Chapter 18: Other Aspects of Aqueous Equilibria

Chapter 18 is mainly a continuation of the topics you discussed in chapter 17. In this chapter they introduce common ions, buffer solutions, acid base titrations, and solubility of salts.

Our first topic will be common ions. This topic is not a relatively difficult one, but it deals with ions that appear in other equilibrium equations. An example of a common ion would be NO2- added to the weak acid HNO2. The added NO2- suppresses the ionization of HNO2 in a solution. This principle of common ions is used to make buffer solutions.

Buffer solutions are the next item on the list. Buffer solutions are solutions that are resistant to pH changes when a strong acid or a base is added. This is a very important principle, because many living things depend on buffer solutions to survive. If you remember back in Intro to Bio, you learned that enzymes have specific pH where they function correctly. If a cell is constantly experiencing pH changes those enzymes would not be able to function. In fact our blood is a buffer solution. Now there are 2 requirements to be a buffer. They are:

  1. Two substances are needed: an acid that is capable of reacting with added OH- ions and a base that can consume added H3O+ ions.
  2. The acid and the base must not react with each other.

Next we will discuss acid-base titrations. This topic is the bulk of your chapter. Now there are 4 types of titrations we will discuss: Strong Acid-Strong Base, Strong Acid-Weak Base, Weak Acid-Strong Base, and Weak Acid-Weak Base. When you titrate a Strong Acid and Strong Base the pH will be around 7. When titrating though, you will notice minimal changes in pH until the equivalence point is reached, then after the equivalence point is reached the pH will drastically increase if excess strong base is added. The same events take place in the other forms of titrations, however they may not have a pH of 7 at their equivalence point.

Now in a titration there are 4 different places in the titration that most chemists talk about. There is the starting point, halfway point, equivalence point, and the point after equivalence has been reached. The pH can be determined at each of these points, and you will be expected to calculate them. The pH at the beginning can simply be calculated by using the Ka and the known concentrations to calculate the H3O+ ion concentration. The pH at the halfway point can be calculated using a special formula known as the H-H equation. The H-H equation is pH = pKa + log[Conjugate base/acid]. This formula is especially important, because it saves a lot of unnecessary work. The pH of the solution at the equivalence point and after the equivalence point is calculated by using your ICE charts and equilibrium equations.

Next our focus moves to the solubility of salts. Back in chapter 5 you were taught by your solubility rules that certain compounds are completely insoluble. Now that may be true in some cases, but usually a small amount of that insoluble compound does in fact dissociate in water. On page 873 and 874 of your textbook provides a good example of this concept. However, I will try and explain it briefly here. According to your solubility chart silver bromide is insoluble in water, but in the experiment it was calculated that 7.35 x 10-7 moles of AgBr dissociated. This can be turned into a equilibrium equation:

A G B R solid yields A G positive aqueous plus B R negative aqueous.
Then an equilibrium expression can be written: 

K sub S P equals the concentration of A G positive multiplied by the concentration of B R negative.

Note that AgBr was not included because it is a solid. Then from the given information it can be plugged into this expression and solved for the equilibrium expression. This equation could then be manipulated to solve other values.

This should provide you with a general overview of this chapter. There are several topics I didn’t cover so don’t substitute this for you text. I hope this helps you guys out.