Question

The Ka of a monoprotic weak acid is 5.91 × 10-3. What is the percent ionization of a 0.162 M solution of this acid?

Answer 1

Lets write the acid as HA

HA dissociates as:

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

0.162 0 0

0.162-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 ((5.91*10^-3)*0.162) = 3.094*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)

5.91*10^-3 = x^2/(0.162-x)

9.574*10^-4 - 5.91*10^-3 *x = x^2

x^2 + 5.91*10^-3 *x-9.574*10^-4 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 5.91*10^-3

c = -9.574*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 = 3.865*10^-3

roots are :

x = 2.813*10^-2 and x = -3.404*10^-2

since x can't be negative, the possible value of x is

x = 2.813*10^-2

% dissociation = (x*100)/c

= 2.813*10^-2*100/0.162

= 17.4 %

Answer: 17.4 %