Question

Consider the titration of 20.0 mL of 0.0800 M H2NNH2 (a weak base; Kb = 1.30e-06) with 0.100 M HCl. 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) 20.8 mL pH = ?

Answer 1

a)when 0.0 mL of HCl is added

H2NNH2 dissociates as:

H2NNH2 +H2O -----> H2NNH3+ + OH-

8*10^-2 0 0

8*10^-2-x x x

Kb = [H2NNH3+][OH-]/[H2NNH2]

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 ((1.8*10^-5)*8*10^-2) = 1.2*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)

1.8*10^-5 = x^2/(8*10^-2-x)

1.44*10^-6 - 1.8*10^-5 *x = x^2

x^2 + 1.8*10^-5 *x-1.44*10^-6 = 0

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

a = 1

b = 1.8*10^-5

c = -1.44*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 = 5.76*10^-6

roots are :

x = 1.191*10^-3 and x = -1.209*10^-3

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

x = 1.191*10^-3

So, [OH-] = x = 1.191*10^-3 M

use:

pOH = -log [OH-]

= -log (1.191*10^-3)

= 2.9241

use:

PH = 14 - pOH

= 14 - 2.9241

= 11.0759

Answer: 11.08

b)when 4.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 4 mL

M(H2NNH2) = 0.08 M

V(H2NNH2) = 20 mL

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

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

mol(H2NNH2) = M(H2NNH2) * V(H2NNH2)

mol(H2NNH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HCl) = 0.4 mmol

mol(H2NNH2) = 1.6 mmol

0.4 mmol of both will react

excess H2NNH2 remaining = 1.2 mmol

Volume of Solution = 4 + 20 = 24 mL

[H2NNH2] = 1.2 mmol/24 mL = 0.05 M

[H2NNH3+] = 0.4 mmol/24 mL = 0.0167 M

They form basic buffer

base is H2NNH2

conjugate acid is H2NNH3+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {1.667*10^-2/5*10^-2}

= 4.268

use:

PH = 14 - pOH

= 14 - 4.2676

= 9.7324

Answer: 9.73

c)when 8.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 8 mL

M(H2NNH2) = 0.08 M

V(H2NNH2) = 20 mL

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

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

mol(H2NNH2) = M(H2NNH2) * V(H2NNH2)

mol(H2NNH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HCl) = 0.8 mmol

mol(H2NNH2) = 1.6 mmol

0.8 mmol of both will react

excess H2NNH2 remaining = 0.8 mmol

Volume of Solution = 8 + 20 = 28 mL

[H2NNH2] = 0.8 mmol/28 mL = 0.0286 M

[H2NNH3+] = 0.8 mmol/28 mL = 0.0286 M

They form basic buffer

base is H2NNH2

conjugate acid is H2NNH3+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {2.857*10^-2/2.857*10^-2}

= 4.745

use:

PH = 14 - pOH

= 14 - 4.7447

= 9.2553

Answer: 9.26

d)when 12.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 12 mL

M(H2NNH2) = 0.08 M

V(H2NNH2) = 20 mL

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

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

mol(H2NNH2) = M(H2NNH2) * V(H2NNH2)

mol(H2NNH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HCl) = 1.2 mmol

mol(H2NNH2) = 1.6 mmol

1.2 mmol of both will react

excess H2NNH2 remaining = 0.4 mmol

Volume of Solution = 12 + 20 = 32 mL

[H2NNH2] = 0.4 mmol/32 mL = 0.0125 M

[H2NNH3+] = 1.2 mmol/32 mL = 0.0375 M

They form basic buffer

base is H2NNH2

conjugate acid is H2NNH3+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {3.75*10^-2/1.25*10^-2}

= 5.222

use:

PH = 14 - pOH

= 14 - 5.2218

= 8.7782

Answer: 8.78

e)when 16.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 16 mL

M(H2NNH2) = 0.08 M

V(H2NNH2) = 20 mL

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

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

mol(H2NNH2) = M(H2NNH2) * V(H2NNH2)

mol(H2NNH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HCl) = 1.6 mmol

mol(H2NNH2) = 1.6 mmol

1.6 mmol of both will react to form H2NNH3+ and H2O

H2NNH3+ here is strong acid

H2NNH3+ formed = 1.6 mmol

Volume of Solution = 16 + 20 = 36 mL

Ka of H2NNH3+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10

concentration ofH2NNH3+,c = 1.6 mmol/36 mL = 0.0444 M

H2NNH3+ + H2O -----> H2NNH2 + H+

4.444*10^-2 0 0

4.444*10^-2-x x x

Ka = [H+][H2NNH2]/[H2NNH3+]

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 ((5.556*10^-10)*4.444*10^-2) = 4.969*10^-6

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

so, x = 4.969*10^-6 M

[H+] = x = 4.969*10^-6 M

use:

pH = -log [H+]

= -log (4.969*10^-6)

= 5.3037

Answer: 5.30

f)when 20.8 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 20.8 mL

M(H2NNH2) = 0.08 M

V(H2NNH2) = 20 mL

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

mol(HCl) = 0.1 M * 20.8 mL = 2.08 mmol

mol(H2NNH2) = M(H2NNH2) * V(H2NNH2)

mol(H2NNH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HCl) = 2.08 mmol

mol(H2NNH2) = 1.6 mmol

1.6 mmol of both will react

excess HCl remaining = 0.48 mmol

Volume of Solution = 20.8 + 20 = 40.8 mL

[H+] = 0.48 mmol/40.8 mL = 0.0118 M

use:

pH = -log [H+]

= -log (1.176*10^-2)

= 1.9294

Answer: 1.93