Part (e): Which of these solutes would you use to prepare a solution with a pH of 5.50 by dissolving the appropriate amount of a solute in water with pH=7.00?

Ammonia ($NH_3$) is a weak base. Dissolving it in water would increase the pH above 7.00, making the solution basic. This is contrary to the goal of achieving a pH of 5.50. Therefore, option (a) is eliminated.

Hydrochloric acid ($HCl$) is a strong acid. It would certainly decrease the pH. However, 12M $HCl$ is an extremely concentrated acid. To achieve a pH of 5.50, it would require a very precise and significant dilution, which might not be implied by "dissolving the appropriate amount" for direct preparation. While technically possible with dilution, it's not the most direct or practical choice among the given options for this specific pH.

Ammonium chloride ($NH_4Cl$) is a salt formed from a weak base ($NH_3$) and a strong acid ($HCl$). When dissolved in water, the ammonium ion ($NH_4^+$) acts as a weak acid, hydrolyzing water to produce hydronium ions ($H_3O^+$), thus making the solution acidic.
The given hydronium ion concentration is $[H_3O^+] = 2.8 \times 10^{-6}$ M.
We can calculate the pH for this solution:

This pH (5.55) is very close to the target pH of 5.50. This means that dissolving $NH_4Cl$ in water can directly yield a solution with a pH very near the desired value.

Based on the elimination process and the direct calculation, $NH_4Cl$ is the most appropriate solute.

The correct solute to use is (c) $NH_4Cl$.