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
Correct Answer: C
Solution :
\[NaC{{l}_{(\text{s})}}\] ⇌ \[Na_{(\text{aq})}^{+}+Cl_{(\text{aq})}^{-}\] \[HCl\]⇌ \[{{H}^{+}}+C{{l}^{-}}\]. 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 \[\ge {{K}_{sp}}\]
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