a)when 0.0 mL of HI is added
(CH3)3N dissociates as:
(CH3)3N +H2O -----> (CH3)3NH+ + OH-
0.254 0 0
0.254-x x x
Kb = [(CH3)3NH+][OH-]/[(CH3)3N]
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 ((6.3*10^-5)*0.254) = 4*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)
6.3*10^-5 = x^2/(0.254-x)
1.6*10^-5 - 6.3*10^-5 *x = x^2
x^2 + 6.3*10^-5 *x-1.6*10^-5 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1
b = 6.3*10^-5
c = -1.6*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 = 6.401*10^-5
roots are :
x = 3.969*10^-3 and x = -4.032*10^-3
since x can't be negative, the possible value of x is
x = 3.969*10^-3
So, [OH-] = x = 3.969*10^-3 M
use:
pOH = -log [OH-]
= -log (3.969*10^-3)
= 2.4013
use:
PH = 14 - pOH
= 14 - 2.4013
= 11.5987
Answer: 11.60
b)when 12.6 mL of HI is added
Given:
M(HI) = 0.254 M
V(HI) = 12.6 mL
M((CH3)3N) = 0.254 M
V((CH3)3N) = 32.8 mL
mol(HI) = M(HI) * V(HI)
mol(HI) = 0.254 M * 12.6 mL = 3.2004 mmol
mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)
mol((CH3)3N) = 0.254 M * 32.8 mL = 8.3312 mmol
We have:
mol(HI) = 3.2004 mmol
mol((CH3)3N) = 8.3312 mmol
3.2004 mmol of both will react
excess (CH3)3N remaining = 5.1308 mmol
Volume of Solution = 12.6 + 32.8 = 45.4 mL
[(CH3)3N] = 5.1308 mmol/45.4 mL = 0.113 M
[(CH3)3NH+] = 3.2004 mmol/45.4 mL = 0.0705 M
They form basic buffer
base is (CH3)3N
conjugate acid is (CH3)3NH+
Kb = 6.3*10^-5
pKb = - log (Kb)
= - log(6.3*10^-5)
= 4.201
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.201+ log {7.049*10^-2/0.113}
= 3.996
use:
PH = 14 - pOH
= 14 - 3.9957
= 10.0043
Answer: 10.00
c)
find the volume of HI used to reach equivalence point
M((CH3)3N)*V((CH3)3N) =M(HI)*V(HI)
0.254 M *32.8 mL = 0.254M *V(HI)
V(HI) = 32.8 mL
At half equivalence point, volume of HI required will be half of the above value
So, volume of HI = 32.8/2 mL
= 16.4 mL
Given:
M(HI) = 0.254 M
V(HI) = 16.4 mL
M((CH3)3N) = 0.254 M
V((CH3)3N) = 32.8 mL
mol(HI) = M(HI) * V(HI)
mol(HI) = 0.254 M * 16.4 mL = 4.1656 mmol
mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)
mol((CH3)3N) = 0.254 M * 32.8 mL = 8.3312 mmol
We have:
mol(HI) = 4.1656 mmol
mol((CH3)3N) = 8.3312 mmol
4.1656 mmol of both will react
excess (CH3)3N remaining = 4.1656 mmol
Volume of Solution = 16.4 + 32.8 = 49.2 mL
[(CH3)3N] = 4.1656 mmol/49.2 mL = 0.0847 M
[(CH3)3NH+] = 4.1656 mmol/49.2 mL = 0.0847 M
They form basic buffer
base is (CH3)3N
conjugate acid is (CH3)3NH+
Kb = 6.3*10^-5
pKb = - log (Kb)
= - log(6.3*10^-5)
= 4.201
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 4.201+ log {8.467*10^-2/8.467*10^-2}
= 4.201
use:
PH = 14 - pOH
= 14 - 4.2007
= 9.7993
Answer: 9.80
d)
find the volume of HI used to reach equivalence point
M((CH3)3N)*V((CH3)3N) =M(HI)*V(HI)
0.254 M *32.8 mL = 0.254M *V(HI)
V(HI) = 32.8 mL
Given:
M(HI) = 0.254 M
V(HI) = 32.8 mL
M((CH3)3N) = 0.254 M
V((CH3)3N) = 32.8 mL
mol(HI) = M(HI) * V(HI)
mol(HI) = 0.254 M * 32.8 mL = 8.3312 mmol
mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)
mol((CH3)3N) = 0.254 M * 32.8 mL = 8.3312 mmol
We have:
mol(HI) = 8.3312 mmol
mol((CH3)3N) = 8.3312 mmol
8.3312 mmol of both will react to form (CH3)3NH+ and H2O
(CH3)3NH+ here is strong acid
(CH3)3NH+ formed = 8.3312 mmol
Volume of Solution = 32.8 + 32.8 = 65.6 mL
Ka of (CH3)3NH+ = Kw/Kb = 1.0E-14/6.3E-5 = 1.587*10^-10
concentration of(CH3)3NH+,c = 8.3312 mmol/65.6 mL = 0.127 M
(CH3)3NH+ + H2O -----> (CH3)3N + H+
0.127 0 0
0.127-x x x
Ka = [H+][(CH3)3N]/[(CH3)3NH+]
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 ((1.587*10^-10)*0.127) = 4.49*10^-6
since c is much greater than x, our assumption is correct
so, x = 4.49*10^-6 M
[H+] = x = 4.49*10^-6 M
use:
pH = -log [H+]
= -log (4.49*10^-6)
= 5.3478
Answer: 5.35
e)when 47.9 mL of HI is added
Given:
M(HI) = 0.254 M
V(HI) = 47.9 mL
M((CH3)3N) = 0.254 M
V((CH3)3N) = 32.8 mL
mol(HI) = M(HI) * V(HI)
mol(HI) = 0.254 M * 47.9 mL = 12.1666 mmol
mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)
mol((CH3)3N) = 0.254 M * 32.8 mL = 8.3312 mmol
We have:
mol(HI) = 12.1666 mmol
mol((CH3)3N) = 8.3312 mmol
8.3312 mmol of both will react
excess HI remaining = 3.8354 mmol
Volume of Solution = 47.9 + 32.8 = 80.7 mL
[H+] = 3.8354 mmol/80.7 mL = 0.0475 M
use:
pH = -log [H+]
= -log (4.753*10^-2)
= 1.3231
Answer: 1.32
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