Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.150 M pyridine, C5H5N(aq) with 0.150 M HBr(aq): (a) before addition of any HBr (b) after addition of 12.5 mL of HBr (c) after addition of 14.0 mL of HBr (d) after addition of 25.0 mL of HBr (e) after addition of 30.0 mL of HBr

Answer 1

Kb of C5H5N = 1.70*10^-9

a)when 0.0 mL of HBr is added

C5H5N dissociates as:

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

0.15 0 0

0.15-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)*0.15) = 1.597*10^-5

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

so, x = 1.597*10^-5 M

So, [OH-] = x = 1.597*10^-5 M

use:

pOH = -log [OH-]

= -log (1.597*10^-5)

= 4.7967

use:

PH = 14 - pOH

= 14 - 4.7967

= 9.2033

Answer: 9.20

b)when 12.5 mL of HBr is added

Given:

M(HBr) = 0.15 M

V(HBr) = 12.5 mL

M(C5H5N) = 0.15 M

V(C5H5N) = 25 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.15 M * 12.5 mL = 1.875 mmol

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

mol(C5H5N) = 0.15 M * 25 mL = 3.75 mmol

We have:

mol(HBr) = 1.875 mmol

mol(C5H5N) = 3.75 mmol

1.875 mmol of both will react

excess C5H5N remaining = 1.875 mmol

Volume of Solution = 12.5 + 25 = 37.5 mL

[C5H5N] = 1.875 mmol/37.5 mL = 0.05 M

[C5H5NH+] = 1.875 mmol/37.5 mL = 0.05 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 {5*10^-2/5*10^-2}

= 8.77

use:

PH = 14 - pOH

= 14 - 8.7696

= 5.2304

Answer: 5.23

c)when 14.0 mL of HBr is added

Given:

M(HBr) = 0.15 M

V(HBr) = 14 mL

M(C5H5N) = 0.15 M

V(C5H5N) = 25 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.15 M * 14 mL = 2.1 mmol

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

mol(C5H5N) = 0.15 M * 25 mL = 3.75 mmol

We have:

mol(HBr) = 2.1 mmol

mol(C5H5N) = 3.75 mmol

2.1 mmol of both will react

excess C5H5N remaining = 1.65 mmol

Volume of Solution = 14 + 25 = 39 mL

[C5H5N] = 1.65 mmol/39 mL = 0.0423 M

[C5H5NH+] = 2.1 mmol/39 mL = 0.0538 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 {5.385*10^-2/4.231*10^-2}

= 8.874

use:

PH = 14 - pOH

= 14 - 8.8743

= 5.1257

Answer: 5.13

d)when 25.0 mL of HBr is added

Given:

M(HBr) = 0.15 M

V(HBr) = 25 mL

M(C5H5N) = 0.15 M

V(C5H5N) = 25 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.15 M * 25 mL = 3.75 mmol

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

mol(C5H5N) = 0.15 M * 25 mL = 3.75 mmol

We have:

mol(HBr) = 3.75 mmol

mol(C5H5N) = 3.75 mmol

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

C5H5NH+ here is strong acid

C5H5NH+ formed = 3.75 mmol

Volume of Solution = 25 + 25 = 50 mL

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

concentration ofC5H5NH+,c = 3.75 mmol/50 mL = 0.075 M

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

7.5*10^-2 0 0

7.5*10^-2-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)*7.5*10^-2) = 6.642*10^-4

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

so, x = 6.642*10^-4 M

[H+] = x = 6.642*10^-4 M

use:

pH = -log [H+]

= -log (6.642*10^-4)

= 3.1777

Answer: 3.18

e)when 30.0 mL of HBr is added

Given:

M(HBr) = 0.15 M

V(HBr) = 30 mL

M(C5H5N) = 0.15 M

V(C5H5N) = 25 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.15 M * 30 mL = 4.5 mmol

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

mol(C5H5N) = 0.15 M * 25 mL = 3.75 mmol

We have:

mol(HBr) = 4.5 mmol

mol(C5H5N) = 3.75 mmol

3.75 mmol of both will react

excess HBr remaining = 0.75 mmol

Volume of Solution = 30 + 25 = 55 mL

[H+] = 0.75 mmol/55 mL = 0.0136 M

use:

pH = -log [H+]

= -log (1.364*10^-2)

= 1.8653

Answer: 1.87