Question

Find the pH and fraction of dissociation (α) of a 0.129 M solution of the weak acid HA with Ka = 1.79 ✕ 10−5.

Answer 1

1)

Let the weak acid be written as HA

HA dissociates as:

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

0.129 0 0

0.129-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 ((1.79*10^-5)*0.129) = 1.52*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)

1.79*10^-5 = x^2/(0.129-x)

2.309*10^-6 - 1.79*10^-5 *x = x^2

x^2 + 1.79*10^-5 *x-2.309*10^-6 = 0

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

a = 1

b = 1.79*10^-5

c = -2.309*10^-6

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 = 9.237*10^-6

roots are :

x = 1.511*10^-3 and x = -1.529*10^-3

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

x = 1.511*10^-3

so.[H+] = x = 1.511*10^-3 M

use:

pH = -log [H+]

= -log (1.511*10^-3)

= 2.82

Answer: 2.82

2)

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

x = 1.511*10^-3

degree of dissociation = x/c

= 1.511*10^-3/0.129

= 0.0117

Answer: 0.0117