Question

Determine the pH of each of the following solutions.

(a) 0.410 M propionic acid (weak acid with Ka = 1.3e-05).

(b) 0.122 M benzoic acid (weak acid with Ka = 6.3e-05).

(c) 0.867 M pyridine (weak base with Kb = 1.7e-09).

Answer 1

a)

C2H5COOH dissociates as:

C2H5COOH -----> H+ + C2H5COO-

0.41 0 0

0.41-x x x

Ka = [H+][C2H5COO-]/[C2H5COOH]

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 ((1.3*10^-5)*0.41) = 2.309*10^-3

since c is much greater than x, our assumption is correct

so, x = 2.309*10^-3 M

So, [H+] = x = 2.309*10^-3 M

use:

pH = -log [H+]

= -log (2.309*10^-3)

= 2.6366

Answer: 2.64

b)

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

0.122 0 0

0.122-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

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 ((6.3*10^-5)*0.122) = 2.772*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)

6.3*10^-5 = x^2/(0.122-x)

7.686*10^-6 - 6.3*10^-5 *x = x^2

x^2 + 6.3*10^-5 *x-7.686*10^-6 = 0

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

a = 1

b = 6.3*10^-5

c = -7.686*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 = 3.075*10^-5

roots are :

x = 2.741*10^-3 and x = -2.804*10^-3

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

x = 2.741*10^-3

So, [H+] = x = 2.741*10^-3 M

use:

pH = -log [H+]

= -log (2.741*10^-3)

= 2.5621

Answer: 2.56

c)

C5H5N dissociates as:

C5H5N +H2O -----> C5H5NH+ + OH-

0.867 0 0

0.867-x x x

Kb = [C5H5NH+][OH-]/[C5H5N]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.7*10^-9)*0.867) = 3.839*10^-5

since c is much greater than x, our assumption is correct

so, x = 3.839*10^-5 M

So, [OH-] = x = 3.839*10^-5 M

use:

pOH = -log [OH-]

= -log (3.839*10^-5)

= 4.4158

use:

PH = 14 - pOH

= 14 - 4.4158

= 9.5842

Answer: 9.58