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6-2.Equilibrium-II (Ionic Equilibrium)
medium
At equilibrium, if to a saturated solution of $NaCl,\,HCl$ is passed, $NaCl$ gets precipitated because
A
$HCl$ is a strong acid
B
Solubility of $NaCl$ decreases
C
Ionic product of $NaCl$ becomes greater than its ${K_{sp}}$
D
$HCl$ is a weak acid
Solution
(c) $NaC{l_{({\rm{s}})}} ⇌ Na_{({\rm{aq}})}^ + + Cl_{({\rm{aq}})}^ – $
$HCl⇌ {H^ + } + C{l^ – }$. The increase in $[C{l^ – }]$brings in an increase in $[N{a^ + }]$$[C{l^ – }]$which will lead for backward reaction because
${K_{sp}}(NaCl) = [N{a^ + }]\,\,[C{l^ – }]$
means Ionic product $ \ge {K_{sp}}$
Standard 11
Chemistry
