Acids and Bases: Buffers and Titration Curves

 

A buffer, by definition, is a solution that resists change in pH. In a buffered solution, adding acid will only result in a small decrease in pH whereas adding the same volume and concentration of acid to a non-buffered solution will cause a much larger change in pH. In order for this to happen, a chemical must be in solution that will neutralize the acid, or a base if that were added.

Luckily, the properties of a weak acid fit this description. In a solution of weak acid, such as acetic acid in water, some of the acid will dissociate, and the rest of it will remain in its acidic form:

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The extent to which the acid dissociates is dependent on the acid, denoted by its corresponding Ka value. However, if we were to add acid to the solution of water and acetic acid, we would increase the amount of H+ in solution. To maintain its equilibrium, the equation will shift to the left (Le Chatelier’s principle), and the hydrogen will re-bond with the acetate (CH3COO-), forming more acetic acid, decreasing the amount of H+, and maintaining a fairly constant pH. The same idea applies if we were to add base to the solution. As OH- is added, it will bond with hydrogen to create water. To maintain equilibrium, the reaction will shift right, forming more acetate and hydrogen ions while decreasing acetic acid.

Buffer systems can be made with many other weak bases and acids, all you need is a chemical that can dissociate or re-associate in an acid/base neutralization. This may be an acid and its conjugate base in salt form (for example, HCOOH and KHCOO, or CH3COOH and KCH3COO).

Problems with Buffers

In almost every quantitative problem involving buffers, you’ll use the Henderson-Hasselbalch equation.

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If you have a question asking about buffers, it’s not a bad idea to just write this down. Using this equation we can see that pH is equal to the pKa of the weak acid, plus a factor that varies based on the amount of protonated and deprotonated acid, or acid and conjugate base, respectively. Thus, when the log portion of the equation is equal to zero, then the pH is equal to the pKa of the acid. Log(1) = 0, so the pH will equal the pKa when there are equal portions acid and conjugate base.

Additionally, the log of a number greater than 1 will be positive, and the log of a number between 0 and 1, or a decimal less than 1, will be a negative value. Therefore, if the concentration of the acid is greater than a conjugate base, the log component will be negative, so the pH will be less than the pKa. On the other hand, if the concentration of the conjugate base is greater than the acid, then the log component will be positive and the pH will be greater than the pKa. This makes sense conceptually as well—if there is more acid than conjugate base, there is more protonated acid than deprotonated, so the pH must be lower, more acidic.

Titration Curves

A titration curve can plot several different things, so the first thing you should do is look at the axes. They can be either volume of acid or base added, pH, or moles of acid or base added. The next step I would recommend is to find the equivalence point. This is the point on the line where the moles of acid is equal to the moles of base. Note that this is not a comparison of molarity or any other property. Only moles.

For a strong acid/base titration, the equivalence point is in the center of the steepest portion. At this point the pH = 7.00, as there are equal portions acid and base.

However, for a weak acid/base titration, there are two points you should identify. The equivalence point as with a strong acid/base titration, but also the ½ equivalence point. Take a graph with pH on the y-axis and volume of strong base added on the x-axis. The equivalence point will still occur at equal parts acid and base, but the pH at this point will be greater than 7.0 (the opposite is true for a weak base titrated with a strong acid, the equivalence point will occur below 7.0). The arguably more important point on these graphs is the ½ equivalence point. This is located halfway down the x-axis from the equivalence point. If there are equal moles of acid and base when 8mL of strong base are added, the ½ equivalence point is at 4mL. The value of the y-axis at this point is the pH as well as the pKa. At the ½ equivalence point, the concentration of acid and conjugate base are equal. Relate this back to the Henderson-Hasselbalch equation, at equal portions acid and conjugate base, the pH = pKa.