Question

Consider the titration of 80.0 mL of 0.0200 M CH₃NH₂ (a weak base; K_b = 0.000440) with 0.100 M HNO₃. Calculate the pH after the following volumes of titrant have been added:
a) 0.0 ml
b) 4.0 ml
c) 8.0 ml
d) 12.0 ml
e) 16.0 ml
f) 24.0 ml

Answer 1

a)when 0.0 mL of HNO3 is added

CH3NH2 dissociates as:

CH3NH2 +H2O -----> CH3NH3+ + OH-

2*10^-2 0 0

2*10^-2-x x x

Kb = [CH3NH3+][OH-]/[CH3NH2]

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 ((4.4*10^-4)*2*10^-2) = 2.966*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

4.4*10^-4 = x^2/(2*10^-2-x)

8.8*10^-6 - 4.4*10^-4 *x = x^2

x^2 + 4.4*10^-4 *x-8.8*10^-6 = 0

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

a = 1

b = 4.4*10^-4

c = -8.8*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 = 3.539*10^-5

roots are :

x = 2.755*10^-3 and x = -3.195*10^-3

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

x = 2.755*10^-3

So, [OH-] = x = 2.755*10^-3 M

use:

pOH = -log [OH-]

= -log (2.755*10^-3)

= 2.5599

use:

PH = 14 - pOH

= 14 - 2.5599

= 11.4401

Answer: 11.44

b)when 4.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 4 mL

M(CH3NH2) = 0.02 M

V(CH3NH2) = 80 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.1 M * 4 mL = 0.4 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 0.4 mmol

mol(CH3NH2) = 1.6 mmol

0.4 mmol of both will react

excess CH3NH2 remaining = 1.2 mmol

Volume of Solution = 4 + 80 = 84 mL

[CH3NH2] = 1.2 mmol/84 mL = 0.0143 M

[CH3NH3+] = 0.4 mmol/84 mL = 0.0048 M

They form basic buffer

base is CH3NH2

conjugate acid is CH3NH3+

Kb = 4.4*10^-4

pKb = - log (Kb)

= - log(4.4*10^-4)

= 3.357

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 3.357+ log {4.762*10^-3/1.429*10^-2}

= 2.879

use:

PH = 14 - pOH

= 14 - 2.8794

= 11.1206

Answer: 11.12

c)when 8.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 8 mL

M(CH3NH2) = 0.02 M

V(CH3NH2) = 80 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.1 M * 8 mL = 0.8 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 0.8 mmol

mol(CH3NH2) = 1.6 mmol

0.8 mmol of both will react

excess CH3NH2 remaining = 0.8 mmol

Volume of Solution = 8 + 80 = 88 mL

[CH3NH2] = 0.8 mmol/88 mL = 0.0091 M

[CH3NH3+] = 0.8 mmol/88 mL = 0.0091 M

They form basic buffer

base is CH3NH2

conjugate acid is CH3NH3+

Kb = 4.4*10^-4

pKb = - log (Kb)

= - log(4.4*10^-4)

= 3.357

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 3.357+ log {9.091*10^-3/9.091*10^-3}

= 3.357

use:

PH = 14 - pOH

= 14 - 3.3565

= 10.6435

Answer: 10.64

d)when 12.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 12 mL

M(CH3NH2) = 0.02 M

V(CH3NH2) = 80 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.1 M * 12 mL = 1.2 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 1.2 mmol

mol(CH3NH2) = 1.6 mmol

1.2 mmol of both will react

excess CH3NH2 remaining = 0.4 mmol

Volume of Solution = 12 + 80 = 92 mL

[CH3NH2] = 0.4 mmol/92 mL = 0.0043 M

[CH3NH3+] = 1.2 mmol/92 mL = 0.013 M

They form basic buffer

base is CH3NH2

conjugate acid is CH3NH3+

Kb = 4.4*10^-4

pKb = - log (Kb)

= - log(4.4*10^-4)

= 3.357

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 3.357+ log {1.304*10^-2/4.348*10^-3}

= 3.834

use:

PH = 14 - pOH

= 14 - 3.8337

= 10.1663

Answer: 10.17

e)when 16.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 16 mL

M(CH3NH2) = 0.02 M

V(CH3NH2) = 80 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.1 M * 16 mL = 1.6 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 1.6 mmol

mol(CH3NH2) = 1.6 mmol

1.6 mmol of both will react to form CH3NH3+ and H2O

CH3NH3+ here is strong acid

CH3NH3+ formed = 1.6 mmol

Volume of Solution = 16 + 80 = 96 mL

Ka of CH3NH3+ = Kw/Kb = 1.0E-14/4.4E-4 = 2.273*10^-11

concentration ofCH3NH3+,c = 1.6 mmol/96 mL = 0.0167 M

CH3NH3+ + H2O -----> CH3NH2 + H+

1.667*10^-2 0 0

1.667*10^-2-x x x

Ka = [H+][CH3NH2]/[CH3NH3+]

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 ((2.273*10^-11)*1.667*10^-2) = 6.155*10^-7

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

so, x = 6.155*10^-7 M

[H+] = x = 6.155*10^-7 M

use:

pH = -log [H+]

= -log (6.155*10^-7)

= 6.2108

Answer: 6.21

f)when 24.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 24 mL

M(CH3NH2) = 0.02 M

V(CH3NH2) = 80 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.1 M * 24 mL = 2.4 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 2.4 mmol

mol(CH3NH2) = 1.6 mmol

1.6 mmol of both will react

excess HNO3 remaining = 0.8 mmol

Volume of Solution = 24 + 80 = 104 mL

[H+] = 0.8 mmol/104 mL = 0.0077 M

use:

pH = -log [H+]

= -log (7.692*10^-3)

= 2.1139

Answer: 2.11