(A) The acid dissociation constant $(K_a)$ values for the given acids are as follows:$A$. Phenol $(C_6H_5OH)$: $K_a \approx 1.0 \times 10^{-10}$ (Matc

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(A) The acid dissociation constant $(K_a)$ values for the given acids are as follows:$A$. Phenol $(C_6H_5OH)$: $K_a \approx 1.0 \times 10^{-10}$ (Match $III$)$B$. Benzoic acid $(C_6H_5COOH)$: $K_a \approx 6.5 \times 10^{-5}$ (Match $IV$)$C$. Hypochlorous acid $(HClO)$: $K_a \approx 3.0 \times 10^{-8}$ (Match $II$)$D$. Acetic acid $(CH_3COOH)$: $K_a \approx 1.75 \times 10^{-5}$ (Match $V$)Thus,the correct matching is $A-III, B-IV, C-II, D-V$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Acids and Bases

Medium

Match the following acids in List-$I$ with their respective acid dissociation constants $(K_a)$ in List-$II$:
List-$I$ (Acid) List-$II$ $(K_a)$
$A$. Phenol $I$. $1 \times 10^{-13}$
$B$. Benzoic acid $II$. $3.0 \times 10^{-8}$
$C$. $HClO$ $III$. $1.0 \times 10^{-10}$
$D$. $CH_3COOH$ $IV$. $6.5 \times 10^{-5}$
$V$. $1.75 \times 10^{-5}$

The correct match is:

List-$I$ (Acid)	List-$II$ $(K_a)$
$A$. Phenol	$I$. $1 \times 10^{-13}$
$B$. Benzoic acid	$II$. $3.0 \times 10^{-8}$
$C$. $HClO$	$III$. $1.0 \times 10^{-10}$
$D$. $CH_3COOH$	$IV$. $6.5 \times 10^{-5}$
	$V$. $1.75 \times 10^{-5}$