Question

Consider the titration of 30.0 mL of 0.0700 M (CH3)3N (a weak base; Kb = 6.400-05) with 0.100 M HBr. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL (b) 5.3 mL (c) 10.5 mL pH = pH = pH = (d) 15.8 ml (e) 21.0 mL (f) 27.3 mL pH = pH = pH =

Answer 1

a)when 0.0 mL of HBr is added

(CH3)3N dissociates as:

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

7*10^-2 0 0

7*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)*7*10^-2) = 2.117*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/(7*10^-2-x)

4.48*10^-6 - 6.4*10^-5 *x = x^2

x^2 + 6.4*10^-5 *x-4.48*10^-6 = 0

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

a = 1

b = 6.4*10^-5

c = -4.48*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 = 1.792*10^-5

roots are :

x = 2.085*10^-3 and x = -2.149*10^-3

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

x = 2.085*10^-3

So, [OH-] = x = 2.085*10^-3 M

use:

pOH = -log [OH-]

= -log (2.085*10^-3)

= 2.6809

use:

PH = 14 - pOH

= 14 - 2.6809

= 11.3191

Answer: 11.32

b)when 5.3 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 5.3 mL

M((CH3)3N) = 0.07 M

V((CH3)3N) = 30 mL

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

mol(HBr) = 0.1 M * 5.3 mL = 0.53 mmol

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

mol((CH3)3N) = 0.07 M * 30 mL = 2.1 mmol

We have:

mol(HBr) = 0.53 mmol

mol((CH3)3N) = 2.1 mmol

0.53 mmol of both will react

excess (CH3)3N remaining = 1.57 mmol

Volume of Solution = 5.3 + 30 = 35.3 mL

[(CH3)3N] = 1.57 mmol/35.3 mL = 0.0445 M

[(CH3)3NH+] = 0.53 mmol/35.3 mL = 0.015 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 {1.501*10^-2/4.448*10^-2}

= 3.722

use:

PH = 14 - pOH

= 14 - 3.7222

= 10.2778

Answer: 10.28

c)when 10.5 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 10.5 mL

M((CH3)3N) = 0.07 M

V((CH3)3N) = 30 mL

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

mol(HBr) = 0.1 M * 10.5 mL = 1.05 mmol

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

mol((CH3)3N) = 0.07 M * 30 mL = 2.1 mmol

We have:

mol(HBr) = 1.05 mmol

mol((CH3)3N) = 2.1 mmol

1.05 mmol of both will react

excess (CH3)3N remaining = 1.05 mmol

Volume of Solution = 10.5 + 30 = 40.5 mL

[(CH3)3N] = 1.05 mmol/40.5 mL = 0.0259 M

[(CH3)3NH+] = 1.05 mmol/40.5 mL = 0.0259 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 {2.593*10^-2/2.593*10^-2}

= 4.194

use:

PH = 14 - pOH

= 14 - 4.1938

= 9.8062

Answer: 9.81

d)when 15.8 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 15.8 mL

M((CH3)3N) = 0.07 M

V((CH3)3N) = 30 mL

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

mol(HBr) = 0.1 M * 15.8 mL = 1.58 mmol

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

mol((CH3)3N) = 0.07 M * 30 mL = 2.1 mmol

We have:

mol(HBr) = 1.58 mmol

mol((CH3)3N) = 2.1 mmol

1.58 mmol of both will react

excess (CH3)3N remaining = 0.52 mmol

Volume of Solution = 15.8 + 30 = 45.8 mL

[(CH3)3N] = 0.52 mmol/45.8 mL = 0.0114 M

[(CH3)3NH+] = 1.58 mmol/45.8 mL = 0.0345 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 {3.45*10^-2/1.135*10^-2}

= 4.676

use:

PH = 14 - pOH

= 14 - 4.6765

= 9.3235

Answer: 9.32

e)when 21.0 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 21 mL

M((CH3)3N) = 0.07 M

V((CH3)3N) = 30 mL

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

mol(HBr) = 0.1 M * 21 mL = 2.1 mmol

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

mol((CH3)3N) = 0.07 M * 30 mL = 2.1 mmol

We have:

mol(HBr) = 2.1 mmol

mol((CH3)3N) = 2.1 mmol

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

(CH3)3NH+ here is strong acid

(CH3)3NH+ formed = 2.1 mmol

Volume of Solution = 21 + 30 = 51 mL

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

concentration of(CH3)3NH+,c = 2.1 mmol/51 mL = 0.0412 M

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

4.118*10^-2 0 0

4.118*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)*4.118*10^-2) = 2.536*10^-6

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

so, x = 2.536*10^-6 M

[H+] = x = 2.536*10^-6 M

use:

pH = -log [H+]

= -log (2.536*10^-6)

= 5.5958

Answer: 5.60

f)when 27.3 mL of HBr is added

Given:

M(HBr) = 0.1 M

V(HBr) = 27.3 mL

M((CH3)3N) = 0.07 M

V((CH3)3N) = 30 mL

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

mol(HBr) = 0.1 M * 27.3 mL = 2.73 mmol

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

mol((CH3)3N) = 0.07 M * 30 mL = 2.1 mmol

We have:

mol(HBr) = 2.73 mmol

mol((CH3)3N) = 2.1 mmol

2.1 mmol of both will react

excess HBr remaining = 0.63 mmol

Volume of Solution = 27.3 + 30 = 57.3 mL

[H+] = 0.63 mmol/57.3 mL = 0.011 M

use:

pH = -log [H+]

= -log (1.099*10^-2)

= 1.9588

Answer: 1.96