The Ka of a monoprotic weak acid is 5.97 × 10-3. What is the percent ionization of a 0.153 M solution of this acid?
let the weak acid be HA
now
HA ---> H+ + A-
using ICE table
initial conc of HA , H+ , A- are 0.153 , 0 , 0
change in conc of HA , H+ , A- are -y , y , y
equilibrium conc of HA ,H+ ,A- are 0.153 - y , y , y
now
Ka = [H+] [A-] / [HA]
so
5.97 x 10-3 = [y] [y] / [ 0.153 - y]
5.97 x 10-3 [0.153 -y] = y2
(9.1341 x 10-4) - ( 5.97 x 10-3 )y = y2
solving
we get
y = 0.027385
now
[H+] = y = 0.027385
now
% ionization = [H+] x 100 / initial [HA]
so
% ionization = 0.027385 x 100 / 0.153
% ionization = 17.9
so
the percent ionization is 17.9%
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