Question

Find the pH of a 0.150 M solution of a weak monoprotic acid having Ka= 0.11. Also, find the percent dissociation of this solution.

Answer 1

1)

HA dissociates as:

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

0.15 0 0

0.15-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 ((0.11)*0.15) = 0.1285

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

0.11 = x^2/(0.15-x)

1.65*10^-2 - 0.11 *x = x^2

x^2 + 0.11 *x-1.65*10^-2 = 0

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

a = 1

b = 0.11

c = -1.65*10^-2

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 = 7.81*10^-2

roots are :

x = 8.473*10^-2 and x = -0.1947

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

x = 8.473*10^-2

So, [H+] = x = 8.473*10^-2 M

use:

pH = -log [H+]

= -log (8.473*10^-2)

= 1.072

Answer: 1.07

b)

% dissociation = x * 100 / initial concentration

= (8.473*10^-2)*100 / 0.11

= 77.0 %

Answer: 77.0 %