Identify the anion present in the following compound:

When barium chloride solution is added to solution E, a white precipitate insoluble in dilute hydrochloric acid is obtained.

Explanation:

• When barium chloride $\left(BaC{l}_2\right)$ solution is added to Sodium sulphate $\left(N{a}_{2}S{O}_4\right)$, a white precipitate of Barium sulphate $\left(BaS{O}_4\right)$ is formed which is insoluble in all mineral acids.
• The chemical reaction is as follows.

$\underset{\mathrm{Barium}\mathrm{chloride}}{{\mathrm{BaCl}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Sodium}\mathrm{sulfate}}{{\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)}\to \underset{\mathrm{Barium}\mathrm{sulphate}}{{\mathrm{BaSO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Sodium}\mathrm{chloride}}{2\mathrm{NaCl}\left(\mathrm{s}\right)}$

Therefore, the anion present in the compound is a sulphate $\left({{\mathrm{SO}}_4}^{2-}\right)$ion.