Question

Consider the titration of 60.0 mL of 0.0400 M (CH₃)₂NH (a weak base; K_b = 0.000540) with 0.100 M HCl. 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) 40.8 mL pH =

Answer 1

a)when 0.0 mL of HCl is added

(CH3)2NH dissociates as:

(CH3)2NH +H2O -----> (CH3)2NH2+ + OH-

4*10^-2 0 0

4*10^-2-x x x

Kb = [(CH3)2NH2+][OH-]/[(CH3)2NH]

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 ((5.4*10^-4)*4*10^-2) = 4.648*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)

5.4*10^-4 = x^2/(4*10^-2-x)

2.16*10^-5 - 5.4*10^-4 *x = x^2

x^2 + 5.4*10^-4 *x-2.16*10^-5 = 0

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

a = 1

b = 5.4*10^-4

c = -2.16*10^-5

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 = 8.669*10^-5

roots are :

x = 4.385*10^-3 and x = -4.925*10^-3

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

x = 4.385*10^-3

So, [OH-] = x = 4.385*10^-3 M

use:

pOH = -log [OH-]

= -log (4.385*10^-3)

= 2.358

use:

PH = 14 - pOH

= 14 - 2.358

= 11.642

Answer: 11.64

b)when 6.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 6 mL

M((CH3)2NH) = 0.04 M

V((CH3)2NH) = 60 mL

mol(HCl) = M(HCl) * V(HCl)

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

mol((CH3)2NH) = M((CH3)2NH) * V((CH3)2NH)

mol((CH3)2NH) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HCl) = 0.6 mmol

mol((CH3)2NH) = 2.4 mmol

0.6 mmol of both will react

excess (CH3)2NH remaining = 1.8 mmol

Volume of Solution = 6 + 60 = 66 mL

[(CH3)2NH] = 1.8 mmol/66 mL = 0.0273 M

[(CH3)2NH2+] = 0.6 mmol/66 mL = 0.0091 M

They form basic buffer

base is (CH3)2NH

conjugate acid is (CH3)2NH2+

Kb = 5.4*10^-4

pKb = - log (Kb)

= - log(5.4*10^-4)

= 3.268

use:

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

= 3.268+ log {9.091*10^-3/2.727*10^-2}

= 2.79

use:

PH = 14 - pOH

= 14 - 2.7905

= 11.2095

Answer: 11.21

c)when 12.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 12 mL

M((CH3)2NH) = 0.04 M

V((CH3)2NH) = 60 mL

mol(HCl) = M(HCl) * V(HCl)

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

mol((CH3)2NH) = M((CH3)2NH) * V((CH3)2NH)

mol((CH3)2NH) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HCl) = 1.2 mmol

mol((CH3)2NH) = 2.4 mmol

1.2 mmol of both will react

excess (CH3)2NH remaining = 1.2 mmol

Volume of Solution = 12 + 60 = 72 mL

[(CH3)2NH] = 1.2 mmol/72 mL = 0.0167 M

[(CH3)2NH2+] = 1.2 mmol/72 mL = 0.0167 M

They form basic buffer

base is (CH3)2NH

conjugate acid is (CH3)2NH2+

Kb = 5.4*10^-4

pKb = - log (Kb)

= - log(5.4*10^-4)

= 3.268

use:

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

= 3.268+ log {1.667*10^-2/1.667*10^-2}

= 3.268

use:

PH = 14 - pOH

= 14 - 3.2676

= 10.7324

Answer: 10.73

d)when 18.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 18 mL

M((CH3)2NH) = 0.04 M

V((CH3)2NH) = 60 mL

mol(HCl) = M(HCl) * V(HCl)

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

mol((CH3)2NH) = M((CH3)2NH) * V((CH3)2NH)

mol((CH3)2NH) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HCl) = 1.8 mmol

mol((CH3)2NH) = 2.4 mmol

1.8 mmol of both will react

excess (CH3)2NH remaining = 0.6 mmol

Volume of Solution = 18 + 60 = 78 mL

[(CH3)2NH] = 0.6 mmol/78 mL = 0.0077 M

[(CH3)2NH2+] = 1.8 mmol/78 mL = 0.0231 M

They form basic buffer

base is (CH3)2NH

conjugate acid is (CH3)2NH2+

Kb = 5.4*10^-4

pKb = - log (Kb)

= - log(5.4*10^-4)

= 3.268

use:

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

= 3.268+ log {2.308*10^-2/7.692*10^-3}

= 3.745

use:

PH = 14 - pOH

= 14 - 3.7447

= 10.2553

Answer: 10.26

e)when 24.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 24 mL

M((CH3)2NH) = 0.04 M

V((CH3)2NH) = 60 mL

mol(HCl) = M(HCl) * V(HCl)

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

mol((CH3)2NH) = M((CH3)2NH) * V((CH3)2NH)

mol((CH3)2NH) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HCl) = 2.4 mmol

mol((CH3)2NH) = 2.4 mmol

2.4 mmol of both will react to form (CH3)2NH2+ and H2O

(CH3)2NH2+ here is strong acid

(CH3)2NH2+ formed = 2.4 mmol

Volume of Solution = 24 + 60 = 84 mL

Ka of (CH3)2NH2+ = Kw/Kb = 1.0E-14/5.4E-4 = 1.852*10^-11

concentration of(CH3)2NH2+,c = 2.4 mmol/84 mL = 0.0286 M

(CH3)2NH2+ + H2O -----> (CH3)2NH + H+

2.857*10^-2 0 0

2.857*10^-2-x x x

Ka = [H+][(CH3)2NH]/[(CH3)2NH2+]

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 ((1.852*10^-11)*2.857*10^-2) = 7.274*10^-7

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

so, x = 7.274*10^-7 M

[H+] = x = 7.274*10^-7 M

use:

pH = -log [H+]

= -log (7.274*10^-7)

= 6.1382

Answer: 6.14

f)when 40.8 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 40.8 mL

M((CH3)2NH) = 0.04 M

V((CH3)2NH) = 60 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 40.8 mL = 4.08 mmol

mol((CH3)2NH) = M((CH3)2NH) * V((CH3)2NH)

mol((CH3)2NH) = 0.04 M * 60 mL = 2.4 mmol

We have:

mol(HCl) = 4.08 mmol

mol((CH3)2NH) = 2.4 mmol

2.4 mmol of both will react

excess HCl remaining = 1.68 mmol

Volume of Solution = 40.8 + 60 = 100.8 mL

[H+] = 1.68 mmol/100.8 mL = 0.0167 M

use:

pH = -log [H+]

= -log (1.667*10^-2)

= 1.7782

Answer: 1.78