Question
The following equilibrium in established when hydrogen
chloride is dissolved in acetic acid.
$$HCl + C{H_3}COOH \rightleftharpoons C{l^ - } + C{H_3}COOH_2^ + $$
The set that characterises the conjugate acid-base pairs is
A.
$$\left( {HCl,C{H_3}COOH} \right){\text{and}}\left( {C{H_2}COOH_2^ + C{l^ - }} \right)$$
B.
$$\left( {HCl,C{H_3}COOH_2^ + } \right){\text{and}}\left( {C{H_3}COOH,C{l^ - }} \right)$$
C.
$$\left( {C{H_3}COOH_2^ + ,HCl} \right){\text{and}}\left( {C{l^ - },C{H_3}COOH} \right)$$
D.
$$\left( {HCl,C{l^ - }} \right)and\left( {C{H_3}COOH_2^ + ,C{H_3}COOH} \right)$$
Answer :
$$\left( {HCl,C{l^ - }} \right)and\left( {C{H_3}COOH_2^ + ,C{H_3}COOH} \right)$$
Solution :
Since $$HCl$$ is stronger than $$C{H_3}COOH$$ hence acts as
acid. On the other hand $$C{l^ - }$$ is a stronger base than $$C{H_3}COOH_2^ + $$ and is the conjugate base of $$HCI.$$
$$\eqalign{
& HCl + C{H_3}COOH \rightleftharpoons C{l^ - } + C{H_3}COOH_2^ + \cr
& {\text{aci}}{{\text{d}}_{\text{1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{bas}}{{\text{e}}_{\text{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{bas}}{{\text{e}}_{\text{1}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{aci}}{{\text{d}}_{\text{2}}} \cr} $$