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?

Question

Brayan Christian

Chemistry

15 March, 05:04

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?

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Ben Hampton · 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