Question

Consider the titration of 100.0 mL of 0.400 M C₅H₅N by 0.200 M HCl.
(K_b for C₅H₅N = 1.7×10^-9)

Part 1
Calculate the pH after 0.0 mL of HCl added.

pH =

Part 2
Calculate the pH after 25.0 mL of HCl added.

pH =

Part 3
Calculate the pH after 75.0 mL of HCl added.

pH =

Part 4
Calculate the pH at the equivalence point.

pH =

Part 5
Calculate the pH after 300.0 mL of HCl added.

pH =

Part 6
At what volume of HCl added, does the pH = 5.23?

________mL

Answer 1

1)when 0.0 mL of HCl is added

C5H5N dissociates as:

C5H5N +H2O -----> C5H5NH+ + OH-

0.4 0 0

0.4-x x x

Kb = [C5H5NH+][OH-]/[C5H5N]

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 ((1.7*10^-9)*0.4) = 2.608*10^-5

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

so, x = 2.608*10^-5 M

So, [OH-] = x = 2.608*10^-5 M

use:

pOH = -log [OH-]

= -log (2.608*10^-5)

= 4.5837

use:

PH = 14 - pOH

= 14 - 4.5837

= 9.4163

Answer: 9.42

2)when 25.0 mL of HCl is added

Given:

M(HCl) = 0.2 M

V(HCl) = 25 mL

M(C5H5N) = 0.4 M

V(C5H5N) = 100 mL

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

mol(HCl) = 0.2 M * 25 mL = 5 mmol

mol(C5H5N) = M(C5H5N) * V(C5H5N)

mol(C5H5N) = 0.4 M * 100 mL = 40 mmol

We have:

mol(HCl) = 5 mmol

mol(C5H5N) = 40 mmol

5 mmol of both will react

excess C5H5N remaining = 35 mmol

Volume of Solution = 25 + 100 = 125 mL

[C5H5N] = 35 mmol/125 mL = 0.28 M

[C5H5NH+] = 5 mmol/125 mL = 0.04 M

They form basic buffer

base is C5H5N

conjugate acid is C5H5NH+

Kb = 1.7*10^-9

pKb = - log (Kb)

= - log(1.7*10^-9)

= 8.77

use:

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

= 8.77+ log {4*10^-2/0.28}

= 7.924

use:

PH = 14 - pOH

= 14 - 7.9245

= 6.0755

Answer: 6.08

3)when 75.0 mL of HCl is added

Given:

M(HCl) = 0.2 M

V(HCl) = 75 mL

M(C5H5N) = 0.4 M

V(C5H5N) = 100 mL

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

mol(HCl) = 0.2 M * 75 mL = 15 mmol

mol(C5H5N) = M(C5H5N) * V(C5H5N)

mol(C5H5N) = 0.4 M * 100 mL = 40 mmol

We have:

mol(HCl) = 15 mmol

mol(C5H5N) = 40 mmol

15 mmol of both will react

excess C5H5N remaining = 25 mmol

Volume of Solution = 75 + 100 = 175 mL

[C5H5N] = 25 mmol/175 mL = 0.1429 M

[C5H5NH+] = 15 mmol/175 mL = 0.0857 M

They form basic buffer

base is C5H5N

conjugate acid is C5H5NH+

Kb = 1.7*10^-9

pKb = - log (Kb)

= - log(1.7*10^-9)

= 8.77

use:

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

= 8.77+ log {8.571*10^-2/0.1429}

= 8.548

use:

PH = 14 - pOH

= 14 - 8.5477

= 5.4523

Answer: 5.45

4)

find the volume of HCl used to reach equivalence point

M(C5H5N)*V(C5H5N) =M(HCl)*V(HCl)

0.4 M *100.0 mL = 0.2M *V(HCl)

V(HCl) = 200 mL

Given:

M(HCl) = 0.2 M

V(HCl) = 200 mL

M(C5H5N) = 0.4 M

V(C5H5N) = 100 mL

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

mol(HCl) = 0.2 M * 200 mL = 40 mmol

mol(C5H5N) = M(C5H5N) * V(C5H5N)

mol(C5H5N) = 0.4 M * 100 mL = 40 mmol

We have:

mol(HCl) = 40 mmol

mol(C5H5N) = 40 mmol

40 mmol of both will react to form C5H5NH+ and H2O

C5H5NH+ here is strong acid

C5H5NH+ formed = 40 mmol

Volume of Solution = 200 + 100 = 300 mL

Ka of C5H5NH+ = Kw/Kb = 1.0E-14/1.7E-9 = 5.882*10^-6

concentration ofC5H5NH+,c = 40 mmol/300 mL = 0.1333 M

C5H5NH+ + H2O -----> C5H5N + H+

0.1333 0 0

0.1333-x x x

Ka = [H+][C5H5N]/[C5H5NH+]

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 ((5.882*10^-6)*0.1333) = 8.856*10^-4

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

so, x = 8.856*10^-4 M

[H+] = x = 8.856*10^-4 M

use:

pH = -log [H+]

= -log (8.856*10^-4)

= 3.0528

Answer: 3.05

Only 4 parts at a time