Question

27. + -/0.1 points 0/4 Submissions Used Determine the pH of each of the following solutions. (a) 0.120 M hydrazoic acid (weak acid with Ka = 1.98-05). (b) 0.146 M phenol (weak acid with Ka = 1.3e-10). (c) 0.534 M pyridine (weak base with Kb = 1.72-09).

Answer 1

a)

HN3 dissociates as:

HN3 -----> H+ + N3-

0.12 0 0

0.12-x x x

Ka = [H+][N3-]/[HN3]

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 ((1.9*10^-5)*0.12) = 1.51*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)

1.9*10^-5 = x^2/(0.12-x)

2.28*10^-6 - 1.9*10^-5 *x = x^2

x^2 + 1.9*10^-5 *x-2.28*10^-6 = 0

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

a = 1

b = 1.9*10^-5

c = -2.28*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 = 9.12*10^-6

roots are :

x = 1.5*10^-3 and x = -1.519*10^-3

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

x = 1.5*10^-3

So, [H+] = x = 1.5*10^-3 M

use:

pH = -log [H+]

= -log (1.5*10^-3)

= 2.8238

Answer: 2.82

b)

C6H5OH dissociates as:

C6H5OH -----> H+ + C6H5O-

0.146 0 0

0.146-x x x

Ka = [H+][C6H5O-]/[C6H5OH]

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^-10)*0.146) = 4.357*10^-6

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

so, x = 4.357*10^-6 M

So, [H+] = x = 4.357*10^-6 M

use:

pH = -log [H+]

= -log (4.357*10^-6)

= 5.3609

Answer: 5.36

C)

C5H5N dissociates as:

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

0.534 0 0

0.534-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)*0.534) = 3.013*10^-5

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

so, x = 3.013*10^-5 M

So, [OH-] = x = 3.013*10^-5 M

use:

pOH = -log [OH-]

= -log (3.013*10^-5)

= 4.521

use:

PH = 14 - pOH

= 14 - 4.521

= 9.479

Answer: 9.48