Animated Solution for Chemistry - Ionic Equilibrium: Arrange the following solutions in the decreasing order of pOH. (A) $0.01\text{ M HCl}$ (B) $0.01\text{ M NaOH}$ (C) $0.01{\text{ M CH}}_3\text{COONa}$ (D) $0.01\text{ M NaCl}$

Arranging Solutions by Acidity and Basicity

When dealing with aqueous solutions, we often talk about pH to measure acidity. However, sometimes a problem flips the script and asks us to evaluate the pOH. Remember the golden rule at $25^{\circ }\text{C}$:

This means that a highly acidic solution (low pH) will have a very high pOH, while a highly basic solution (high pH) will have a very low pOH. Let's analyze our four $0.01\text{ M}$ solutions to arrange them in decreasing order of pOH.

Let's start with the extremes. Solution A is $0.01\text{ M HCl}$. Hydrochloric acid is a strong acid, meaning it dissociates completely in water.

The hydrogen ion concentration is $[{\text{H}}^{+}]=10^{-2}\text{ M}$. Taking the negative logarithm gives us a pH of $2$. But we need the pOH! Subtracting the pH from $14$, we get a pOH of $12$. This is our most acidic solution, so it will have the highest pOH.

On the other end of the spectrum, we have Solution B: $0.01\text{ M NaOH}$. Sodium hydroxide is a strong base, completely dissociating to give a hydroxide ion concentration of $[{\text{OH}}^{-}]=10^{-2}\text{ M}$.

The pOH is simply the negative logarithm of this concentration, which is exactly $2$. This highly basic solution will have the lowest pOH.

Now let's look at the salts. Solution D is $0.01\text{ M NaCl}$. Sodium chloride is formed from a strong acid ($\text{HCl}$) and a strong base ($\text{NaOH}$).

Because both parent components are strong, neither the ${\text{Na}}^{+}$ nor the ${\text{Cl}}^{-}$ ions undergo hydrolysis. The solution remains perfectly neutral, meaning both its pH and pOH are exactly $7$.

Finally, we have Solution C: $0.01{\text{ M CH}}_3\text{COONa}$. Sodium acetate is a salt of a weak acid (acetic acid) and a strong base ($\text{NaOH}$).

Here, the acetate ion undergoes anionic hydrolysis, reacting with water to produce hydroxide ions. This makes the solution basic, pushing its pH above $7$. Consequently, its pOH must be less than $7$. While it is basic, it is not as strongly basic as pure $\text{NaOH}$, so its pOH lies somewhere between $2$ and $7$.

We now have a clear picture of where each solution stands on the pOH scale:

A (HCl): $\text{pOH}=12$ D (NaCl): $\text{pOH}=7$ C (CH${}_3$COONa): $\text{pOH}<7$ B (NaOH): $\text{pOH}=2$

Arranging these in decreasing order, we get A > D > C > B. This logical breakdown of dissociation and hydrolysis makes comparing any set of solutions a breeze!