Determining the pH During a Strong/Strong Titration

A strong/strong titration, that is a titration using a strong acid and a strong base, is really an exercise in stoichiometry and limiting reagents as described here. (A weak/strong titration problem is done here.)

It is important to remember that stoich is done with moles, and the problem we are trying to solve (shown below) gives us volume and molarity. We will need to deal with that by remembering that:

\(moles = Molarity (\frac{moles}{liter}) \cdot Volume (L)\)

Here's the problem we are going to look at:

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?

The first point is easy. We have 2.00 M strong acid and we know that the \([H_3O^{+1}]\) of a strong acid is just the concentration of the acid. Remembering that pH is the -log of \([H_3O^{+1}]\) means we can do the following math:

\(-log[H_3O^{+1}] = -log(2.00) = -0.301\)

To solve the rest of the problem, we are going to need to keep track of the following:

• volume of base added
• total volume of the solution
• moles of acid (\(M_a \cdot V_a\))
• moles of base (\(M_b \cdot V_b\))

We can do all this by putting the relevant information into a table, where each column will be one of the bits of information listed above and each row of the table will be a step (a different volume of base) in the problem.

Let's tackle the first addition of base. We can start by filling out what we know:

Then we know that strong acids and bases react ~100%, so we can simply subtract to determine the amount of acid that remains:

Lastly, we can find the concentration of the acid (\(\frac{moles}{liters}\)) and then pH.

Each step before the equivalence point works the same way

and

and

Suddenly, things change dramatically -- at the equivalence point the initial reaction leaves neither acid or base in the solution.

After the equivalence point, things change again. In this case, we have used up all of the acid, and have extra base in solution. 

This means that we can calculate the \([XOH]\). Since this is a strong base, the \([OH^{-1}]\) will be the same as \([XOH]\). We can then take the -log of \([OH^{-1}]\), which will be the pOH, and subtract that from 14 to get the pH.

Now that you've been through the mathematics of this process it is worth looking at how the pH changes graphically.