Question

Identify the anion present in each of the following compounds.

(i) A salt M on treatment with concentrated Sulphuric acid produces a gas which fumes in moist air gives dense fumes with Ammonia.

(ii) A salt D on treatment with dilute Sulphuric acid produces a gas which turns lime water milky but has no effect on acidified Potassium dichromate solution

(iii) When Barium chloride is added to salt solution E, a white precipitate insoluble in dilute Hydrochloric acid is obtained.

Answer 1

Part 1: ${\mathrm{Cl}}^{-}$ ion. is present:-

1. A salt M on treatment with concentrated sulphuric acid produces a gas which fumes in moist air gives dense fumes with ammonia.
2. ${\mathrm{Cl}}^{-}$ is the anion present. When Sodium chloride $\left(\mathrm{NaCl}\right)$ is added to a concentrated Sulphuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ solution and the solution is heated, a colourless gas, $\mathrm{HCl}$, and ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is produced . $$\begin{gathered} 2\mathrm{NaCl}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)\to {\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{g}\right)+2\mathrm{HCl}\left(\mathrm{g}\right) \\ (\mathrm{Sodium}(\mathrm{sulphuric}(\mathrm{sodium}(\mathrm{hydrochloric}\mathrm{acid}) \\ \mathrm{chloride})\mathrm{acid})\mathrm{sulphate}) \end{gathered}$$ When an Ammonia-soaked glass rod is brought close to this vessel, copious white vapours result from the production of ${\mathrm{NH}}_4\mathrm{Cl}$.
3. This is a confirmation test for the presence of ${\mathrm{Cl}}^{-}$ ion.

Part 2:- Sodium carbonate is the salt:-

1. A salt D on treatment with dilute Sulphuric acid produces a gas that turns lime water milky but has no effect on acidified Potassium dichromate solution
2. When diluted Sulphuric acid is heated with Sodium carbonate, colourless, odourless gas is produced. ${\mathrm{CO}}_2$ is the gas in question. When ${\mathrm{CO}}_2$ is carried through lime water, Calcium carbonate is formed, which causes it to turn milky. Also, when this gas is pushed through a ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$solution, it has no effect. $\underset{\mathrm{Sodium}\mathrm{carbonate}}{{\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right)}+\underset{\mathrm{Sulphuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Sodium}\mathrm{sulphate}}{{\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{carbon}\mathrm{dioxide}}{{\mathrm{CO}}_2\left(\mathrm{g}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}$

Part 3: Sodium sulphtate is the salt solution:-

1. When Barium chloride ${\mathrm{BaCl}}_2$ is added to salt solution E, a white precipitate insoluble in dilute hydrochloric acid is obtained.
2. Sodium sulphate ${\mathrm{Na}}_2{\mathrm{SO}}_4$ is the salt solution in this case. Barium sulphate ${\mathrm{BaSO}}_4$ white precipitates occur when ${\mathrm{BaCl}}_2$ solution is introduced to ${\mathrm{Na}}_2{\mathrm{SO}}_4$. In dil. $\mathrm{HCl}$, this whitish precipitate is insoluble. $$\begin{gathered} {\mathrm{BaCl}}_2\left(\mathrm{aq}\right)+{\mathrm{Na}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{BaSO}}_4\left(\mathrm{s}\right)(\downarrow )+\mathrm{NaCl}\left(\mathrm{aq}\right) \\ (\mathrm{barium}(\mathrm{sodium}(\mathrm{barium}(\mathrm{sodium} \\ \mathrm{chloride})\mathrm{sulphate})\mathrm{sulphate})\mathrm{chloride}) \end{gathered}$$
3. The sulphate $\left({{\mathrm{SO}}_4}^{2-}\right)$ radical is identified using this procedure.

Therefore,

(i) A salt M is sodium chloride on treatment with concentrated Sulphuric acid produces a gas that fumes in moist air and gives dense fumes with Ammonia.

(ii) A salt D is sodium carbonate on treatment with dilute Sulphuric acid produces a gas that turns lime-water milky but has no effect on acidified Potassium dichromate solution.

(iii) When Barium chloride is added to salt solution E is sodium sulphate, a white precipitate insoluble in dilute Hydrochloric acid is obtained.