percent ionization?

Question

(a) Calculate the percent ionization of 0.0040 M butanoic acid (Ka = 1.5 10-5).
%

(b) Calculate the percent ionization of 0.0040 M butanoic acid in a solution containing 0.085 M sodium butanoate.
%

Answer 1

Butanoic acid is CH3CH2CH2COOH.
I'll just write it HA for simplicity

.. .. .. .. .. .. .. HA <=> H+ +A-
Initial .. .. .. .. C
Dissociate .. x
Produce .. .. .. .. .. .. .. x .. .. x
At Equil .. .. C-x .. .. .. x .. .. x

Ka= [H+][A-] / [HA] = x^2/(C-x) (1)
By definition the degree of dissociation is a=x/C => x=aC (2)

Substitute x in (1) using (2)
Ka = a^2 *C^2 / (C-aC) =>
Ka = a^2C/(1-a) =>
Ca^2 =Ka-Ka*a =>
Ca^2 + Ka a -Ka=0. Substitute the values for Ka and C

0.004 a^2 + 1.5*10^-5 a - 1.5*10^-5 =0
a1= 0.059 =5.9%
a2= -0.063<0 rejected

In the presence of sodium butanoate you have initial A- = 0.085


.. .. .. .. .. .. .. HA <=> H+ + A-
Initial .. .. ..0.004 .. .. .. .. .. 0.085
Dissociate .. x
Produce .. .. .. .. .. .. .. x .. .. x
At Equil .0.004-x .. .. .. x .. .x+0.085

Ka= x(x+0.085)/(0.004-x) =1.5*10-5 =>
x^2+0.085x = 6*10^-8 -1.5*10^-5x =>
x^2 + 0.085015 x -6*10^-8=0
x1= 7.1 *10^-7
x2= -0.085016 <0 rejected

a= x/C = (7.1/0.004)*10^-7 = 0.000178 = 0.18%
Note that we could have solved it as above, getting a quadratic for a instead of x

Answer 2

a.) Ka = 1.5*10^-5 = (x^2)/(0.004-x), where x is the amount of butanoic acid dissociated into butane and hydrogen. Solve for x.

% Ionization = (x/0.004) * 100


b.) Ka = 1.5*10^-5 = (0.085 + x)(x)/(0.004 - x), with x being the same thing as above. Again solve for x.

% Ionization = (x/0.004) * 100