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JAMB Chemistry 2018 Paper

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Question : 37

Total: 87

The salt that reacts with dilute hydrochloric acid to produce a pungent smelling gas which decolourizes acidified purple potassium tetraoxomanganate (VII) solution is

Na $_2$ SO $_4$
Na $_2$ SO $_3$
Na $_2$ S
Na $_2$ CO $_3$

Solution:

Concept:

Sodium sulphite (

\mathrm{Na_2SO_3}

) with dilute hydrochloric acid liberates sulphur dioxide,

\mathrm{SO_2}

, which is pungent and decolourizes acidified

\mathrm{KMnO_4}

.

Explanation:

\mathrm{Na_2SO_3} + 2\mathrm{HCl} \rightarrow 2\mathrm{NaCl} + \mathrm{H_2O} + \mathrm{SO_2}

The

\mathrm{SO_2}

gas is a reducing agent and reduces purple

\mathrm{MnO_4^-}

ions to colourless

\mathrm{Mn^{2+}}

ions in acidic solution, hence the acidified purple

\mathrm{KMnO_4}

is decolourized.

Answer:

B.

\mathrm{Na_2SO_3}

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