Question

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?

Answer 1

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%