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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Ответ:
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
Ответ:
I dont have a phone lol sry
Explanation: