
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
163.2k+ views
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.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
