Question

Consider the titration of 80.0 mL of 0.0200 M (CH3)2NH (a weak base; Kb = 0.000540) with 0.100 M HBrO4. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL

pH =

(b) 4.0 mL

pH =

(c) 8.0 mL

pH =

(d) 12.0 mL

pH =

(e) 16.0 mL

pH =

(f) 30.4 mL

pH =

please help :) I'm super confused

Answer 1

1)when 0.0 mL of HBrO4 is added

(CH3)2NH dissociates as:

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

2*10^-2 0 0

2*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)*2*10^-2) = 3.286*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/(2*10^-2-x)

1.08*10^-5 - 5.4*10^-4 *x = x^2

x^2 + 5.4*10^-4 *x-1.08*10^-5 = 0

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

a = 1

b = 5.4*10^-4

c = -1.08*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 = 4.349*10^-5

roots are :

x = 3.027*10^-3 and x = -3.567*10^-3

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

x = 3.027*10^-3

So, [OH-] = x = 3.027*10^-3 M

use:

pOH = -log [OH-]

= -log (3.027*10^-3)

= 2.5189

use:

PH = 14 - pOH

= 14 - 2.5189

= 11.4811

Answer: 11.48

2)when 4.0 mL of HBrO4 is added

Given:

M(HBrO4) = 0.1 M

V(HBrO4) = 4 mL

M((CH3)2NH) = 0.02 M

V((CH3)2NH) = 80 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.1 M * 4 mL = 0.4 mmol

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

mol((CH3)2NH) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HBrO4) = 0.4 mmol

mol((CH3)2NH) = 1.6 mmol

0.4 mmol of both will react

excess (CH3)2NH remaining = 1.2 mmol

Volume of Solution = 4 + 80 = 84 mL

[(CH3)2NH] = 1.2 mmol/84 mL = 0.0143 M

[(CH3)2NH2+] = 0.4 mmol/84 mL = 0.0048 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 {4.762*10^-3/1.429*10^-2}

= 2.79

use:

PH = 14 - pOH

= 14 - 2.7905

= 11.2095

Answer: 11.21

3)when 8.0 mL of HBrO4 is added

Given:

M(HBrO4) = 0.1 M

V(HBrO4) = 8 mL

M((CH3)2NH) = 0.02 M

V((CH3)2NH) = 80 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.1 M * 8 mL = 0.8 mmol

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

mol((CH3)2NH) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HBrO4) = 0.8 mmol

mol((CH3)2NH) = 1.6 mmol

0.8 mmol of both will react

excess (CH3)2NH remaining = 0.8 mmol

Volume of Solution = 8 + 80 = 88 mL

[(CH3)2NH] = 0.8 mmol/88 mL = 0.0091 M

[(CH3)2NH2+] = 0.8 mmol/88 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^-3/9.091*10^-3}

= 3.268

use:

PH = 14 - pOH

= 14 - 3.2676

= 10.7324

Answer: 10.73

4)when 12.0 mL of HBrO4 is added

Given:

M(HBrO4) = 0.1 M

V(HBrO4) = 12 mL

M((CH3)2NH) = 0.02 M

V((CH3)2NH) = 80 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.1 M * 12 mL = 1.2 mmol

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

mol((CH3)2NH) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HBrO4) = 1.2 mmol

mol((CH3)2NH) = 1.6 mmol

1.2 mmol of both will react

excess (CH3)2NH remaining = 0.4 mmol

Volume of Solution = 12 + 80 = 92 mL

[(CH3)2NH] = 0.4 mmol/92 mL = 0.0043 M

[(CH3)2NH2+] = 1.2 mmol/92 mL = 0.013 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 {1.304*10^-2/4.348*10^-3}

= 3.745

use:

PH = 14 - pOH

= 14 - 3.7447

= 10.2553

Answer: 10.26

5)when 16.0 mL of HBrO4 is added

Given:

M(HBrO4) = 0.1 M

V(HBrO4) = 16 mL

M((CH3)2NH) = 0.02 M

V((CH3)2NH) = 80 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.1 M * 16 mL = 1.6 mmol

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

mol((CH3)2NH) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HBrO4) = 1.6 mmol

mol((CH3)2NH) = 1.6 mmol

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

(CH3)2NH2+ here is strong acid

(CH3)2NH2+ formed = 1.6 mmol

Volume of Solution = 16 + 80 = 96 mL

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

concentration of(CH3)2NH2+,c = 1.6 mmol/96 mL = 0.0167 M

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

1.667*10^-2 0 0

1.667*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)*1.667*10^-2) = 5.556*10^-7

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

so, x = 5.556*10^-7 M

[H+] = x = 5.556*10^-7 M

use:

pH = -log [H+]

= -log (5.556*10^-7)

= 6.2553

Answer: 6.26

6)when 30.4 mL of HBrO4 is added

Given:

M(HBrO4) = 0.1 M

V(HBrO4) = 30.4 mL

M((CH3)2NH) = 0.02 M

V((CH3)2NH) = 80 mL

mol(HBrO4) = M(HBrO4) * V(HBrO4)

mol(HBrO4) = 0.1 M * 30.4 mL = 3.04 mmol

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

mol((CH3)2NH) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HBrO4) = 3.04 mmol

mol((CH3)2NH) = 1.6 mmol

1.6 mmol of both will react

excess HBrO4 remaining = 1.44 mmol

Volume of Solution = 30.4 + 80 = 110.4 mL

[H+] = 1.44 mmol/110.4 mL = 0.013 M

use:

pH = -log [H+]

= -log (1.304*10^-2)

= 1.8846

Answer: 1.88