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(D) $I$. $0.01 \ M \ HCl$ is a strong acid: $[H^+] = 10^{-2} \ M$,so $pH = 2$ and $pOH = 14 - 2 = 12$.$II$. $0.01 \ M \ NaOH$ is a strong base: $[OH^-

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$A > D > C > B$

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(D) $I$. $0.01 \ M \ HCl$ is a strong acid: $[H^+] = 10^{-2} \ M$,so $pH = 2$ and $pOH = 14 - 2 = 12$.$II$. $0.01 \ M \ NaOH$ is a strong base: $[OH^-] = 10^{-2} \ M$,so $pOH = -\log(10^{-2}) = 2$.$III$. $0.01 \ M \ CH_3COONa$ is a salt of a weak acid and strong base,which undergoes anionic hydrolysis: $[OH^-] > 10^{-7} \ M$,so $pOH $IV$. $0.01 \ M \ NaCl$ is a salt of a strong acid and strong base,which is neutral: $[OH^-] = 10^{-7} \ M$,so $pOH = 7$.Comparing the $pOH$ values: $A (12) > D (7) > C ( B (2)$.Thus,the decreasing order is $A > D > C > B$.

List-$I$	List-$II$
$A$. $Aq.$ solution of $AlCl_3$	$I$. Basic
$B$. $Aq.$ solution of $CH_3COONa$	$II$. Acidic
$C$. $Aq.$ solution of $KCl$	$III$. Highly conductive
$D$. $Al_2O_3$	$IV$. Strongly basic
	$V$. Amphoteric

Arrange the following solutions in the decreasing order of $pOH$:
$A$. $0.01 \ M \ HCl$
$B$. $0.01 \ M \ NaOH$
$C$. $0.01 \ M \ CH_3COONa$
$D$. $0.01 \ M \ NaCl$

Similar Questions

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Match the following:
List-$I$ List-$II$
$A$. $Aq.$ solution of $AlCl_3$ $I$. Basic
$B$. $Aq.$ solution of $CH_3COONa$ $II$. Acidic
$C$. $Aq.$ solution of $KCl$ $III$. Highly conductive
$D$. $Al_2O_3$ $IV$. Strongly basic
$V$. Amphoteric

The correct match is:

If $0.1 \, M, 100 \, mL$ $NaOH$ solution is mixed with $0.05 \, M, 100 \, mL$ $H_2SO_4$ solution,the $pH$ of the resulting solution is:

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Arrange the following solutions in the decreasing order of $pOH$:$A$. $0.01 \ M \ HCl$$B$. $0.01 \ M \ NaOH$$C$. $0.01 \ M \ CH_3COONa$$D$. $0.01 \ M \ NaCl$