Question

Distinguish between the following pair of compounds using the test given with in brackets: Sodium nitrate and sodium sulphite (using dilute sulphuric acid)

Answer 1

Hint: In this question we have to differentiate between the sodium nitrate, and sodium sulphite compounds using the dilute sulphuric acid. Write down the chemical reactions for both cases. In both the cases there will be evolution of gases as the by-products.

Complete step by step answer:
Sodium nitrate which is a chemical compound with the formula \[NaN{{O}_{3}}\] . It is a white solid and highly water soluble. Sodium nitrate combines with sulphuric acid to give nitric acid.
Let us see the chemical reaction in which Sodium nitrate react with dilute sulphuric acid
\[NaNO_{3}^{{}}+dil.{{H}_{2}}S{{O}_{4}}\xrightarrow{>{{200}^{0}}C}N{{a}_{2}}S{{O}_{4}}+HN{{O}_{3}}\]
When sodium nitrate reacts with dilute sulphuric acid it will liberate colourless vapors of nitric acid which then condenses to form nitric acid.
Whereas, Sodium sulphite when reacts with dilute sulphuric acid it will give colourless gas with smell of burning Sulphur which is acidic in nature and that means sulphur dioxide \[\left( S{{O}_{2}} \right)\]
Now the reaction between sodium sulphite and dilute sulphuric acid occurs in two steps which is also called double replacement reaction.
Now let us see the first step of double replacement reaction:
In first step, sodium sulfite reacts with dilute sulphuric acid to give sodium sulphate and sulphuric acid:
\[N{{a}_{2}}S{{O}_{3}}+dil.{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}S{{O}_{3}}\]
And chemical reaction for the second step:
\[{{H}_{_{2}}}S{{O}_{3}}\to S{{O}_{3}}+{{H}_{2}}O\]
Therefore, overall reaction written as,
\[N{{a}_{2}}S{{O}_{3}}+dil{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O+S{{O}_{2}}\]
Sodium sulphite is an inorganic compound with formula \[N{{a}_{2}}S{{O}_{3}}\]. It is pale yellow in colour and water soluble solid.
It is widely used as an antioxidant and preservatives.

Note: As we can see sodium sulfate as a final product in each reaction but do not confuse here because in the second reaction it occurs in two steps which is double replacement reaction (It occurs when parts of two ionic compounds are exchanged to making new compounds) where sulfurous acid decomposes spontaneously because of its high instability and give sodium sulfate, water, and sulfur dioxide as the final product.