Question

In: Chemistry

A 101.6 mL sample of 0.115 M methylamine (CH3NH2;Kb=3.7×10^−4) is titrated with 0.265 M HNO3. Calculate...

A 101.6 mL sample of 0.115 M methylamine (CH3NH2;Kb=3.7×10^−4) is titrated with 0.265 M HNO3. Calculate the pH after the addition of each of the following volumes of acid.

a) 0.0 mL b) 22.0 mL c) 44.1 mL d) 66.1 mL

Solutions

Expert Solution

1)when 0.0 mL of HNO3 is added

CH3NH2 dissociates as:

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

0.115 0 0

0.115-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)*0.115) = 6.523*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/(0.115-x)

4.255*10^-5 - 3.7*10^-4 *x = x^2

x^2 + 3.7*10^-4 *x-4.255*10^-5 = 0

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

a = 1

b = 3.7*10^-4

c = -4.255*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 = 1.703*10^-4

roots are :

x = 6.341*10^-3 and x = -6.711*10^-3

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

x = 6.341*10^-3

So, [OH-] = x = 6.341*10^-3 M

use:

pOH = -log [OH-]

= -log (6.341*10^-3)

= 2.1979

use:

PH = 14 - pOH

= 14 - 2.1979

= 11.8021

2)when 22.0 mL of HNO3 is added

Given:

M(HNO3) = 0.265 M

V(HNO3) = 22 mL

M(CH3NH2) = 0.115 M

V(CH3NH2) = 101.6 mL

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

mol(HNO3) = 0.265 M * 22 mL = 5.83 mmol

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

mol(CH3NH2) = 0.115 M * 101.6 mL = 11.684 mmol

We have:

mol(HNO3) = 5.83 mmol

mol(CH3NH2) = 11.684 mmol

5.83 mmol of both will react

excess CH3NH2 remaining = 5.854 mmol

Volume of Solution = 22 + 101.6 = 123.6 mL

[CH3NH2] = 5.854 mmol/123.6 mL = 0.0474 M

[CH3NH3+] = 5.83 mmol/123.6 mL = 0.0472 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 {4.717*10^-2/4.736*10^-2}

= 3.43

use:

PH = 14 - pOH

= 14 - 3.43

= 10.57

3)when 44.1 mL of HNO3 is added

Given:

M(HNO3) = 0.265 M

V(HNO3) = 44.1 mL

M(CH3NH2) = 0.115 M

V(CH3NH2) = 101.6 mL

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

mol(HNO3) = 0.265 M * 44.1 mL = 11.6865 mmol

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

mol(CH3NH2) = 0.115 M * 101.6 mL = 11.684 mmol

We have:

mol(HNO3) = 11.6865 mmol

mol(CH3NH2) = 11.684 mmol

11.684 mmol of both will react

excess HNO3 remaining = 0.0025 mmol

Volume of Solution = 44.1 + 101.6 = 145.7 mL

[H+] = 0.0025 mmol/145.7 mL = 0 M

use:

pH = -log [H+]

= -log (1.716*10^-5)

= 4.7655

4)when 66.1 mL of HNO3 is added

Given:

M(HNO3) = 0.265 M

V(HNO3) = 66.1 mL

M(CH3NH2) = 0.115 M

V(CH3NH2) = 101.6 mL

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

mol(HNO3) = 0.265 M * 66.1 mL = 17.5165 mmol

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

mol(CH3NH2) = 0.115 M * 101.6 mL = 11.684 mmol

We have:

mol(HNO3) = 17.5165 mmol

mol(CH3NH2) = 11.684 mmol

11.684 mmol of both will react

excess HNO3 remaining = 5.8325 mmol

Volume of Solution = 66.1 + 101.6 = 167.7 mL

[H+] = 5.8325 mmol/167.7 mL = 0.0348 M

use:

pH = -log [H+]

= -log (3.478*10^-2)

= 1.4587


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