Question

Consider the titration of 60.0 mL of 0.0400 M (CH3)3N (a weak base; Kb = 6.40e-05) with 0.100 M HBr. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =

(b) 6.0 mL pH =

(c) 12.0 mL pH =

(d) 18.0 mL pH =

(e) 24.0 mL pH =

(f) 38.4 mL pH =

Answer 1

a)when 0.0 mL of HBr is added

(CH3)3N dissociates as:

(CH3)3N +H2O -----> (CH3)3NH+ + OH-

4*10^-2 0 0

4*10^-2-x x x

Kb = [(CH3)3NH+][OH-]/[(CH3)3N]

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 ((6.4*10^-5)*4*10^-2) = 1.6*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)

6.4*10^-5 = x^2/(4*10^-2-x)

2.56*10^-6 - 6.4*10^-5 *x = x^2

x^2 + 6.4*10^-5 *x-2.56*10^-6 = 0

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

a = 1

b = 6.4*10^-5

c = -2.56*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 = 1.024*10^-5

roots are :

x = 1.568*10^-3 and x = -1.632*10^-3

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

x = 1.568*10^-3

So, [OH-] = x = 1.568*10^-3 M

use:

pOH = -log [OH-]

= -log (1.568*10^-3)

= 2.8046

use:

PH = 14 - pOH

= 14 - 2.8046

= 11.1954

Answer: 11.20

b)when 6.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 6 mL

M((CH3)3N) = 0.04 M

V((CH3)3N) = 60 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.1 M * 6 mL = 0.6 mmol

mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)

mol((CH3)3N) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HBr) = 0.6 mmol

mol((CH3)3N) = 2.4 mmol

0.6 mmol of both will react

excess (CH3)3N remaining = 1.8 mmol

Volume of Solution = 6 + 60 = 66 mL

[(CH3)3N] = 1.8 mmol/66 mL = 0.0273 M

[(CH3)3NH+] = 0.6 mmol/66 mL = 0.0091 M

They form basic buffer

base is (CH3)3N

conjugate acid is (CH3)3NH+

Kb = 6.4*10^-5

pKb = - log (Kb)

= - log(6.4*10^-5)

= 4.194

use:

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

= 4.194+ log {9.091*10^-3/2.727*10^-2}

= 3.717

use:

PH = 14 - pOH

= 14 - 3.7167

= 10.2833

Answer: 10.28

c)when 12.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 12 mL

M((CH3)3N) = 0.04 M

V((CH3)3N) = 60 mL

mol(HBr) = M(HBr) * V(HBr)

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

mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)

mol((CH3)3N) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HBr) = 1.2 mmol

mol((CH3)3N) = 2.4 mmol

1.2 mmol of both will react

excess (CH3)3N remaining = 1.2 mmol

Volume of Solution = 12 + 60 = 72 mL

[(CH3)3N] = 1.2 mmol/72 mL = 0.0167 M

[(CH3)3NH+] = 1.2 mmol/72 mL = 0.0167 M

They form basic buffer

base is (CH3)3N

conjugate acid is (CH3)3NH+

Kb = 6.4*10^-5

pKb = - log (Kb)

= - log(6.4*10^-5)

= 4.194

use:

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

= 4.194+ log {1.667*10^-2/1.667*10^-2}

= 4.194

use:

PH = 14 - pOH

= 14 - 4.1938

= 9.8062

Answer: 9.81

d)when 18.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 18 mL

M((CH3)3N) = 0.04 M

V((CH3)3N) = 60 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.1 M * 18 mL = 1.8 mmol

mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)

mol((CH3)3N) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HBr) = 1.8 mmol

mol((CH3)3N) = 2.4 mmol

1.8 mmol of both will react

excess (CH3)3N remaining = 0.6 mmol

Volume of Solution = 18 + 60 = 78 mL

[(CH3)3N] = 0.6 mmol/78 mL = 0.0077 M

[(CH3)3NH+] = 1.8 mmol/78 mL = 0.0231 M

They form basic buffer

base is (CH3)3N

conjugate acid is (CH3)3NH+

Kb = 6.4*10^-5

pKb = - log (Kb)

= - log(6.4*10^-5)

= 4.194

use:

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

= 4.194+ log {2.308*10^-2/7.692*10^-3}

= 4.671

use:

PH = 14 - pOH

= 14 - 4.6709

= 9.3291

Answer: 9.33

e)when 24.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 24 mL

M((CH3)3N) = 0.04 M

V((CH3)3N) = 60 mL

mol(HBr) = M(HBr) * V(HBr)

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

mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)

mol((CH3)3N) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HBr) = 2.4 mmol

mol((CH3)3N) = 2.4 mmol

2.4 mmol of both will react to form (CH3)3NH+ and H2O

(CH3)3NH+ here is strong acid

(CH3)3NH+ formed = 2.4 mmol

Volume of Solution = 24 + 60 = 84 mL

Ka of (CH3)3NH+ = Kw/Kb = 1.0E-14/6.4E-5 = 1.563*10^-10

concentration of(CH3)3NH+,c = 2.4 mmol/84 mL = 0.0286 M

(CH3)3NH+ + H2O -----> (CH3)3N + H+

2.857*10^-2 0 0

2.857*10^-2-x x x

Ka = [H+][(CH3)3N]/[(CH3)3NH+]

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 ((1.563*10^-10)*2.857*10^-2) = 2.113*10^-6

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

so, x = 2.113*10^-6 M

[H+] = x = 2.113*10^-6 M

use:

pH = -log [H+]

= -log (2.113*10^-6)

= 5.6751

Answer: 5.68

f)when 38.4 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 38.4 mL

M((CH3)3N) = 0.04 M

V((CH3)3N) = 60 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.1 M * 38.4 mL = 3.84 mmol

mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)

mol((CH3)3N) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HBr) = 3.84 mmol

mol((CH3)3N) = 2.4 mmol

2.4 mmol of both will react

excess HBr remaining = 1.44 mmol

Volume of Solution = 38.4 + 60 = 98.4 mL

[H+] = 1.44 mmol/98.4 mL = 0.0146 M

use:

pH = -log [H+]

= -log (1.463*10^-2)

= 1.8346

Answer: 1.83