Consider the titration of 60.0 mL of 0.0400 M C6H5NH2 (a weak base; Kb = 4.30e-10) with 0.100 M HBr. Calculate the pH after the following volumes of titrant have been added:
a)when 0.0 mL of HBr is added
C6H5NH2 dissociates as:
C6H5NH2 +H2O
-----> C6H5NH3+
+ OH-
4*10^-2
0 0
4*10^-2-x
x x
Kb = [C6H5NH3+][OH-]/[C6H5NH2]
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 ((4.3*10^-10)*4*10^-2) = 4.147*10^-6
since c is much greater than x, our assumption is correct
so, x = 4.147*10^-6 M
So, [OH-] = x = 4.147*10^-6 M
use:
pOH = -log [OH-]
= -log (4.147*10^-6)
= 5.3822
use:
PH = 14 - pOH
= 14 - 5.3822
= 8.6178
Answer: 8.62
b)when 6.0 mL of HBr is added
Given:
M(HBr) = 0.1 M
V(HBr) = 6 mL
M(C6H5NH2) = 0.04 M
V(C6H5NH2) = 60 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.1 M * 6 mL = 0.6 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.04 M * 60 mL = 2.4 mmol
We have:
mol(HBr) = 0.6 mmol
mol(C6H5NH2) = 2.4 mmol
0.6 mmol of both will react
excess C6H5NH2 remaining = 1.8 mmol
Volume of Solution = 6 + 60 = 66 mL
[C6H5NH2] = 1.8 mmol/66 mL = 0.0273 M
[C6H5NH3+] = 0.6 mmol/66 mL = 0.0091 M
They form basic buffer
base is C6H5NH2
conjugate acid is C6H5NH3+
Kb = 4.3*10^-10
pKb = - log (Kb)
= - log(4.3*10^-10)
= 9.367
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 9.367+ log {9.091*10^-3/2.727*10^-2}
= 8.889
use:
PH = 14 - pOH
= 14 - 8.8894
= 5.1106
Answer: 5.11
c)when 12.0 mL of HBr is added
Given:
M(HBr) = 0.1 M
V(HBr) = 12 mL
M(C6H5NH2) = 0.04 M
V(C6H5NH2) = 60 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.1 M * 12 mL = 1.2 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.04 M * 60 mL = 2.4 mmol
We have:
mol(HBr) = 1.2 mmol
mol(C6H5NH2) = 2.4 mmol
1.2 mmol of both will react
excess C6H5NH2 remaining = 1.2 mmol
Volume of Solution = 12 + 60 = 72 mL
[C6H5NH2] = 1.2 mmol/72 mL = 0.0167 M
[C6H5NH3+] = 1.2 mmol/72 mL = 0.0167 M
They form basic buffer
base is C6H5NH2
conjugate acid is C6H5NH3+
Kb = 4.3*10^-10
pKb = - log (Kb)
= - log(4.3*10^-10)
= 9.367
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 9.367+ log {1.667*10^-2/1.667*10^-2}
= 9.367
use:
PH = 14 - pOH
= 14 - 9.3665
= 4.6335
Answer: 4.63
d)when 18.0 mL of HBr is added
Given:
M(HBr) = 0.1 M
V(HBr) = 18 mL
M(C6H5NH2) = 0.04 M
V(C6H5NH2) = 60 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.1 M * 18 mL = 1.8 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.04 M * 60 mL = 2.4 mmol
We have:
mol(HBr) = 1.8 mmol
mol(C6H5NH2) = 2.4 mmol
1.8 mmol of both will react
excess C6H5NH2 remaining = 0.6 mmol
Volume of Solution = 18 + 60 = 78 mL
[C6H5NH2] = 0.6 mmol/78 mL = 0.0077 M
[C6H5NH3+] = 1.8 mmol/78 mL = 0.0231 M
They form basic buffer
base is C6H5NH2
conjugate acid is C6H5NH3+
Kb = 4.3*10^-10
pKb = - log (Kb)
= - log(4.3*10^-10)
= 9.367
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 9.367+ log {2.308*10^-2/7.692*10^-3}
= 9.844
use:
PH = 14 - pOH
= 14 - 9.8437
= 4.1563
Answer: 4.16
e)when 24.0 mL of HBr is added
Given:
M(HBr) = 0.1 M
V(HBr) = 24 mL
M(C6H5NH2) = 0.04 M
V(C6H5NH2) = 60 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.1 M * 24 mL = 2.4 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.04 M * 60 mL = 2.4 mmol
We have:
mol(HBr) = 2.4 mmol
mol(C6H5NH2) = 2.4 mmol
2.4 mmol of both will react to form C6H5NH3+ and H2O
C6H5NH3+ here is strong acid
C6H5NH3+ formed = 2.4 mmol
Volume of Solution = 24 + 60 = 84 mL
Ka of C6H5NH3+ = Kw/Kb = 1.0E-14/4.3E-10 = 2.326*10^-5
concentration ofC6H5NH3+,c = 2.4 mmol/84 mL = 0.0286 M
C6H5NH3+ + H2O
-----> C6H5NH2 +
H+
2.857*10^-2
0 0
2.857*10^-2-x
x x
Ka = [H+][C6H5NH2]/[C6H5NH3+]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((2.326*10^-5)*2.857*10^-2) = 8.151*10^-4
since x is comparable c, our assumption is not correct
we need to solve this using Quadratic equation
Ka = x*x/(c-x)
2.326*10^-5 = x^2/(2.857*10^-2-x)
6.645*10^-7 - 2.326*10^-5 *x = x^2
x^2 + 2.326*10^-5 *x-6.645*10^-7 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 2.326*10^-5
c = -6.645*10^-7
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 = 2.658*10^-6
roots are :
x = 8.036*10^-4 and x = -8.268*10^-4
since x can't be negative, the possible value of x is
x = 8.036*10^-4
[H+] = x = 8.036*10^-4 M
use:
pH = -log [H+]
= -log (8.036*10^-4)
= 3.095
Answer: 3.10
f)when 31.2 mL of HBr is added
Given:
M(HBr) = 0.1 M
V(HBr) = 31.2 mL
M(C6H5NH2) = 0.04 M
V(C6H5NH2) = 60 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.1 M * 31.2 mL = 3.12 mmol
mol(C6H5NH2) = M(C6H5NH2) * V(C6H5NH2)
mol(C6H5NH2) = 0.04 M * 60 mL = 2.4 mmol
We have:
mol(HBr) = 3.12 mmol
mol(C6H5NH2) = 2.4 mmol
2.4 mmol of both will react
excess HBr remaining = 0.72 mmol
Volume of Solution = 31.2 + 60 = 91.2 mL
[H+] = 0.72 mmol/91.2 mL = 0.0079 M
use:
pH = -log [H+]
= -log (7.895*10^-3)
= 2.1027
Answer: 2.10
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