Question
The Ka of a monoprotic weak acid is 4.8 x 10^-3. What is the percent ionization of a .198M solution of this acid?
(I got 16.8% and it's wrong)
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Explanation / Answer
Let be the dissociation of the weak acid
HA <---> H + + A-
initial conc. c 0 0
change -c +c +c
Equb. conc. c(1-) c c
Dissociation constant , Ka = c x c / ( c(1-)
= c 2 / (1-)
In the case of weak acids is very small so 1- is taken as 1
So Ka = c2
==> = ( Ka / c )
Given Ka = 4.8x10-3
c = concentration = 0.198 M
Plug the values we get = 0.156
% dissociation = 0.156 x 100 = 15.6 %
