Question

Be sure to answer all parts.

Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH₃CH₂)₃N

(

K_b = 5.2

×

10⁻⁴

)

,

with 0.1000 M HCl solution after the following additions of titrant.

(a) 10.00 mL:

pH =

(b) 20.50 mL:

pH =

(c) 25.00 mL:

pH =

Answer 1

1)when 10.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 10 mL

M((CH3CH2)3N) = 0.1 M

V((CH3CH2)3N) = 20 mL

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

mol(HCl) = 0.1 M * 10 mL = 1 mmol

mol((CH3CH2)3N) = M((CH3CH2)3N) * V((CH3CH2)3N)

mol((CH3CH2)3N) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 1 mmol

mol((CH3CH2)3N) = 2 mmol

1 mmol of both will react

excess (CH3CH2)3N remaining = 1 mmol

Volume of Solution = 10 + 20 = 30 mL

[(CH3CH2)3N] = 1 mmol/30 mL = 0.0333 M

[(CH3CH2)3NH+] = 1 mmol/30 mL = 0.0333 M

They form basic buffer

base is (CH3CH2)3N

conjugate acid is (CH3CH2)3NH+

Kb = 5.2*10^-4

pKb = - log (Kb)

= - log(5.2*10^-4)

= 3.284

use:

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

= 3.284+ log {3.333*10^-2/3.333*10^-2}

= 3.284

use:

PH = 14 - pOH

= 14 - 3.284

= 10.716

2)when 20.5 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 20.5 mL

M((CH3CH2)3N) = 0.1 M

V((CH3CH2)3N) = 20 mL

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

mol(HCl) = 0.1 M * 20.5 mL = 2.05 mmol

mol((CH3CH2)3N) = M((CH3CH2)3N) * V((CH3CH2)3N)

mol((CH3CH2)3N) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 2.05 mmol

mol((CH3CH2)3N) = 2 mmol

2 mmol of both will react

excess HCl remaining = 0.05 mmol

Volume of Solution = 20.5 + 20 = 40.5 mL

[H+] = 0.05 mmol/40.5 mL = 0.0012 M

use:

pH = -log [H+]

= -log (1.235*10^-3)

= 2.9085

3)when 25.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 25 mL

M((CH3CH2)3N) = 0.1 M

V((CH3CH2)3N) = 20 mL

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

mol(HCl) = 0.1 M * 25 mL = 2.5 mmol

mol((CH3CH2)3N) = M((CH3CH2)3N) * V((CH3CH2)3N)

mol((CH3CH2)3N) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 2.5 mmol

mol((CH3CH2)3N) = 2 mmol

2 mmol of both will react

excess HCl remaining = 0.5 mmol

Volume of Solution = 25 + 20 = 45 mL

[H+] = 0.5 mmol/45 mL = 0.0111 M

use:

pH = -log [H+]

= -log (1.111*10^-2)

= 1.9542