How pH changes during a weak titration - the big ideas

Let's look at how the pH changes during the titration of a weak acid with a strong base. For this, we'll assume that you've already mastered strong/strong titration theory, math and graphs.

So, for the discussion on weak acids, we’ll focus our thoughts on this problem, which is a slightly altered version of the strong/strong titration problem we did before.: 

50.00 mL of 2.00 M HA (an imaginary weak acid with a \(K_a=4.23\cdot10^{-6}\)) are titrated with 1.00M XOH (an imaginary strong 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?

First, let’s consider what is happening in the flask. At the beginning, there is only weak acid in the flask, so the pH is relatively low. Determining that actual pH will require us to do an equilibrium problem (with an ICE table) using \(K_a\) and the reaction below:

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

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.

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

\(HA + XOH \rightarrow HOH + X^{+1} + A^{-1}\)

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 relatively low. We can again do an equilibrium problem using \(K_a\) to find the \([H_3O^{+1}]\) and then the pH. However, it has now become more complicated. Since the acid (HA) is weak, it's conjugate base (\(A^{-1}\)) cannot be ignored. It is easy to find that amount, since it is produced in a 1:1 ratio with the amount of acid that was used up in the reaction. So, 

 \(2~moles~HA \cdot \frac{1~A^{-1}}{1~HA} = 2~moles~A^{-1}~produced\)

This amount of \(A^{-1}\) will also appear in the "I" row of the ICE table.

For all points up to (but not including) the equivalence point of the titration this will be our situation: we will react the strong base completely. That reaction will leave some of the weak acid and will create some of the conjugate base. Solving for pH will require us to do an equilibrium problem using an ICE table that initially contains both the remaining acid and the created conjugate base. 

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 conjugate base (in this case \(A^{-1}\)). The pH will not be 7 because the solution contains a base. 

To determine the pH we will need to rethink what can happen in the solution. Our previous reaction can't occur any more since we have no HA present. Instead, the base will react with water like this

\(A^{-1} + H_2O \rightleftharpoons HA + OH^{-1}\)

We know the amount of base present (from our simple stoichiometry) so we can use that in the "I" row of an ICE table but, of course, we need an appropriate K value. Since this is a basic reaction, we need a \(K_b\). Remembering that 

\(K_w=K_a \cdot K_b\) 

we can rearrange to solve for \(K_b\)

\(K_b=\frac{K_w}{K_a}\)

All of that will allow us to find the \([OH^{-1}]\). From that we can calculate the pOH and then the pH. 

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. And, more importantly, since the base is strong, we will be able to find \([OH^{-1}]\) easily since, for strong bases \([OH^{-1}] = [XOH]\)