How pH Changes During a Titration

Titration, as first discussed here, involves adding one solution to another. Most commonly, these are acid and base solutions and we generally add a basic solution to an acidic one. We are going to work through these ideas using that as our model, but it is important to know that acids can be added to bases and that titration can be done with solutions that are not acidic or basic. 

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

50.00 mL of 2.00 M HA (an imaginary acid) 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?

How this plays out, and as a result, how we deal with it depends on whether the acid and base are strong or weak. On this page, we will assume that both the acid and the base are strong. On this page we will look at the situation if the acid is weak (and the base is strong).

First, let’s consider what is happening in the flask. At the beginning, there is only strong acid in the flask, so the pH is low.

When we start to add the base, a reaction will occur between the acid and base. Since both are strong, that reaction will (effectively) go 100%. This makes for a simple limiting reagent problem.

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. 

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

and

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

Since unreacted acid remains in the flask, the pH will still be low. This will be the situation up to (but not including) the equivalence point of the titration. 

At the equivalence point, things will change. Again, let's imagine that we started with 10 moles of the acid, but now we have added 10 moles of the base. The same reaction occurs, again ~100%. This time, however, we use up all of the acid and all of the base. That leaves us with only the ionic product of the reaction (in this case XA). the pH of this solution will be 7. (The reason is explained here.)

Once we pass the equivalence point, things change again. Now the acid becomes the limiting reagent. Again, let's imagine that we started with 10 moles of the acid, but now we have added 12 moles of the base. Since the acid and base react 1:1, we will use all of the acid and 10 moles of the base, leaving 2 moles of base unreacted. 

\(10~moles~HA \cdot \frac{1~XOH}{1~HA} = 10~moles~XOH~used\) 

and

\(12~moles~XOH~added~-10~moles~XOH~used = 2~moles~XOH~remaining\)

Since there is extra base in the flask, the pH will be high.

Let's look at how this plays out in the context of our problem above.