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Consider the titration of 100 ml of 0.1 M HCl with 0.1 M NaOH. Calculate the pH of the solution at the following amounts of NaOH in the titration:

1. At the beginning, when no NaOH is added
2. When 50 ml of 0.1 M NaOH is added
3. When 98 ml of 0.1 M NaOH is added
4. When 99.9 ml of 0.1 M NaOH is added
5. When 100 ml of 0.1 M NaOH is added
6. When 100.1 ml of 0.1 M NaOH is added

Answer :

The pH of a strong acid-strong base titration can be calculated at various stages: initially the pH is acidic, mid-titration is neutral at the equivalence point, and just beyond the equivalence point, the pH becomes slightly basic depending on the excess base added.

In the titration of a strong acid with a strong base, the pH at various stages can be calculated by considering the mole balance between the acid and base.

Initial pH

1. Before NaOH is added, the solution is just 0.1 M HCl. Since HCl is a strong acid, it dissociates completely in water, therefore pH = -log[0.1] = 1.

Mid-Titration

2. When 50 ml of 0.1 M NaOH is added, we've added 0.005 moles of NaOH which neutralize 0.005 moles of HCl, leaving no excess acid or base. Thus, the pH at this point is neutral, pH = 7.

3. At 98 ml, there is still a slight excess of HCl, only 0.002 moles of acid remain unneutralized, so the pH will be slightly acidic, just above 3.

4. At 99.9 ml, the amount of unneutralized HCl is very small, so the pH will be very close to 7.

Equivalence Point

5. With 100 ml of 0.1 M NaOH, all the HCl is neutralized, and the pH will be 7, as it is a neutralization of strong acid and base.

Beyond Equivalence Point

6. If 100.1 ml of NaOH is added, there's a tiny excess of NaOH, leading to a slightly basic pH, just over 7.

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