Question

Sodium hydroxide reacts with $ S{O_2} $ in aqueous medium to give:
(A) $ NaHS{O_3} $
(B) $ N{a_2}{S_2}{O_3} $
(C) $ NaHS{O_4} $
(D) $ N{a_2}S{O_3} $

Answer 1

Hint: sodium sulfite is an inorganic compound. It can be formed by treating sodium hydroxide with sulfur dioxide in aqueous medium. It is a pale yellow, water-soluble solid commercially used as an antioxidant and preservative. It can also be prepared industrially by treating sulfur dioxide with a solution of sodium carbonate.

Complete step by step answer:
When a solution of sodium hydroxide ( $ NaOH $ ) is treated with sulfur dioxide ( $ S{O_2} $ ) in aqueous medium, $ N{a_2}S{O_3} $ first precipitates as a yellow solid. With more $ S{O_2} $ , the solid dissolves to give the disulfite, which crystallizes upon cooling.
Here is the balanced chemical reaction:
 $ S{O_2} + 2NaOH \to N{a_2}S{O_3} + {H_2}O $
Hence, option (D) is correct.
Though, if excess sulfur dioxide is supplied, then the sodium sulfite formed reacts with excess sulfur dioxide to form sodium metabisulfite ( $ N{a_2}{S_2}{O_5} $ ). The balanced chemical reaction is given below:
 $ N{a_2}S{O_3} + S{O_2} \to N{a_2}{S_2}{O_5} $
On the other hand, if water is also supplied (provided sulfur dioxide is in excess), sodium sulfite reacts with water and $ S{O_2} $ to give sodium hydrogen sulfite ( $ NaHS{O_3} $ ). The balanced chemical reaction is given below:
 $ N{a_2}S{O_3} + {H_2}O + S{O_2} \to 2NaHS{O_3} $
Therefore, option (D) is correct.

Note:
A heptahydrate of sodium sulfite is also known but is less useful because of its greater susceptibility towards oxidation by air. Sodium sulfite has reducing properties. Sodium sulfite exhibits bleaching, desulfurizing, and dechlorinating activities. This agent was used by the food industry to help maintain the fresh appearance of food products. It is also a component in many drugs, which helps to maintain their potency and stability. It is moderately toxic and dissolves slowly in water.