a)
find the volume of HCl used to reach equivalence point
M(C2H5NH2)*V(C2H5NH2) =M(HCl)*V(HCl)
0.087 M *35.0 mL = 0.15M *V(HCl)
V(HCl) = 20.3 mL
Given:
M(HCl) = 0.15 M
V(HCl) = 20.3 mL
M(C2H5NH2) = 0.087 M
V(C2H5NH2) = 35 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.15 M * 20.3 mL = 3.045 mmol
mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)
mol(C2H5NH2) = 0.087 M * 35 mL = 3.045 mmol
We have:
mol(HCl) = 3.045 mmol
mol(C2H5NH2) = 3.045 mmol
3.045 mmol of both will react to form C2H5NH3+ and H2O
C2H5NH3+ here is strong acid
C2H5NH3+ formed = 3.045 mmol
Volume of Solution = 20.3 + 35 = 55.3 mL
Ka of C2H5NH3+ = Kw/Kb = 1.0E-14/5.6E-4 = 1.786*10^-11
concentration ofC2H5NH3+,c = 3.045 mmol/55.3 mL = 0.0551 M
C2H5NH3+ + H2O -----> C2H5NH2 + H+
5.506*10^-2 0 0
5.506*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 ((1.786*10^-11)*5.506*10^-2) = 9.916*10^-7
since c is much greater than x, our assumption is correct
so, x = 9.916*10^-7 M
[H+] = x = 9.916*10^-7 M
use:
pH = -log [H+]
= -log (9.916*10^-7)
= 6.0037
Answer: 6.00
i)when 0.0 mL of HCl is added
C2H5NH2 dissociates as:
C2H5NH2 +H2O -----> C2H5NH3+ + OH-
8.7*10^-2 0 0
8.7*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 ((5.6*10^-4)*8.7*10^-2) = 6.98*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)
5.6*10^-4 = x^2/(8.7*10^-2-x)
4.872*10^-5 - 5.6*10^-4 *x = x^2
x^2 + 5.6*10^-4 *x-4.872*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 5.6*10^-4
c = -4.872*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 = 1.952*10^-4
roots are :
x = 6.706*10^-3 and x = -7.266*10^-3
since x can't be negative, the possible value of x is
x = 6.706*10^-3
So, [OH-] = x = 6.706*10^-3 M
use:
pOH = -log [OH-]
= -log (6.706*10^-3)
= 2.1736
use:
PH = 14 - pOH
= 14 - 2.1736
= 11.8264
Answer: 11.83
ii)when 10.15 mL of HCl is added
Given:
M(HCl) = 0.15 M
V(HCl) = 10.15 mL
M(C2H5NH2) = 0.087 M
V(C2H5NH2) = 35 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.15 M * 10.15 mL = 1.5225 mmol
mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)
mol(C2H5NH2) = 0.087 M * 35 mL = 3.045 mmol
We have:
mol(HCl) = 1.5225 mmol
mol(C2H5NH2) = 3.045 mmol
1.5225 mmol of both will react
excess C2H5NH2 remaining = 1.5225 mmol
Volume of Solution = 10.15 + 35 = 45.15 mL
[C2H5NH2] = 1.5225 mmol/45.15 mL = 0.0337 M
[C2H5NH3+] = 1.5225 mmol/45.15 mL = 0.0337 M
They form basic buffer
base is C2H5NH2
conjugate acid is C2H5NH3+
Kb = 5.6*10^-4
pKb = - log (Kb)
= - log(5.6*10^-4)
= 3.252
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.252+ log {3.372*10^-2/3.372*10^-2}
= 3.252
use:
PH = 14 - pOH
= 14 - 3.2518
= 10.7482
Answer: 10.75
iii)when 14.5 mL of HCl is added
Given:
M(HCl) = 0.15 M
V(HCl) = 14.5 mL
M(C2H5NH2) = 0.087 M
V(C2H5NH2) = 35 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.15 M * 14.5 mL = 2.175 mmol
mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)
mol(C2H5NH2) = 0.087 M * 35 mL = 3.045 mmol
We have:
mol(HCl) = 2.175 mmol
mol(C2H5NH2) = 3.045 mmol
2.175 mmol of both will react
excess C2H5NH2 remaining = 0.87 mmol
Volume of Solution = 14.5 + 35 = 49.5 mL
[C2H5NH2] = 0.87 mmol/49.5 mL = 0.0176 M
[C2H5NH3+] = 2.175 mmol/49.5 mL = 0.0439 M
They form basic buffer
base is C2H5NH2
conjugate acid is C2H5NH3+
Kb = 5.6*10^-4
pKb = - log (Kb)
= - log(5.6*10^-4)
= 3.252
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.252+ log {4.394*10^-2/1.758*10^-2}
= 3.65
use:
PH = 14 - pOH
= 14 - 3.6498
= 10.3502
Answer: 10.35
iv)Given:
M(HCl) = 0.15 M
V(HCl) = 26.3 mL
M(C2H5NH2) = 0.087 M
V(C2H5NH2) = 35 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.15 M * 26.3 mL = 3.945 mmol
mol(C2H5NH2) = M(C2H5NH2) * V(C2H5NH2)
mol(C2H5NH2) = 0.087 M * 35 mL = 3.045 mmol
We have:
mol(HCl) = 3.945 mmol
mol(C2H5NH2) = 3.045 mmol
3.045 mmol of both will react
excess HCl remaining = 0.9 mmol
Volume of Solution = 26.3 + 35 = 61.3 mL
[H+] = 0.9 mmol/61.3 mL = 0.0147 M
use:
pH = -log [H+]
= -log (1.468*10^-2)
= 1.8332
Answer: 1.83
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