Question

The Ka of a monoprotic weak acid is 5.12 × 10-3. What is the percent ionization of a 0.185 M solution of this acid?

Answer 1

Lets write the acid as HA
HA dissociates as:

HA          ----->     H+   + A-
0.185                 0         0
0.185-x               x         x


Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes

Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((5.12*10^-3)*0.185) = 3.078*10^-2

since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
5.12*10^-3 = x^2/(0.185-x)
9.472*10^-4 - 5.12*10^-3 *x = x^2
x^2 + 5.12*10^-3 *x-9.472*10^-4 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 5.12*10^-3
c = -9.472*10^-4

Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.815*10^-3

roots are :
x = 2.832*10^-2 and x = -3.344*10^-2

since x can't be negative, the possible value of x is
x = 2.832*10^-2

% dissociation = (x*100)/c
= 2.832*10^-2*100/0.185
= 15.3 %

Answer: 15.3 %