Arrange the following solutions in the decreasing order of pOH:
(A) 0.01 M HCl
(B) 0.01 M NaOH
(C) 0.01 M $$\text{CH}_3\text{COONa}$$
(D) 0.01 M NaCl

First recall the definition of pOH: $$\text{pOH} = -\log[\text{OH}^-]$$. A larger numerical pOH means a smaller concentration of hydroxide ions.

We examine each 0.01 M solution one by one.

(A) 0.01 M HCl

HCl is a strong acid, so it dissociates completely: $$\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-$$

This gives $$[\text{H}^+] = 0.01\ \text{M} = 10^{-2}\ \text{M}$$.

Using the ionic product of water $$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14},$$ we write

$$[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1.0 \times 10^{-14}}{10^{-2}} = 1.0 \times 10^{-12}\ \text{M}.$$

Hence $$\text{pOH}_A = -\log(1.0 \times 10^{-12}) = 12.$$

(B) 0.01 M NaOH

NaOH is a strong base, so it dissociates completely: $$\text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-$$

This gives $$[\text{OH}^-] = 0.01\ \text{M} = 10^{-2}\ \text{M}.$$

Therefore $$\text{pOH}_B = -\log(10^{-2}) = 2.$$

(C) 0.01 M CH$$_3$$COONa

CH$$_3$$COONa is a salt of a weak acid (CH$$_3$$COOH) and a strong base (NaOH). Only the acetate ion undergoes hydrolysis:

$$\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$$

The base-hydrolysis constant is obtained from $$K_b = \frac{K_w}{K_a}.$$ Using $$K_w = 1.0 \times 10^{-14}$$ and $$K_a(\text{CH}_3\text{COOH}) = 1.8 \times 10^{-5},$$

$$K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}.$$

For a weak base of concentration $$C$$ the hydroxide-ion concentration is $$[\text{OH}^-] = \sqrt{K_b\,C}.$$ Putting $$C = 0.01\ \text{M} = 1.0 \times 10^{-2}\ \text{M},$$

$$[\text{OH}^-] = \sqrt{5.56 \times 10^{-10} \times 1.0 \times 10^{-2}} = \sqrt{5.56 \times 10^{-12}}.$$

Break the square root into mantissa and power:

$$\sqrt{5.56} \approx 2.36,\qquad \sqrt{10^{-12}} = 10^{-6}.$$

So $$[\text{OH}^-] = 2.36 \times 10^{-6}\ \text{M}.$$

Now

$$\text{pOH}_C = -\log(2.36 \times 10^{-6}) = -( \log 2.36 + \log 10^{-6} ) = -(0.373 - 6) = 5.627 \approx 5.63.$$

(D) 0.01 M NaCl

NaCl is the salt of a strong acid and a strong base; its aqueous solution is neutral. Thus

$$[\text{OH}^-] = 1.0 \times 10^{-7}\ \text{M}, \qquad \text{pOH}_D = 7.$$

We now list the four pOH values:

$$(A): 12, \quad (D): 7, \quad (C): 5.63, \quad (B): 2.$$

Since we must arrange them in decreasing order (largest pOH first), we get

$$(A) \gt (D) \gt (C) \gt (B).$$

Hence, the correct answer is Option B.