Question

A 109.2 mL sample of 0.105 M methylamine (CH3NH2;Kb=3.7×10−4) is titrated with 0.250 M HNO3. Calculate the pH after the addition of each of the following volumes of acid.

a.0.0 mL Express the pH to two decimal places.

b. 22.9 mL

c.45.9ml

d.68.8ml

Answer 1

a)when 0.0 mL of HNO3 is added

CH3NH2 dissociates as:

CH3NH2 +H2O -----> CH3NH3+ + OH-

0.105 0 0

0.105-x x x

Kb = [CH3NH3+][OH-]/[CH3NH2]

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 ((3.7*10^-4)*0.105) = 6.233*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)

3.7*10^-4 = x^2/(0.105-x)

3.885*10^-5 - 3.7*10^-4 *x = x^2

x^2 + 3.7*10^-4 *x-3.885*10^-5 = 0

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

a = 1

b = 3.7*10^-4

c = -3.885*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 = 1.555*10^-4

roots are :

x = 6.051*10^-3 and x = -6.421*10^-3

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

x = 6.051*10^-3

So, [OH-] = x = 6.051*10^-3 M

use:

pOH = -log [OH-]

= -log (6.051*10^-3)

= 2.2182

use:

PH = 14 - pOH

= 14 - 2.2182

= 11.7818

Answer: 11.78

b)when 22.9 mL of HNO3 is added

Given:

M(HNO3) = 0.25 M

V(HNO3) = 22.9 mL

M(CH3NH2) = 0.105 M

V(CH3NH2) = 109.2 mL

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

mol(HNO3) = 0.25 M * 22.9 mL = 5.725 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.105 M * 109.2 mL = 11.466 mmol

We have:

mol(HNO3) = 5.725 mmol

mol(CH3NH2) = 11.466 mmol

5.725 mmol of both will react

excess CH3NH2 remaining = 5.741 mmol

Volume of Solution = 22.9 + 109.2 = 132.1 mL

[CH3NH2] = 5.741 mmol/132.1 mL = 0.0435 M

[CH3NH3+] = 5.725 mmol/132.1 mL = 0.0433 M

They form basic buffer

base is CH3NH2

conjugate acid is CH3NH3+

Kb = 3.7*10^-4

pKb = - log (Kb)

= - log(3.7*10^-4)

= 3.432

use:

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

= 3.432+ log {4.334*10^-2/4.346*10^-2}

= 3.431

use:

PH = 14 - pOH

= 14 - 3.4306

= 10.5694

Answer: 10.57

c)when 45.9 mL of HNO3 is added

Given:

M(HNO3) = 0.25 M

V(HNO3) = 45.9 mL

M(CH3NH2) = 0.105 M

V(CH3NH2) = 109.2 mL

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

mol(HNO3) = 0.25 M * 45.9 mL = 11.475 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.105 M * 109.2 mL = 11.466 mmol

We have:

mol(HNO3) = 11.475 mmol

mol(CH3NH2) = 11.466 mmol

11.466 mmol of both will react

excess HNO3 remaining = 0.009 mmol

Volume of Solution = 45.9 + 109.2 = 155.1 mL

[H+] = 0.009 mmol/155.1 mL = 0.0001 M

use:

pH = -log [H+]

= -log (5.803*10^-5)

= 4.2364

Answer: 4.24

d)when 68.8 mL of HNO3 is added

Given:

M(HNO3) = 0.25 M

V(HNO3) = 68.8 mL

M(CH3NH2) = 0.105 M

V(CH3NH2) = 109.2 mL

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

mol(HNO3) = 0.25 M * 68.8 mL = 17.2 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)

mol(CH3NH2) = 0.105 M * 109.2 mL = 11.466 mmol

We have:

mol(HNO3) = 17.2 mmol

mol(CH3NH2) = 11.466 mmol

11.466 mmol of both will react

excess HNO3 remaining = 5.734 mmol

Volume of Solution = 68.8 + 109.2 = 178 mL

[H+] = 5.734 mmol/178 mL = 0.0322 M

use:

pH = -log [H+]

= -log (3.221*10^-2)

= 1.492

Answer: 1.49