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6-2.Equilibrium-II (Ionic Equilibrium)
hard
Why pure $NaCl$ is precipitated when $HCl$ gas is passed in a saturated solution of $NaCl$
A
Impurities dissolves in $HCl$
B
The value of $[N{a^ + }]$and $[C{l^ - }]$ becomes smaller than ${K_{sp}}$ of $NaCl$
C
The value of $[N{a^ + }]$and $[C{l^ - }]$ becomes greater than ${K_{sp}}$ of $NaCl$
D
$HCl$ dissolves in the water
Solution
(c) $NaC{l_{(s)}} \rightleftharpoons N{a^ + }_{({\rm{aq}})} + C{l^ – }_{({\rm{aq}})}$
$HCl \rightleftharpoons {H^ + } + C{l^ – }$
The increase in $[C{l^ – }]$ brings in an increase in $[N{a^ + }]\,\,$$[C{l^ – }]$ which will lead for backward reaction because ${K_{sp}}\,NaCl = [N{a^ + }]\,\,[C{l^ – }]$.
Standard 11
Chemistry
