Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0646 M HNO, to a 50.0 mL solution of 0.0750 M aziridine. The pk, of aziridinium is 8.04. What is the pH of the solution after the addition of 0.00 mL HNO,? pH = What is the pH of the solution after the addition of 5.07 mL HNO,? pH = What is the pH of the solution after the addition of a volume of HNO, equal to half the equivalence point volume? pH = What is the pH of the solution after the addition of 53.5 mL HNO,? pH = What is the pH of the solution after the addition of a volume of HNO, equal to the equivalence point? What is the pH of the solution after the addition of 62.8 mL HNO,? pH =

Answer 1

use:

pKa = -log Ka

8.04 = -log Ka

Ka = 9.12*10^-9

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/9.12*10^-9

Kb = 1.096*10^-6

1)when 0.0 mL of HNO3 is added

B dissociates as:

B +H2O -----> BH+ + OH-

7.5*10^-2 0 0

7.5*10^-2-x x x

Kb = [BH+][OH-]/[B]

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 ((1.096*10^-6)*7.5*10^-2) = 2.867*10^-4

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

so, x = 2.867*10^-4 M

So, [OH-] = x = 2.867*10^-4 M

use:

pOH = -log [OH-]

= -log (2.867*10^-4)

= 3.5426

use:

PH = 14 - pOH

= 14 - 3.5426

= 10.4574

Answer: 10.46

2)when 5.07 mL of HNO3 is added

Given:

M(HNO3) = 0.0646 M

V(HNO3) = 5.07 mL

M(B) = 0.075 M

V(B) = 50 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.0646 M * 5.07 mL = 0.3275 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.075 M * 50 mL = 3.75 mmol

We have:

mol(HNO3) = 0.3275 mmol

mol(B) = 3.75 mmol

0.3275 mmol of both will react

excess B remaining = 3.4225 mmol

Volume of Solution = 5.07 + 50 = 55.07 mL

[B] = 3.4225 mmol/55.07 mL = 0.0621 M

[BH+] = 0.3275 mmol/55.07 mL = 0.0059 M

They form basic buffer

base is B

conjugate acid is BH+

Kb = 1.096*10^-6

pKb = - log (Kb)

= - log(1.096*10^-6)

= 5.96

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 5.96+ log {5.947*10^-3/6.215*10^-2}

= 4.941

use:

PH = 14 - pOH

= 14 - 4.9411

= 9.0589

Answer: 9.06

3)

find the volume of HNO3 used to reach equivalence point

M(B)*V(B) =M(HNO3)*V(HNO3)

0.075 M *50.0 mL = 0.0646M *V(HNO3)

V(HNO3) = 58.0495 mL

At half equivalence point, volume of HNO3 required will be half of the above value

So, volume of HNO3 = 58.049535603715164/2 mL

= 29.024767801857582 mL

Given:

M(HNO3) = 0.0646 M

V(HNO3) = 29.0248 mL

M(B) = 0.075 M

V(B) = 50 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.0646 M * 29.0248 mL = 1.875 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.075 M * 50 mL = 3.75 mmol

We have:

mol(HNO3) = 1.875 mmol

mol(B) = 3.75 mmol

1.875 mmol of both will react

excess B remaining = 1.875 mmol

Volume of Solution = 29.0248 + 50 = 79.0248 mL

[B] = 1.875 mmol/79.0248 mL = 0.0237 M

[BH+] = 1.875 mmol/79.0248 mL = 0.0237 M

They form basic buffer

base is B

conjugate acid is BH+

Kb = 1.096*10^-6

pKb = - log (Kb)

= - log(1.096*10^-6)

= 5.96

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 5.96+ log {2.373*10^-2/2.373*10^-2}

= 5.96

use:

PH = 14 - pOH

= 14 - 5.9602

= 8.0398

Answer: 8.04

4)when 53.5 mL of HNO3 is added

Given:

M(HNO3) = 0.0646 M

V(HNO3) = 53.5 mL

M(B) = 0.075 M

V(B) = 50 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.0646 M * 53.5 mL = 3.4561 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.075 M * 50 mL = 3.75 mmol

We have:

mol(HNO3) = 3.4561 mmol

mol(B) = 3.75 mmol

3.4561 mmol of both will react

excess B remaining = 0.2939 mmol

Volume of Solution = 53.5 + 50 = 103.5 mL

[B] = 0.2939 mmol/103.5 mL = 0.0028 M

[BH+] = 3.4561 mmol/103.5 mL = 0.0334 M

They form basic buffer

base is B

conjugate acid is BH+

Kb = 1.096*10^-6

pKb = - log (Kb)

= - log(1.096*10^-6)

= 5.96

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 5.96+ log {3.339*10^-2/2.84*10^-3}

= 7.031

use:

PH = 14 - pOH

= 14 - 7.0306

= 6.9694

Answer: 6.97

5)

find the volume of HNO3 used to reach equivalence point

M(B)*V(B) =M(HNO3)*V(HNO3)

0.075 M *50.0 mL = 0.0646M *V(HNO3)

V(HNO3) = 58.0495 mL

Given:

M(HNO3) = 0.0646 M

V(HNO3) = 58.0495 mL

M(B) = 0.075 M

V(B) = 50 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.0646 M * 58.0495 mL = 3.75 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.075 M * 50 mL = 3.75 mmol

We have:

mol(HNO3) = 3.75 mmol

mol(B) = 3.75 mmol

3.75 mmol of both will react to form BH+ and H2O

BH+ here is strong acid

BH+ formed = 3.75 mmol

Volume of Solution = 58.0495 + 50 = 108.0495 mL

Ka of BH+ = Kw/Kb = 1.0E-14/1.096E-6 = 9.124*10^-9

concentration ofBH+,c = 3.75 mmol/108.0495 mL = 0.0347 M

BH+ + H2O -----> B + H+

3.471*10^-2 0 0

3.471*10^-2-x x x

Ka = [H+][B]/[BH+]

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 ((9.124*10^-9)*3.471*10^-2) = 1.78*10^-5

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

so, x = 1.78*10^-5 M

[H+] = x = 1.78*10^-5 M

use:

pH = -log [H+]

= -log (1.78*10^-5)

= 4.7497

Answer: 4.75

6)when 62.8 mL of HNO3 is added

Given:

M(HNO3) = 0.0646 M

V(HNO3) = 62.8 mL

M(B) = 0.075 M

V(B) = 50 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.0646 M * 62.8 mL = 4.0569 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.075 M * 50 mL = 3.75 mmol

We have:

mol(HNO3) = 4.0569 mmol

mol(B) = 3.75 mmol

3.75 mmol of both will react

excess HNO3 remaining = 0.3069 mmol

Volume of Solution = 62.8 + 50 = 112.8 mL

[H+] = 0.3069 mmol/112.8 mL = 0.0027 M

use:

pH = -log [H+]

= -log (2.721*10^-3)

= 2.5653

Answer: 2.57