Question

A a0.G0mb sample of daon triethyhbmine, SIOis titrated w 544M HCs Kb= a calculate tha pH before any addition of Hcl Calculate the et at tha abivalence point C) caculate the PH whun 20mL of Hc added 4A solutionn is o.oloon in Poions. f O.l03 mol of salid NaI isadded to 1.00 L of this solotion, what percentage af the vemainin 5olution? 2+ jons Ko &.7

Answer 1

3)

a)when 0.0 mL of HCl is added

(CH3)3N dissociates as:

(CH3)3N +H2O -----> (CH3)3NH+ + OH-

0.22 0 0

0.22-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 ((5.2*10^-4)*0.22) = 1.07*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

5.2*10^-4 = x^2/(0.22-x)

1.144*10^-4 - 5.2*10^-4 *x = x^2

x^2 + 5.2*10^-4 *x-1.144*10^-4 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 5.2*10^-4

c = -1.144*10^-4

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 = 4.579*10^-4

roots are :

x = 1.044*10^-2 and x = -1.096*10^-2

since x can't be negative, the possible value of x is

x = 1.044*10^-2

So, [OH-] = x = 1.044*10^-2 M

use:

pOH = -log [OH-]

= -log (1.044*10^-2)

= 1.9813

use:

PH = 14 - pOH

= 14 - 1.9813

= 12.0187

Answer: 12.02

b)

find the volume of HCl used to reach equivalence point

M((CH3)3N)*V((CH3)3N) =M(HCl)*V(HCl)

0.22 M *20.0 mL = 0.544M *V(HCl)

V(HCl) = 8.0882 mL

Given:

M(HCl) = 0.544 M

V(HCl) = 8.0882 mL

M((CH3)3N) = 0.22 M

V((CH3)3N) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.544 M * 8.0882 mL = 4.4 mmol

mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)

mol((CH3)3N) = 0.22 M * 20 mL = 4.4 mmol

We have:

mol(HCl) = 4.4 mmol

mol((CH3)3N) = 4.4 mmol

4.4 mmol of both will react to form (CH3)3NH+ and H2O

(CH3)3NH+ here is strong acid

(CH3)3NH+ formed = 4.4 mmol

Volume of Solution = 8.0882 + 20 = 28.0882 mL

Ka of (CH3)3NH+ = Kw/Kb = 1.0E-14/5.2E-4 = 1.923*10^-11

concentration of(CH3)3NH+,c = 4.4 mmol/28.0882 mL = 0.1566 M

(CH3)3NH+ + H2O -----> (CH3)3N + H+

0.1566 0 0

0.1566-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 ((1.923*10^-11)*0.1566) = 1.736*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.736*10^-6 M

[H+] = x = 1.736*10^-6 M

use:

pH = -log [H+]

= -log (1.736*10^-6)

= 5.7605

Answer: 5.76

c)when 20.0 mL of HCl is added

Given:

M(HCl) = 0.544 M

V(HCl) = 20 mL

M((CH3)3N) = 0.22 M

V((CH3)3N) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.544 M * 20 mL = 10.88 mmol

mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)

mol((CH3)3N) = 0.22 M * 20 mL = 4.4 mmol

We have:

mol(HCl) = 10.88 mmol

mol((CH3)3N) = 4.4 mmol

4.4 mmol of both will react

excess HCl remaining = 6.48 mmol

Volume of Solution = 20 + 20 = 40 mL

[H+] = 6.48 mmol/40 mL = 0.162 M

use:

pH = -log [H+]

= -log (0.162)

= 0.7905

Answer: 0.791

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