Question

38.

Consider the titration of 80.0 mL of 0.0200 M C₆H₅NH₂ (a weak base; K_b = 4.30e-10) with 0.100 M HNO₃. 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) 24.0 mL pH = _________

Answer 1

a)when 0.0 mL of HNO3 is added

C6H5NH2 dissociates as:

C6H5NH2 +H2O -----> C6H5NH3+ + OH-

2*10^-2 0 0

2*10^-2-x x x

Kb = [C6H5NH3+][OH-]/[C6H5NH2]

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 ((4.3*10^-10)*2*10^-2) = 2.933*10^-6

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

so, x = 2.933*10^-6 M

So, [OH-] = x = 2.933*10^-6 M

use:

pOH = -log [OH-]

= -log (2.933*10^-6)

= 5.5328

use:

PH = 14 - pOH

= 14 - 5.5328

= 8.4672

Answer: 8.47

b)when 4.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 4 mL

M(C6H5NH2) = 0.02 M

V(C6H5NH2) = 80 mL

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

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

mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)

mol(C6H5NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 0.4 mmol

mol(C6H5NH2) = 1.6 mmol

0.4 mmol of both will react

excess C6H5NH2 remaining = 1.2 mmol

Volume of Solution = 4 + 80 = 84 mL

[C6H5NH2] = 1.2 mmol/84 mL = 0.0143 M

[C6H5NH3+] = 0.4 mmol/84 mL = 0.0048 M

They form basic buffer

base is C6H5NH2

conjugate acid is C6H5NH3+

Kb = 4.3*10^-10

pKb = - log (Kb)

= - log(4.3*10^-10)

= 9.367

use:

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

= 9.367+ log {4.762*10^-3/1.429*10^-2}

= 8.889

use:

PH = 14 - pOH

= 14 - 8.8894

= 5.1106

Answer: 5.11

c)when 8.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 8 mL

M(C6H5NH2) = 0.02 M

V(C6H5NH2) = 80 mL

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

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

mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)

mol(C6H5NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 0.8 mmol

mol(C6H5NH2) = 1.6 mmol

0.8 mmol of both will react

excess C6H5NH2 remaining = 0.8 mmol

Volume of Solution = 8 + 80 = 88 mL

[C6H5NH2] = 0.8 mmol/88 mL = 0.0091 M

[C6H5NH3+] = 0.8 mmol/88 mL = 0.0091 M

They form basic buffer

base is C6H5NH2

conjugate acid is C6H5NH3+

Kb = 4.3*10^-10

pKb = - log (Kb)

= - log(4.3*10^-10)

= 9.367

use:

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

= 9.367+ log {9.091*10^-3/9.091*10^-3}

= 9.367

use:

PH = 14 - pOH

= 14 - 9.3665

= 4.6335

Answer: 4.63

d)when 12.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 12 mL

M(C6H5NH2) = 0.02 M

V(C6H5NH2) = 80 mL

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

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

mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)

mol(C6H5NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 1.2 mmol

mol(C6H5NH2) = 1.6 mmol

1.2 mmol of both will react

excess C6H5NH2 remaining = 0.4 mmol

Volume of Solution = 12 + 80 = 92 mL

[C6H5NH2] = 0.4 mmol/92 mL = 0.0043 M

[C6H5NH3+] = 1.2 mmol/92 mL = 0.013 M

They form basic buffer

base is C6H5NH2

conjugate acid is C6H5NH3+

Kb = 4.3*10^-10

pKb = - log (Kb)

= - log(4.3*10^-10)

= 9.367

use:

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

= 9.367+ log {1.304*10^-2/4.348*10^-3}

= 9.844

use:

PH = 14 - pOH

= 14 - 9.8437

= 4.1563

Answer: 4.16

e)when 16.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 16 mL

M(C6H5NH2) = 0.02 M

V(C6H5NH2) = 80 mL

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

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

mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)

mol(C6H5NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 1.6 mmol

mol(C6H5NH2) = 1.6 mmol

1.6 mmol of both will react to form C6H5NH3+ and H2O

C6H5NH3+ here is strong acid

C6H5NH3+ formed = 1.6 mmol

Volume of Solution = 16 + 80 = 96 mL

Ka of C6H5NH3+ = Kw/Kb = 1.0E-14/4.3E-10 = 2.326*10^-5

concentration ofC6H5NH3+,c = 1.6 mmol/96 mL = 0.0167 M

C6H5NH3+ + H2O -----> C6H5NH2 + H+

1.667*10^-2 0 0

1.667*10^-2-x x x

Ka = [H+][C6H5NH2]/[C6H5NH3+]

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 ((2.326*10^-5)*1.667*10^-2) = 6.226*10^-4

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

2.326*10^-5 = x^2/(1.667*10^-2-x)

3.876*10^-7 - 2.326*10^-5 *x = x^2

x^2 + 2.326*10^-5 *x-3.876*10^-7 = 0

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

a = 1

b = 2.326*10^-5

c = -3.876*10^-7

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 = 1.551*10^-6

roots are :

x = 6.111*10^-4 and x = -6.343*10^-4

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

x = 6.111*10^-4

[H+] = x = 6.111*10^-4 M

use:

pH = -log [H+]

= -log (6.111*10^-4)

= 3.2139

Answer: 3.21

f)when 24.0 mL of HNO3 is added

Given:

M(HNO3) = 0.1 M

V(HNO3) = 24 mL

M(C6H5NH2) = 0.02 M

V(C6H5NH2) = 80 mL

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

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

mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)

mol(C6H5NH2) = 0.02 M * 80 mL = 1.6 mmol

We have:

mol(HNO3) = 2.4 mmol

mol(C6H5NH2) = 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