Question

A 0.122 M monoprotic acid has a K_a of 5.7 x 10^-4. What is the percent dissociation of the acid? Report to two decimal places but do not include a percent sign with your answer.

Answer 1

let the acid be written as HA

HA dissociates as:

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

0.122 0 0

0.122-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 ((5.7*10^-4)*0.122) = 8.339*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

5.7*10^-4 = x^2/(0.122-x)

6.954*10^-5 - 5.7*10^-4 *x = x^2

x^2 + 5.7*10^-4 *x-6.954*10^-5 = 0

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

a = 1

b = 5.7*10^-4

c = -6.954*10^-5

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 = 2.785*10^-4

roots are :

x = 8.059*10^-3 and x = -8.629*10^-3

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

x = 8.059*10^-3

% dissociation = (x*100)/c

= 0.0081*100/0.122

= 6.61 %

Answer: 6.61