Question

Review Topics [References] TUTOR Analysis of a Weak Base-Strong Acid Titration Curve Determine the pH during the titration of 32.8 mL of 0.254 M trimethylamine ((CH3)3N, K-6.3x105) by 0.254 M HI at the following points. (Assume the titration is done at 25 °C.) Note that state symbols are not shown for species in this problem. (a) Before the addition of any HI (b) After the addition of 12.6 mL of HI (e) At the titration midpoint (d) At the equivalence point (e) After adding 47.9 mL of HI

Answer 1

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