A student titrates 25.00 mL of 0.100 M acetic acid with 0.100 M NaOH while monitoring pH. Selected pH values: at 0.00 mL added, pH = 2.87; at 12.50 mL (half-equivalence), pH = 4.74; at 25.00 mL (equivalence), pH = 8.72.
What is the Ka of acetic acid from these data?
- A
4.74
- B
5.5e-10
- C
1.0e-7
- Dcheck_circle
1.8e-5
Explanation
At half-equivalence, pH = pKa = 4.74, so Ka = 10^-4.74 = 1.8e-5.