Step-by-step explanation:
The titration of acetic acid with KOH is a weak acid-strong base titration. At the beginning of the titration (part a), we have only acetic acid in the solution, and its concentration is 0.200 M. As we add KOH, it reacts with acetic acid to form acetate and water:
CH3COOH + KOH → CH3COOK + H2O
The acetate ion is the conjugate base of acetic acid and can be considered a weak base. We can use the following equation to calculate the pH of the resulting solution at each point of the titration:
pH = pKa + log([A^-]/[HA])
where pKa is the acid dissociation constant of acetic acid (1.8 × 10^-5), [A^-] is the concentration of acetate ion, and [HA] is the concentration of undissociated acetic acid.
a) At the beginning of the titration (0.0 mL of KOH added), the solution contains only acetic acid. Therefore, [HA] = 0.200 M and [A^-] = 0 M.
pH = pKa + log([A^-]/[HA])
pH = -log(1.8 × 10^-5) + log(0/0.200)
pH = 2.40
The pH of the solution is 2.40.
b) When 50.0 mL of 0.100 M KOH is added, we have added 5.00 mmol of KOH. This amount of KOH reacts with 5.00 mmol of acetic acid, and the remaining 0.050 mol - 0.005 mol = 0.045 mol of acetic acid remains in the solution. At the same time, 0.005 mol of acetate ion is formed.
[HA] = 0.045 mol / 0.100 L = 0.450 M
[A^-] = 0.005 mol / 0.100 L = 0.050 M
pH = pKa + log([A^-]/[HA])
pH = -log(1.8 × 10^-5) + log(0.050/0.450)
pH = 4.41
The pH of the solution is 4.41.
c) When 100.0 mL of 0.100 M KOH is added, we have added 10.00 mmol of KOH. This amount of KOH reacts with 10.00 mmol of acetic acid, and there is no acetic acid remaining in the solution. At the same time, 0.010 mol of acetate ion is formed.
[HA] = 0 mol / 0.100 L = 0 M
[A^-] = 0.010 mol / 0.100 L = 0.100 M
pH = pKa + log([A^-]/[HA])
pH = -log(1.8 × 10^-5) + log(0.100/0)
pH = 4.74
The pH of the solution is 4.74.
d) When 110.0 mL of 0.100 M KOH is added, we have added 11.00 mmol of KOH. This amount of KOH reacts with 10.00 mmol of acetic acid, and there is an excess of 1.00 mmol of KOH in the solution. This excess KOH completely dissociates to give 1.00 mmol of OH^- ion. At