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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 $A > D > C > B$.

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

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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$. $0.01 \ M \ NaCl$