Consider the titration of a 60.0 mL of 0.367 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

HA <==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA]

4.2x10^-6 = (x)(x) / 0.367 - x (assume x is small relative to 0.367 and ignore it in denominator)

4.2x10^-6 = x ² / 0.367

x² = 1.54x10^-6

x = 1.24x10^-3 (note this is small relative to 0.367 so above assumption was valid)

[H⁺] = 1.24x10^-3 M

pH = -log [H⁺] = - log 1.24x10^-3

pH = 2.91