Question

A 112.6 mL sample of 0.110 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. You may want to reference (Pages 682 - 686)Section 16.9 while completing this problem.

Part A 0.0 mL Express the pH to two decimal places. pH = nothing SubmitRequest Answer Part B 23.4 mL Express the pH to two decimal places. pH = nothing SubmitRequest Answer Part C 46.7 mL Express the pH to two decimal places. pH = nothing SubmitRequest Answer Part D 70.1 mL Express the pH to two decimal places.

Answer 1

A)when 0.0 mL of HNO3 is added
CH3NH2 dissociates as:

CH3NH2 +H2O     ----->     CH3NH3+   +   OH-
0.11                   0         0
0.11-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)*0.11) = 6.38*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/(0.11-x)
4.07*10^-5 - 3.7*10^-4 *x = x^2
x^2 + 3.7*10^-4 *x-4.07*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 3.7*10^-4
c = -4.07*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 = 1.629*10^-4

roots are :
x = 6.197*10^-3 and x = -6.567*10^-3

since x can't be negative, the possible value of x is
x = 6.197*10^-3

So, [OH-] = x = 6.197*10^-3 M


use:
pOH = -log [OH-]
= -log (6.197*10^-3)
= 2.2078


use:
PH = 14 - pOH
= 14 - 2.2078
= 11.7922
Answer: 11.79

B)when 23.4 mL of HNO3 is added
Given:
M(HNO3) = 0.265 M
V(HNO3) = 23.4 mL
M(CH3NH2) = 0.11 M
V(CH3NH2) = 112.6 mL


mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.265 M * 23.4 mL = 6.201 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)
mol(CH3NH2) = 0.11 M * 112.6 mL = 12.386 mmol



We have:
mol(HNO3) = 6.201 mmol
mol(CH3NH2) = 12.386 mmol

6.201 mmol of both will react
excess CH3NH2 remaining = 6.185 mmol
Volume of Solution = 23.4 + 112.6 = 136 mL
[CH3NH2] = 6.185 mmol/136 mL = 0.0455 M
[CH3NH3+] = 6.201 mmol/136 mL = 0.0456 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 {4.56*10^-2/4.548*10^-2}
= 3.433

use:
PH = 14 - pOH
= 14 - 3.4329
= 10.5671

Answer: 10.57

C)when 46.7 mL of HNO3 is added
Given:
M(HNO3) = 0.265 M
V(HNO3) = 46.7 mL
M(CH3NH2) = 0.11 M
V(CH3NH2) = 112.6 mL


mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.265 M * 46.7 mL = 12.3755 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)
mol(CH3NH2) = 0.11 M * 112.6 mL = 12.386 mmol



We have:
mol(HNO3) = 12.3755 mmol
mol(CH3NH2) = 12.386 mmol

12.3755 mmol of both will react
excess CH3NH2 remaining = 0.0105 mmol
Volume of Solution = 46.7 + 112.6 = 159.3 mL
[CH3NH2] = 0.0105 mmol/159.3 mL = 0.0001 M
[CH3NH3+] = 12.3755 mmol/159.3 mL = 0.0777 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 {7.769*10^-2/6.591*10^-5}
= 6.503

use:
PH = 14 - pOH
= 14 - 6.5032
= 7.4968

Answer: 7.50

D)when 70.1 mL of HNO3 is added
Given:
M(HNO3) = 0.265 M
V(HNO3) = 70.1 mL
M(CH3NH2) = 0.11 M
V(CH3NH2) = 112.6 mL


mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.265 M * 70.1 mL = 18.5765 mmol

mol(CH3NH2) = M(CH3NH2) * V(CH3NH2)
mol(CH3NH2) = 0.11 M * 112.6 mL = 12.386 mmol



We have:
mol(HNO3) = 18.5765 mmol
mol(CH3NH2) = 12.386 mmol

12.386 mmol of both will react
excess HNO3 remaining = 6.1905 mmol
Volume of Solution = 70.1 + 112.6 = 182.7 mL
[H+] = 6.1905 mmol/182.7 mL = 0.0339 M

use:
pH = -log [H+]
= -log (3.388*10^-2)
= 1.47
Answer: 1.47