1. Recall the relation between pH and pOH

At $25^\circ C$, $pH + pOH = 14$

So, to arrange in decreasing order of pOH, we can compare how acidic/basic each solution is.

2. Find pOH of each solution

(A) $0.01\,M\ HCl$

$HCl$ is a strong acid, so it dissociates completely: $[H^+] = 0.01 = 10^{-2}$ $pH = 2$ Hence, $pOH = 14 - 2 = 12$

(B) $0.01\,M\ NaOH$

$NaOH$ is a strong base, so it dissociates completely: $[OH^-] = 0.01 = 10^{-2}$ $pOH = 2$

(C) $0.01\,M\ CH_3COONa$

$CH_3COONa$ is a salt of weak acid ($CH_3COOH$) and strong base ($NaOH$), so its solution is basic due to hydrolysis.

For acetate ion: $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$

Since the solution is basic, $pH > 7 \Rightarrow pOH < 7$

More precisely, $K_b = \frac{K_w}{K_a}$ For acetic acid, $K_a \approx 1.8 \times 10^{-5}$, so $K_b = \frac{10^{-14}}{1.8\times 10^{-5}} \approx 5.6\times 10^{-10}$

For concentration $C=10^{-2}$, $[OH^-] \approx \sqrt{K_b C} = \sqrt{5.6\times10^{-10}\times10^{-2}}$ $= \sqrt{5.6\times10^{-12}} \approx 2.37\times10^{-6}$ So, $pOH \approx -\log(2.37\times10^{-6}) \approx 5.63$

(D) $0.01\,M\ NaCl$

$NaCl$ is a salt of strong acid and strong base, so its solution is neutral. $pH = 7$ $pOH = 7$

3. Compare pOH values

• (A) $pOH = 12$
• (D) $pOH = 7$
• (C) $pOH \approx 5.63$
• (B) $pOH = 2$

Thus, decreasing order of pOH is: $(A) > (D) > (C) > (B)$

4. Match with options

This corresponds to Option B.