Question

Consider the titration of 70.0 mL of 0.0300 M C₂H₅NH₂ (a weak base; K_b = 0.000640) with 0.100 M HI. Calculate the pH after the following volumes of titrant have been added:

(a) 0.0 mL pH =	(b) 5.3 mL pH =	(c) 10.5 mL pH =
(d) 15.8 mL pH =	(e) 21.0 mL pH =	(f) 23.1 mL pH =

Answer 1

1)when 0.0 mL of HI is added
C2H5NH2 dissociates as:

C2H5NH2 +H2O     ----->     C2H5NH3+   +   OH-
3*10^-2                   0         0
3*10^-2-x                 x         x


Kb = [C2H5NH3+][OH-]/[C2H5NH2]
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 ((6.4*10^-4)*3*10^-2) = 4.382*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)
6.4*10^-4 = x^2/(3*10^-2-x)
1.92*10^-5 - 6.4*10^-4 *x = x^2
x^2 + 6.4*10^-4 *x-1.92*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 6.4*10^-4
c = -1.92*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 = 7.721*10^-5

roots are :
x = 4.073*10^-3 and x = -4.713*10^-3

since x can't be negative, the possible value of x is
x = 4.073*10^-3

So, [OH-] = x = 4.073*10^-3 M


use:
pOH = -log [OH-]
= -log (4.073*10^-3)
= 2.39


use:
PH = 14 - pOH
= 14 - 2.39
= 11.61
Answer: 11.61

2)when 5.3 mL of HI is added
Given:
M(HI) = 0.1 M
V(HI) = 5.3 mL
M(C2H5NH2) = 0.03 M
V(C2H5NH2) = 70 mL


mol(HI) = M(HI) * V(HI)
mol(HI) = 0.1 M * 5.3 mL = 0.53 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)
mol(C2H5NH2) = 0.03 M * 70 mL = 2.1 mmol



We have:
mol(HI) = 0.53 mmol
mol(C2H5NH2) = 2.1 mmol

0.53 mmol of both will react
excess C2H5NH2 remaining = 1.57 mmol
Volume of Solution = 5.3 + 70 = 75.3 mL
[C2H5NH2] = 1.57 mmol/75.3 mL = 0.0208 M
[C2H5NH3+] = 0.53 mmol/75.3 mL = 0.007 M

They form basic buffer
base is C2H5NH2
conjugate acid is C2H5NH3+

Kb = 6.4*10^-4

pKb = - log (Kb)
= - log(6.4*10^-4)
= 3.194

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.194+ log {7.039*10^-3/2.085*10^-2}
= 2.722

use:
PH = 14 - pOH
= 14 - 2.7222
= 11.2778
Answer: 11.28


3)when 10.5 mL of HI is added
Given:
M(HI) = 0.1 M
V(HI) = 10.5 mL
M(C2H5NH2) = 0.03 M
V(C2H5NH2) = 70 mL


mol(HI) = M(HI) * V(HI)
mol(HI) = 0.1 M * 10.5 mL = 1.05 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)
mol(C2H5NH2) = 0.03 M * 70 mL = 2.1 mmol



We have:
mol(HI) = 1.05 mmol
mol(C2H5NH2) = 2.1 mmol

1.05 mmol of both will react
excess C2H5NH2 remaining = 1.05 mmol
Volume of Solution = 10.5 + 70 = 80.5 mL
[C2H5NH2] = 1.05 mmol/80.5 mL = 0.013 M
[C2H5NH3+] = 1.05 mmol/80.5 mL = 0.013 M

They form basic buffer
base is C2H5NH2
conjugate acid is C2H5NH3+

Kb = 6.4*10^-4

pKb = - log (Kb)
= - log(6.4*10^-4)
= 3.194

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.194+ log {1.304*10^-2/1.304*10^-2}
= 3.194

use:
PH = 14 - pOH
= 14 - 3.1938
= 10.8062
Answer: 10.81


4)when 15.8 mL of HI is added
Given:
M(HI) = 0.1 M
V(HI) = 15.8 mL
M(C2H5NH2) = 0.03 M
V(C2H5NH2) = 70 mL


mol(HI) = M(HI) * V(HI)
mol(HI) = 0.1 M * 15.8 mL = 1.58 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)
mol(C2H5NH2) = 0.03 M * 70 mL = 2.1 mmol



We have:
mol(HI) = 1.58 mmol
mol(C2H5NH2) = 2.1 mmol

1.58 mmol of both will react
excess C2H5NH2 remaining = 0.52 mmol
Volume of Solution = 15.8 + 70 = 85.8 mL
[C2H5NH2] = 0.52 mmol/85.8 mL = 0.0061 M
[C2H5NH3+] = 1.58 mmol/85.8 mL = 0.0184 M

They form basic buffer
base is C2H5NH2
conjugate acid is C2H5NH3+

Kb = 6.4*10^-4

pKb = - log (Kb)
= - log(6.4*10^-4)
= 3.194

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.194+ log {1.841*10^-2/6.061*10^-3}
= 3.676

use:
PH = 14 - pOH
= 14 - 3.6765
= 10.3235
Answer: 10.32


5)when 21.0 mL of HI is added
Given:
M(HI) = 0.1 M
V(HI) = 21 mL
M(C2H5NH2) = 0.03 M
V(C2H5NH2) = 70 mL


mol(HI) = M(HI) * V(HI)
mol(HI) = 0.1 M * 21 mL = 2.1 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)
mol(C2H5NH2) = 0.03 M * 70 mL = 2.1 mmol



We have:
mol(HI) = 2.1 mmol
mol(C2H5NH2) = 2.1 mmol

2.1 mmol of both will react to form C2H5NH3+ and H2O
C2H5NH3+ here is strong acid
C2H5NH3+ formed = 2.1 mmol
Volume of Solution = 21 + 70 = 91 mL
Ka of C2H5NH3+ = Kw/Kb = 1.0E-14/6.4E-4 = 1.562*10^-11
concentration ofC2H5NH3+,c = 2.1 mmol/91 mL = 0.0231 M


C2H5NH3+      + H2O ----->     C2H5NH2   +   H+
2.308*10^-2                    0         0
2.308*10^-2-x                  x         x


Ka = [H+][C2H5NH2]/[C2H5NH3+]
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 ((1.562*10^-11)*2.308*10^-2) = 6.005*10^-7

since c is much greater than x, our assumption is correct
so, x = 6.005*10^-7 M



[H+] = x = 6.005*10^-7 M

use:
pH = -log [H+]
= -log (6.005*10^-7)
= 6.2215
Answer: 6.22

6)when 23.1 mL of HI is added
Given:
M(HI) = 0.1 M
V(HI) = 23.1 mL
M(C2H5NH2) = 0.03 M
V(C2H5NH2) = 70 mL


mol(HI) = M(HI) * V(HI)
mol(HI) = 0.1 M * 23.1 mL = 2.31 mmol

mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)
mol(C2H5NH2) = 0.03 M * 70 mL = 2.1 mmol



We have:
mol(HI) = 2.31 mmol
mol(C2H5NH2) = 2.1 mmol

2.1 mmol of both will react
excess HI remaining = 0.21 mmol
Volume of Solution = 23.1 + 70 = 93.1 mL
[H+] = 0.21 mmol/93.1 mL = 0.0023 M

use:
pH = -log [H+]
= -log (2.256*10^-3)
= 2.6467
Answer: 2.65