How Weak Titration Differs from Strong Titration

Here, we are assuming that you have already looked at the mathematical discussion for a strong titration (that is, a titration with a strong acid and a strong base). As on that page, we will be looking at the titration of an acid with a base, but in this case, we will be using a weak acid (with a known \K_a value) and a strong base. It is, of course, also possible to use a weak base and a strong acid. The math is very similar in that case. Although, in the lab, you can also add a weak base to a weak acid, the mathematics of that situation is significantly more complex and beyond the scope of this text. 

For this discussion, we’ll focus our thoughts on this problem:

50.00 mL of 2.00 M HA (an imaginary acid with \(K_a = 2.56x10^{-5}\) ) are titrated with 1.00M XOH (an imaginary base). What will the pH be before any base is added and after additions of 1.00 mL, 10.00 mL, 50.00 mL, 98.00 mL, 100.00 mL, and 120.00 mL of the base?

Again, let’s consider what is happening in the flask. At the beginning, there is only acid in the flask, so the pH is relatively low, although to figure out what it actually is, we will need to solve an equilibrium problem.

When we start to add the base, a reaction will occur between the acid and base. Since the base is strong, that reaction will (effectively) go 100%. This makes for a simple limiting reagent problem. However, after that simple process, we will be left with some weak acid and some conjugate base. To find the pH, we will again need to do an equilibrium problem.

For example, say we have 10 moles of acid in the flask and add 2 moles of base. The following reaction will occur:

\(HA + XOH \rightarrow HOH + XA\)

Since the acid and base react in a 1:1 ratio, we will use up 2 moles of the acid, leaving 8.

\(2~moles~XOH \cdot \frac{1~HA}{1~XOH} = 2~moles~HA~used\)

\(10~moles~HA~originally~present~-2~moles~HA~used = 8~moles~HA~remaining\)

In this case, however, we need to ALSO keep track of the amount of conjugate base produced. Fortunately, this is just equal to the amount of acid used, so in our example, 2 moles \(A^{-1}\)

We will then need to solve for the \([H_3O^+]\) using the equilibrium reaction:

\(HA + H_2O \rightarrow H_3O^+ + A^{-1}\)

In our ICE table, the “I” row would contain the following:

\([HA] = \frac{8 ~moles~ HA}{total ~volume ~in ~liters}\)

\([A^{-1}] = \frac{2 ~moles~ A^{-1}}{total ~volume~ in~ liters}\)

Let’s take a hard look at the math.

Buffers

There is one other important difference between strong and weak titration. Weak titrations involve the production of a buffer solution in in the region between the beginning of the titration and the equivalence point. 

It is possible to do all of the titration calculations without ever thinking about this, but it also means that you can do some of the calculations using the Henderson/Hasselbalch equation.