carbonate})}{{\mathrm{CaCO}}_3\left(\mathrm{s}\right)}+\underset{\left(\mathrm{Water}\right)}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$	The gas evolved i.e.; Hydrogen sulphide $\left({\mathrm{H}}_2\mathrm{S}\right)$ can be confirmed by reacting moist lead acetate paper with it. The paper turns black on reacting with Hydrogen sulphide. $\underset{(\mathrm{Lead}\mathrm{acetate})}{{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\mathrm{Pb}\left(\mathrm{aq}\right)}+\underset{(\mathrm{Hydrogen}\mathrm{sulphide})}{{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)}\to \underset{\underset{\left(\mathrm{black}\right)}{(\mathrm{Lead}\mathrm{sulphide})}}{\mathrm{PbS}\left(\mathrm{s}\right)}+\underset{(\mathrm{Acetic}\mathrm{acid})}{2{\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)}$

Therefore, it can be concluded that if:

1	Colourless and odourless gas which turns limewater milky is observed	Magnesium carbonate
2	Rotten egg like smelling gas which turns moist lead acetate paper black is observed	Magnesium sulphide