Question

The following equilibrium in established when hydrogen chloride is dissolved in acetic acid.
$HCl+C{H}_{3}COOH\rightleftharpoons C{l}^{-}+C{H}_{3}COO{H}_2^{+}$
The set that characterises the conjugate acid-base pairs is

• A. $(HCl,C{H}_{3}COOH)\text{and}(C{H}_{2}COO{H}_2^{+},C{l}^{-})$
• B. $(HCl,C{H}_{3}COO{H}_2^{+})\text{and}(C{H}_{3}COOH,C{l}^{-})$
• C. $(C{H}_{3}COO{H}_2^{+},HCl)\text{and}\left(C{l}^{-}C{H}_{3}COOH\right)$
• D. $(HCl,C{l}^{-})\text{and}(C{H}_{3}COO{H}_2^{+},C{H}_{3}COOH)$

Answer 1

The correct option is D $(HCl,C{l}^{-})\text{and}(C{H}_{3}COO{H}_2^{+},C{H}_{3}COOH)$

$\begin{array}{c}HCl+C{H}_{3}COOH\rightleftharpoons C{l}^{-}+C{H}_{3}COO{H}_2^{+}\\ \text{acid1}\text{base2}\text{base1}\text{acid2}\end{array}$

HCl is stronger acid than $C{H}_{3}COOH$ and hence releases proton. HCl and $C{l}^{-}$ differ by a proton. HCl is the acid and $C{l}^{-}$ is its conjugate base.
Since acetic acid accepts proton from HCl, it is a base and
$[C{H}_{3}COO{H}_2{]}^{+}$ is the conjugate acid.