Consider the titration of a 60.0 mL of 0.365 M weak acid HA (Ka = 4.2 x 10⁻⁶) with 0.400 M KOH. What is the pH before any base has been added?

Answer 1

Since this is a weak acid, we can simply use the Ka expression:

HA ==> H⁺ + A^-

Ka = [H⁺][A^-]/[HA]

4.2x10^-6 = (x)(x) / 0.400 -x (we can ignore x in denominator assuming it is small relative to 0.400 M)

x² = 1.68x10^-6

x = 1.3x10^-3 M = [H⁺] (note; our assumption above was correct as this is only 0.3% of the 0.400 M value)

pH = -log [H+] = -log 1.3x10-3

pH = 2.89