The goal of this example is to help students become fluent in calculating pH/pOH of a solution during a titration of a strong acid (HCl) and a strong base (NaOH). This problem includes calculations for titrations of increasing volumes.

Given Problem: Calculate the pH of a titration of 25.0 mL of HCl (.100 M) with .100 M NaOH after 12.5 mL, 25.0 mL, 37.5 mL, and 50.0 mL have been added. Start with calculating your initial pH. 

1. Find your initial pH and Molarity of the solution. You can use given equations such as pH=-log([H+]) and Molarity= (moles/Liter).

2. After 12.5 mL of NaOH is added: first, you can calculate the molarity of the volume of the NaOH added and the HCl left. This will show you respectively the moles of [H+]/[OH-] in the solution. This is because NaOH and HCl are both strong and dissociate readily. They are also monoprotic (contains one proton, or H+) so the reaction is one step, either protonation or deprotonation.

3. After you calculate moles of the base added and the acid left, you can find the pH by using the [H+] value. This is derived from the last step by finding the molarity. You use the moles of acid left and divide it by the total volume. In this case, the total volume is the 25.0 mL of the starting acid + 12.5 mL of the additional base. Always make sure to put this in Liters! 

After you find the Molarity (mol/L) of the acid in the solution, you can use the pH= -log([H+]). This is your pH after 12.5 mL is added.

4. After 25.0 mL of NaOH is titrated, we know that the moles of base= moles of acid. This is because essentially we have equal parts of acid and base that are equal Molarities. This means we have and equal pH and pOH. Since the scale runs from 0-14, and we know they are equal, then the pH and pOH are both 7.

5. This is the same process as the 12.5 step. Only in this case the total volume is .0375 L. From this we can find the pH to equal 12.30. In thisDept. step I showed another way to find the pH using the pH+pOH=14 equation.

6. After 50.00 mL NaOH is added: your final pH for this step is 12.52.

As always, check my numbers! I could have messed up somewhere.

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