
Why is pure $NaCl$ precipitated when hydrochloric acid gas is passed in a saturated solution of sodium chloride?
A. Impurities dissolve in HCl
B. The value of \[[N{{a}^{+}}]~\] and \[[C{{l}^{-}}]~\] products becomes smaller than $Ksp$ of $NaCl$.
C. The value of \[[N{{a}^{+}}]~\] and \[[C{{l}^{-}}]~\] products becomes higher than $Ksp$ of $NaCl$.
D. HCl dissolves in water
Answer
163.2k+ views
Hint: A salt forms precipitate only when the product of the concentration of its cation and anion is greater than its solubility product. Thus sodium chloride will precipitate only when the concentration of the ion is greater than the solubility product.
Complete Step by Step Answer:
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:
$K_{sp}=[N{{a}^{+}}][C{{l}^{-}}]$
Hydrochloric acid is a strong acid. Thus 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.
Thus precipitate is formed when the value of \[[N{{a}^{+}}]~\] and \[[C{{l}^{-}}]~\] products becomes smaller than $Ksp$ of $NaCl$.
Thus the correct option is C.
Note: Precipitate is an insoluble solid substance that is formed from a solution. The process of forming precipitate is called precipitation. It is only formed when the concentration of the species in solution is higher in amount or the solubility of the species decreases due to common ion effect.
Complete Step by Step Answer:
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:
$K_{sp}=[N{{a}^{+}}][C{{l}^{-}}]$
Hydrochloric acid is a strong acid. Thus 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.
Thus precipitate is formed when the value of \[[N{{a}^{+}}]~\] and \[[C{{l}^{-}}]~\] products becomes smaller than $Ksp$ of $NaCl$.
Thus the correct option is C.
Note: Precipitate is an insoluble solid substance that is formed from a solution. The process of forming precipitate is called precipitation. It is only formed when the concentration of the species in solution is higher in amount or the solubility of the species decreases due to common ion effect.
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