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Question: A white precipitate insoluble in \[HN{{O}_{3}}\] is formed when the aqueous solution of x in NaOH is...

A white precipitate insoluble in HNO3HN{{O}_{3}} is formed when the aqueous solution of x in NaOH is treated with barium chloride and bromine water. So x is :
A. SO3S{{O}_{3}}
B. SO2S{{O}_{2}}
C. CO2C{{O}_{2}}
D. None of the above

Explanation

Solution

It is a colourless, pungent smelling gas and is highly soluble in water. It is used to make sulphurous acid (H2SO3{{H}_{2}}S{{O}_{3}}), sulphuric acid(H2SO4{{H}_{2}}S{{O}_{4}}). It behaves as a reducing agent when moist.

Complete-step- by- step answer:
When the aqueous solution of SO2S{{O}_{2}} ​in NaOH ​NaOH ​when treated with barium chloride and bromine water, a white precipitate is formed which is insoluble in conc.HNO3HN{{O}_{3}}.
We can see in the reaction that SO2S{{O}_{2}} readily reacts with sodium hydroxide solution to form sodium sulfite.
SO2+2NaOHNa2SO3+H2OS{{O}_{2}}+2NaOH\to N{{a}_{2}}S{{O}_{3}}+{{H}_{2}}O

The sodium sulphite is then treated with barium chloride to form barium sulfite BaSO3BaS{{O}_{3}}
Na2SO3+BaCl2BaSO3+2NaClN{{a}_{2}}S{{O}_{3}}+BaC{{l}_{2}}\to BaS{{O}_{3}}+2NaCl

Now the barium sulfite is lastly treated with bromine water . the bromine water contains nascent oxygen [O]\left[ O \right]
Br2+H2O2HBr+[O]B{{r}_{2}}+{{H}_{2}}O\to 2HBr+\left[ O \right]
This nascent oxygen oxidises the barium sulfite to give a white precipitate which is insoluble in conc.HNO3HN{{O}_{3}} it is barium sulphate BaSO4BaS{{O}_{4}}.
BaSO3+[O]BaSO4BaS{{O}_{3}}+\left[ O \right]\to BaS{{O}_{4}}
Sulphur (S) is a p- block element of group 16. Its main oxides are SO2S{{O}_{2}} and SO3S{{O}_{3}}.
Sulphur dioxide is a colourless gas, we can easily identify it due its pungent smell. It has an angular shape and the bond length is the same. It forms a resonance hybrid of the two canonical forms.


Sulphur dioxide has many different uses such as refining petroleum and sugar, bleaching wool and silk and as an anti – chloral, preservative and disinfectant.
Therefore we can conclude that the answer is option B. SO2S{{O}_{2}}.

Note: -In the question we are given an important reaction of sulphur dioxide with NaOHNaOH, its reaction with barium chloride and bromine water, it is also used to test the presence ofSO2S{{O}_{2}}.
Keep in mind the basic properties of a gas to determine its presence as for example sulphur dioxide is a colourless and pungent smelling gas.$$$$