Question

2. A 35.00-ml sample of a 0.0870 M solution of ethylamine, C2H5NH2 (Kb = 5.6 x 10), is titrated with 0.150 M HCI. (a) What is the equivalence volume of HCI? (b) Calculate the pH of the solution: (1) prior to the start of the titration. (ii) after addition of 10.15 mL of the 0.150 M HCI. (iii) after addition of 14.50 mL of the 0.150 M HCI. (iv) after addition of 6.00 mL of the HCl beyond the equivalence point.

Answer 1

a)

find the volume of HCl used to reach equivalence point

M(C2H5NH2)*V(C2H5NH2) =M(HCl)*V(HCl)

0.087 M *35.0 mL = 0.15M *V(HCl)

V(HCl) = 20.3 mL

Given:

M(HCl) = 0.15 M

V(HCl) = 20.3 mL

M(C2H5NH2) = 0.087 M

V(C2H5NH2) = 35 mL

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

mol(HCl) = 0.15 M * 20.3 mL = 3.045 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)

mol(C2H5NH2) = 0.087 M * 35 mL = 3.045 mmol

We have:

mol(HCl) = 3.045 mmol

mol(C2H5NH2) = 3.045 mmol

3.045 mmol of both will react to form C2H5NH3+ and H2O

C2H5NH3+ here is strong acid

C2H5NH3+ formed = 3.045 mmol

Volume of Solution = 20.3 + 35 = 55.3 mL

Ka of C2H5NH3+ = Kw/Kb = 1.0E-14/5.6E-4 = 1.786*10^-11

concentration ofC2H5NH3+,c = 3.045 mmol/55.3 mL = 0.0551 M

C2H5NH3+ + H2O -----> C2H5NH2 + H+

5.506*10^-2 0 0

5.506*10^-2-x x x

Ka = [H+][C2H5NH2]/[C2H5NH3+]

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 ((1.786*10^-11)*5.506*10^-2) = 9.916*10^-7

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

so, x = 9.916*10^-7 M

[H+] = x = 9.916*10^-7 M

use:

pH = -log [H+]

= -log (9.916*10^-7)

= 6.0037

Answer: 6.00

i)when 0.0 mL of HCl is added

C2H5NH2 dissociates as:

C2H5NH2 +H2O -----> C2H5NH3+ + OH-

8.7*10^-2 0 0

8.7*10^-2-x x x

Kb = [C2H5NH3+][OH-]/[C2H5NH2]

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 ((5.6*10^-4)*8.7*10^-2) = 6.98*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)

5.6*10^-4 = x^2/(8.7*10^-2-x)

4.872*10^-5 - 5.6*10^-4 *x = x^2

x^2 + 5.6*10^-4 *x-4.872*10^-5 = 0

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

a = 1

b = 5.6*10^-4

c = -4.872*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 = 1.952*10^-4

roots are :

x = 6.706*10^-3 and x = -7.266*10^-3

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

x = 6.706*10^-3

So, [OH-] = x = 6.706*10^-3 M

use:

pOH = -log [OH-]

= -log (6.706*10^-3)

= 2.1736

use:

PH = 14 - pOH

= 14 - 2.1736

= 11.8264

Answer: 11.83

ii)when 10.15 mL of HCl is added

Given:

M(HCl) = 0.15 M

V(HCl) = 10.15 mL

M(C2H5NH2) = 0.087 M

V(C2H5NH2) = 35 mL

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

mol(HCl) = 0.15 M * 10.15 mL = 1.5225 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)

mol(C2H5NH2) = 0.087 M * 35 mL = 3.045 mmol

We have:

mol(HCl) = 1.5225 mmol

mol(C2H5NH2) = 3.045 mmol

1.5225 mmol of both will react

excess C2H5NH2 remaining = 1.5225 mmol

Volume of Solution = 10.15 + 35 = 45.15 mL

[C2H5NH2] = 1.5225 mmol/45.15 mL = 0.0337 M

[C2H5NH3+] = 1.5225 mmol/45.15 mL = 0.0337 M

They form basic buffer

base is C2H5NH2

conjugate acid is C2H5NH3+

Kb = 5.6*10^-4

pKb = - log (Kb)

= - log(5.6*10^-4)

= 3.252

use:

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

= 3.252+ log {3.372*10^-2/3.372*10^-2}

= 3.252

use:

PH = 14 - pOH

= 14 - 3.2518

= 10.7482

Answer: 10.75

iii)when 14.5 mL of HCl is added

Given:

M(HCl) = 0.15 M

V(HCl) = 14.5 mL

M(C2H5NH2) = 0.087 M

V(C2H5NH2) = 35 mL

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

mol(HCl) = 0.15 M * 14.5 mL = 2.175 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)

mol(C2H5NH2) = 0.087 M * 35 mL = 3.045 mmol

We have:

mol(HCl) = 2.175 mmol

mol(C2H5NH2) = 3.045 mmol

2.175 mmol of both will react

excess C2H5NH2 remaining = 0.87 mmol

Volume of Solution = 14.5 + 35 = 49.5 mL

[C2H5NH2] = 0.87 mmol/49.5 mL = 0.0176 M

[C2H5NH3+] = 2.175 mmol/49.5 mL = 0.0439 M

They form basic buffer

base is C2H5NH2

conjugate acid is C2H5NH3+

Kb = 5.6*10^-4

pKb = - log (Kb)

= - log(5.6*10^-4)

= 3.252

use:

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

= 3.252+ log {4.394*10^-2/1.758*10^-2}

= 3.65

use:

PH = 14 - pOH

= 14 - 3.6498

= 10.3502

Answer: 10.35

iv)Given:

M(HCl) = 0.15 M

V(HCl) = 26.3 mL

M(C2H5NH2) = 0.087 M

V(C2H5NH2) = 35 mL

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

mol(HCl) = 0.15 M * 26.3 mL = 3.945 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)

mol(C2H5NH2) = 0.087 M * 35 mL = 3.045 mmol

We have:

mol(HCl) = 3.945 mmol

mol(C2H5NH2) = 3.045 mmol

3.045 mmol of both will react

excess HCl remaining = 0.9 mmol

Volume of Solution = 26.3 + 35 = 61.3 mL

[H+] = 0.9 mmol/61.3 mL = 0.0147 M

use:

pH = -log [H+]

= -log (1.468*10^-2)

= 1.8332

Answer: 1.83