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 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
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