Arrange the following solutions in the decreasing order of pOH:

(A) $0.01 \mathrm{MHCl}$

$\left[\mathrm{H}^{+}\right]=10^{-2}, \mathrm{pH}=-\log 10^{-2}=2$

$\mathrm{pOH}=14-2=12$

(B) $0.01 \mathrm{MNaOH}$

$\left[\mathrm{OH}^{-}\right]=10^{-2}, \mathrm{pOH}=-\log [\mathrm{OH}]=2$

(C) $0.01 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}$

$\mathrm{pH}=7+\frac{1}{2}\left[\mathrm{p} K_{\mathrm{a}}+\log 0.01\right]$

(D) $0.01 \mathrm{M} \mathrm{NaCl}, \mathrm{pH}=7, \mathrm{pOH}=7$

Decreasing order of $\mathrm{pOH}$ value is,

$(\mathrm{A})>(\mathrm{D})>(\mathrm{C})>(\mathrm{B})$.