Final answer:
At the halfway point of titration, the pH for a strong acid titrated with a strong base (0.1 M HCl with 0.1 M NaOH) and vice versa will be 7.00. For a weak acid with a strong base (0.1 M CH3COOH with 0.1 M NaOH), the pH will be equal to the pKa of the weak acid, which is 4.76. Similarly, for a weak base with a strong acid (0.1 M NH3 with 0.1 M HCl), the pH will be equal to the pKa of the conjugate acid, which is 9.25.
Step-by-step explanation:
Calculating the pH at the halfway point for titrations involves different considerations depending on the nature of the acid and base involved in the titration.
Strong Acid with Strong Base
For the titration of a strong acid with a strong base, such as 0.1 M HCl titrated with 0.1 M NaOH, at the halfway point the number of moles of H+ will equal the number of moles of OH−, creating a neutral solution at pH 7.00.
Weak Acid with Strong Base
For the titration of a weak acid, for example, 0.1 M CH3COOH with 0.1 M NaOH, at the halfway point there will be equal amounts of the acid and its conjugate base, resulting in a buffer system. The pH can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A−]/[HA]). The pKa for acetic acid, CH3COOH, is approximately 4.76, so at the halfway point, the pH will also be 4.76 since [A−]/[HA] = 1.
Weak Base with Strong Acid
For the titration of a weak base like 0.1 M NH3 with a strong acid such as 0.1 M HCl, the pH will be determined by the amount of NH4+ formed. At the halfway point, pH = pKa + log([B]/[BH+]), where pKa is the negative log of the Kb of NH3. Because at the halfway point, [B] = [BH+], the pH = pKa. The pKa for NH4+ is 9.25, so at the halfway point, the pH is 9.25.
Strong Base with Strong Acid
In the case of a titration of a strong base with a strong acid, such as 0.1 M NaOH with 0.1 M HCl, the same principle as the strong acid with strong base applies, and at the halfway point, the pH will be 7.00.