Question

(a) Calculate the percent ionization of 0.00250 M acetic acid (Ka = 1.8e-05).

% ionization = %

(b) Calculate the percent ionization of 0.00250 M acetic acid in a solution containing 0.0110 M sodium acetate.

% ionization = %

Answer 1

a) for a weak acid the degree of ionisation x is calculated as

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

C 0 0 initial concentration

C(1-x) Cx Cx equilibrium concentrations

Then Ka = Cx.Cx/C(1-x)   

As x is very small compared to C, we approximate C(1-x) as C

Thus Ka = Cx²  or x = (Ka/C)^1/2

  Thus percent ionization of acetic acid x = (1.8x10^-5 /0.0025)^1/2

   = 8.48x10^-2

% ionisation = 8.48x10^-2 x100

= 8.48%

b)Due to common ion effect the ionisation of acetic acid is further decreases.

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

0.00250 0 0.0110 initial concentrataions

0.0025-x x 0.011 +x

As x is very small 0.11 + x is taken as 0.011 M

NOw Ka = x (0.011)/ (0.0025-x) = 1.8x10^-5

solving for x , we get x= 4.084 x10^-6 M

hence in 0.0025M solution ionisation is 4.084 x10^-6 M  

then % ionisation is = 4.084 x10^-6 x100

= 4.084 x10^-4 %