The Ka of a monoprotic weak acid is 558 x103 What is the per

The Ka of a monoprotic weak acid is 5.58 x103. What is the percent ionization of a 0.145 M solution of this acid? Number 0 0

Solution

Let a be the dissociation of the weak acid
HA <---> H + + A-
initial conc. c 0 0
change -ca +ca +ca
Equb. conc. c(1-a) ca ca

Dissociation constant , Ka = ca x ca / ( c(1-a)
= c a^2 / (1-a)
In the case of weak acids a is very small so 1-a is taken as 1
So Ka = ca^2
==> a = ( Ka / c )
Given Ka = 5.58x10^-3
c = concentration = 0.145 M
Plug the values we get a = 0.196
% dissociation = 0.196 x 100 = 19.6%