Question

Consider the titration of 20.0 mL of 0.0800 M C6H5NH2 (a weak base; Kb = 4.30e-10) with 0.100 M HIO4. 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) 22.4 mL

pH=

Answer 1

a)when 0.0 mL of HIO4 is added

C6H5NH2 dissociates as:

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

8*10^-2 0 0

8*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)*8*10^-2) = 5.865*10^-6

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

so, x = 5.865*10^-6 M

So, [OH-] = x = 5.865*10^-6 M

use:

pOH = -log [OH-]

= -log (5.865*10^-6)

= 5.2317

use:

PH = 14 - pOH

= 14 - 5.2317

= 8.7683

Answer: 8.77

b)when 4.0 mL of HIO4 is added

Given:

M(HIO4) = 0.1 M

V(HIO4) = 4 mL

M(C6H5NH2) = 0.08 M

V(C6H5NH2) = 20 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.1 M * 4 mL = 0.4 mmol

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

mol(C6H5NH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HIO4) = 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 + 20 = 24 mL

[C6H5NH2] = 1.2 mmol/24 mL = 0.05 M

[C6H5NH3+] = 0.4 mmol/24 mL = 0.0167 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 {1.667*10^-2/5*10^-2}

= 8.889

use:

PH = 14 - pOH

= 14 - 8.8894

= 5.1106

Answer: 5.11

c)when 8.0 mL of HIO4 is added

Given:

M(HIO4) = 0.1 M

V(HIO4) = 8 mL

M(C6H5NH2) = 0.08 M

V(C6H5NH2) = 20 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.1 M * 8 mL = 0.8 mmol

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

mol(C6H5NH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HIO4) = 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 + 20 = 28 mL

[C6H5NH2] = 0.8 mmol/28 mL = 0.0286 M

[C6H5NH3+] = 0.8 mmol/28 mL = 0.0286 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 {2.857*10^-2/2.857*10^-2}

= 9.367

use:

PH = 14 - pOH

= 14 - 9.3665

= 4.6335

Answer: 4.63

d)when 12.0 mL of HIO4 is added

Given:

M(HIO4) = 0.1 M

V(HIO4) = 12 mL

M(C6H5NH2) = 0.08 M

V(C6H5NH2) = 20 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.1 M * 12 mL = 1.2 mmol

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

mol(C6H5NH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HIO4) = 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 + 20 = 32 mL

[C6H5NH2] = 0.4 mmol/32 mL = 0.0125 M

[C6H5NH3+] = 1.2 mmol/32 mL = 0.0375 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 {3.75*10^-2/1.25*10^-2}

= 9.844

use:

PH = 14 - pOH

= 14 - 9.8437

= 4.1563

Answer: 4.16

e)when 16.0 mL of HIO4 is added

Given:

M(HIO4) = 0.1 M

V(HIO4) = 16 mL

M(C6H5NH2) = 0.08 M

V(C6H5NH2) = 20 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.1 M * 16 mL = 1.6 mmol

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

mol(C6H5NH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HIO4) = 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 + 20 = 36 mL

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

concentration ofC6H5NH3+,c = 1.6 mmol/36 mL = 0.0444 M

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

4.444*10^-2 0 0

4.444*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)*4.444*10^-2) = 1.017*10^-3

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/(4.444*10^-2-x)

1.034*10^-6 - 2.326*10^-5 *x = x^2

x^2 + 2.326*10^-5 *x-1.034*10^-6 = 0

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

a = 1

b = 2.326*10^-5

c = -1.034*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 = 4.135*10^-6

roots are :

x = 1.005*10^-3 and x = -1.028*10^-3

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

x = 1.005*10^-3

[H+] = x = 1.005*10^-3 M

use:

pH = -log [H+]

= -log (1.005*10^-3)

= 2.9978

Answer: 3.00

f)when 22.4 mL of HIO4 is added

Given:

M(HIO4) = 0.1 M

V(HIO4) = 22.4 mL

M(C6H5NH2) = 0.08 M

V(C6H5NH2) = 20 mL

mol(HIO4) = M(HIO4) * V(HIO4)

mol(HIO4) = 0.1 M * 22.4 mL = 2.24 mmol

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

mol(C6H5NH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HIO4) = 2.24 mmol

mol(C6H5NH2) = 1.6 mmol

1.6 mmol of both will react

excess HIO4 remaining = 0.64 mmol

Volume of Solution = 22.4 + 20 = 42.4 mL

[H+] = 0.64 mmol/42.4 mL = 0.0151 M

use:

pH = -log [H+]

= -log (1.509*10^-2)

= 1.8212

Answer: 1.82