
Pure $NaCl$ is prepared by saturating a cold saturated solution of common salt in water with $HCl$ gas. The principle used is
A. Le Chatelier's principle
B. Displacement law
C. Common ion effect
D. Fractional distillation
Answer
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Hint: Pure $NaCl$ is precipitated by saturating a cold saturated solution of common salt in water with $HCl$ gas due to presence of common chloride ion in both sodium chloride and hydrochloric acid. This is called the common ion effect.
Complete Step by Step Answer:
According to the common ion effect, the presence of the same ion in a solution of a salt decreases the solubility of the salt and forms precipitate easily.
Sodium chloride is a salt of hydrochloric acid and sodium hydroxide. Thus it dissociates in sodium ions and chloride ions. The dissociation is given as follows:
$NaCl\rightleftharpoons N{{a}^{+}}+C{{l}^{-}}$
Thus solubility product is given as follows:
$Ksp=[N{{a}^{+}}][C{{l}^{-}}]$
Thus when hydrochloric acid is added it dissociates in hydrogen ions and chloride ions. The dissociation is given as follows:
$HCl\rightleftharpoons {{H}^{+}}+C{{l}^{-}}$
Thus sodium chloride and hydrochloric acid both form chloride ions. Thus due to the common ion effect the solubility product of the sodium chloride decreases and ultimately the concentration of the corresponding ions increases. So, the sodium chloride is precipitated and obtained easily.
Thus the correct option is C.
Note: The solubility product is the product of the concentration of the cation and anion formed from a species. Due to common ion effect the value of the solubility product decreases and helps to form precipitate.
Complete Step by Step Answer:
According to the common ion effect, the presence of the same ion in a solution of a salt decreases the solubility of the salt and forms precipitate easily.
Sodium chloride is a salt of hydrochloric acid and sodium hydroxide. Thus it dissociates in sodium ions and chloride ions. The dissociation is given as follows:
$NaCl\rightleftharpoons N{{a}^{+}}+C{{l}^{-}}$
Thus solubility product is given as follows:
$Ksp=[N{{a}^{+}}][C{{l}^{-}}]$
Thus when hydrochloric acid is added it dissociates in hydrogen ions and chloride ions. The dissociation is given as follows:
$HCl\rightleftharpoons {{H}^{+}}+C{{l}^{-}}$
Thus sodium chloride and hydrochloric acid both form chloride ions. Thus due to the common ion effect the solubility product of the sodium chloride decreases and ultimately the concentration of the corresponding ions increases. So, the sodium chloride is precipitated and obtained easily.
Thus the correct option is C.
Note: The solubility product is the product of the concentration of the cation and anion formed from a species. Due to common ion effect the value of the solubility product decreases and helps to form precipitate.
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