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JEE Main levelSingle correct optionDifficulty 2 of 3sb177

In the Solvay process, sodium hydrogencarbonate separates from the cold ammoniacal brine saturated with $CO X_{2}$ mainly because

Ait is completely insoluble in water at all temperatures
Bit is the only salt whose ions are present in the solution
Cit is the least soluble of the salts that can form in this solution under these conditions
Dit is formed as a gas that is trapped by brine

Answer: C

The ions

Na X^{+}

,

Cl X^{-}

,

NH X_{4} X^{+}

and

HCO X_{3} X^{-}

are all present. Of the four possible salts,

NaHCO X_{3}

has the lowest solubility in cold solution, so it crystallises while

NH X_{4} Cl

stays dissolved. It is not completely insoluble.

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