Question

(a) Sodium hydroxide solution can be used to distinguish between (i) Iron (II) sulphate solution and (ii) iron (III) sulphate solution; because these solutions give different coloured precipitates with sodium hydroxide solution. Give the colour of the precipitate formed with each of the solution.
(b) What will you observe when barium chloride solution is added to iron (II) sulphate solution?
(c) How will the action of dilute hydrochloric acid on sodium carbonate and sodium sulphite enables you to distinguish between these two compounds?

Answer 1

(a)

Action of sodium hydroxide solution
Iron (II) chloride	Precipitate formed	Iron (III) chloride	Precipitate formed
Iron (II) chloride (pale green colour) solution ${\mathrm{FeCl}}_2+2\mathrm{NaOH}\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_2+2\mathrm{NaCl}$	Dirty green precipitate of ferrous hydroxide is formed	Iron (III) chloride (yellow colour) solution ${\mathrm{FeCl}}_3+3\mathrm{NaOH}\to \mathrm{Fe}{\left(\mathrm{OH}\right)}_3+3\mathrm{NaCl}$	Reddish brown precipitate of ferric hydroxide is formed

(b) When colourless barium chloride is added to pale green iron (II) sulphate solution, barium, displaces iron from iron sulphate and thus forms a white precipitate of barium sulphate and pale green solution of iron chloride.

${\mathrm{BaCl}}_2+{\mathrm{FeSO}}_4\underset{}{\to }{\mathrm{FeCl}}_2+{\mathrm{BaSO}}_4$

(c) Action of dilute hydrochloric acid helps in distinguishing between sodium carbonate and sodium sulphite.

Action of diluted hydrochloric acid
Sodium carbonate	Sodium sulphite
When sodium carbonate is warmed with diluted HCl, it liberates carbon dioxide gas, which turns the lime water milky. $$\begin{gathered} {\mathrm{Na}}_2{\mathrm{CO}}_3+2\mathrm{HCl}\stackrel{}{\to }2\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2 \\ {\mathrm{CO}}_2+\mathrm{Ca}{\left(\mathrm{OH}\right)}_2\to {\mathrm{CaCO}}_3+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$	When sodium sulphite is warmed with diluted HCl, it liberates sulphur dioxide gas, which turns dichromate solution (orange colour) green. $$\begin{gathered} {\mathrm{Na}}_2{\mathrm{SO}}_3+2\mathrm{HCl}\stackrel{}{\to }2\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_2 \\ {\mathrm{SO}}_2+{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$