Question

500 ml of .14 m naoh is added to 535 ml of .2 m weak acid (ka = 7.12 x 10^-5). what is the resulting ph?

Answer 1

final concentration of NaOH = 0.14*500/(500+535) = 0.0676 M ( = concentration base)

final concentration of acid = 0.2*535/(500+535) = 0.1034 M - 0.0676 M = 0.0358

Therefore pH is:

pH = -log(7.12 x 10^-5) + log(0.0676/0.0358)

pH = 4.42