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JEE Main levelSingle correct optionDifficulty 2 of 3sb143

In the Solvay process $NaHCO X_{3}$ crystallises from the cold ammoniacal brine, rather than $NH X_{4} HCO X_{3}$ , $NaCl$ or $NH X_{4} Cl$ , because:

Ait is the only salt formed in the reaction
Bit has the highest lattice enthalpy of all four salts
Cit is the only one that is not hydrated
Dit has the lowest solubility of the four possible salts under these conditions

Answer: D

The solution contains

Na X^{+}

,

NH X_{4} X^{+}

,

Cl X^{-}

and

HCO X_{3} X^{-}

, so four salts are possible in principle. Of these,

NaHCO X_{3}

is the least soluble in the cold, and so it precipitates first and is filtered off.

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