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
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
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