Arrange the following solutions in the decreasing order of ${\mathrm{P}}^{\mathrm{OH}}$:
a) 0 .0 1 M HCl                      b) 0 .0 1 M NaOH 
c) 0 .0 1 M CH₃ COONa       d) 0 .0 1 M NaCl 

$$\begin{gathered} 0.01\mathrm{M} \mathrm{HCl}; \left[{\mathrm{H}}^{+}\right]={10}^{-2}, {\mathrm{P}}^{\mathrm{H}}=-\log {10}^{-2}=2; {\mathrm{P}}^{\mathrm{OH}}=14-2=12 \\ 0.01\mathrm{M} \mathrm{NaOH}; \left[{\mathrm{OH}}^{-}\right]={10}^{-2}, {\mathrm{P}}^{\mathrm{OH}}=-\log \left[{\mathrm{OH}}^{-}\right]=2 \\ 0.01{\mathrm{MCH}}_3\mathrm{COONa} \\ {\mathrm{P}}^{\mathrm{H}}=7+\frac{1}{2}\left[{\mathrm{P}}^{\mathrm{Ka}}+\log 0.01\right] {\mathrm{P}}^{\mathrm{H}}>7\Rightarrow {\mathrm{P}}^{\mathrm{OH}}<7 \end{gathered}$$

$0.01\mathrm{M} \mathrm{NaCl}; {\mathrm{P}}^{\mathrm{H}}=7,{\mathrm{P}}^{\mathrm{OH}}=7$ Decreasing order of ${\mathrm{P}}^{\mathrm{OH}}$ value is $\mathrm{a}>\mathrm{d}>\mathrm{c}>\mathrm{b}$