A 0.0900 M solution of a monoprotic acid is 11.5 percent ionized. Calculate the ionization constant...

Question

Question

A 0.0900 M solution of a monoprotic acid is 11.5 percent ionized. Calculate the ionization constant of the acid.

science chemistry

Answer 1

Answer #1

% dissociation = (x*100)/c

11.5= x*100/0.09

x = 0.0103

HA dissociates as:

HA -----> H+ + A-

9*10^-2 0 0

9*10^-2-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Ka = 1.035*10^-2*1.035*10^-2/(0.09-1.035*10^-2)

Ka = 1.345*10^-3

Answer: 1.35*10^-3

Answer 2

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