Question

A 25.0 mL sample of a 0.3400 M solution of aqueous trimethylamine is titrated with a 0.4250 M solution of HCl. Calculate the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added; pKb of (CH3)3N = 4.19 at 25°C.

Answer 1

use:

pKb = -log Kb

4.19= -log Kb

Kb = 6.457*10^-5

1)when 10.0 mL of HNO3 is added

Given:

M(HNO3) = 0.425 M

V(HNO3) = 10 mL

M((CH3)3N) = 0.4 M

V((CH3)3N) = 25 mL

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

mol(HNO3) = 0.425 M * 10 mL = 4.25 mmol

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

mol((CH3)3N) = 0.4 M * 25 mL = 10 mmol

We have:

mol(HNO3) = 4.25 mmol

mol((CH3)3N) = 10 mmol

4.25 mmol of both will react

excess (CH3)3N remaining = 5.75 mmol

Volume of Solution = 10 + 25 = 35 mL

[(CH3)3N] = 5.75 mmol/35 mL = 0.1643 M

[(CH3)3NH+] = 4.25 mmol/35 mL = 0.1214 M

They form basic buffer

base is (CH3)3N

conjugate acid is (CH3)3NH+

Kb = 6.457*10^-5

pKb = - log (Kb)

= - log(6.457*10^-5)

= 4.19

use:

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

= 4.19+ log {0.1214/0.1643}

= 4.059

use:

PH = 14 - pOH

= 14 - 4.0587

= 9.9413

Answer: 9.94

2)when 20.0 mL of HNO3 is added

Given:

M(HNO3) = 0.425 M

V(HNO3) = 20 mL

M((CH3)3N) = 0.4 M

V((CH3)3N) = 25 mL

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

mol(HNO3) = 0.425 M * 20 mL = 8.5 mmol

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

mol((CH3)3N) = 0.4 M * 25 mL = 10 mmol

We have:

mol(HNO3) = 8.5 mmol

mol((CH3)3N) = 10 mmol

8.5 mmol of both will react

excess (CH3)3N remaining = 1.5 mmol

Volume of Solution = 20 + 25 = 45 mL

[(CH3)3N] = 1.5 mmol/45 mL = 0.0333 M

[(CH3)3NH+] = 8.5 mmol/45 mL = 0.1889 M

They form basic buffer

base is (CH3)3N

conjugate acid is (CH3)3NH+

Kb = 6.457*10^-5

pKb = - log (Kb)

= - log(6.457*10^-5)

= 4.19

use:

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

= 4.19+ log {0.1889/3.333*10^-2}

= 4.943

use:

PH = 14 - pOH

= 14 - 4.9433

= 9.0567

Answer: 9.06

3)when 30.0 mL of HNO3 is added

Given:

M(HNO3) = 0.425 M

V(HNO3) = 30 mL

M((CH3)3N) = 0.4 M

V((CH3)3N) = 25 mL

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

mol(HNO3) = 0.425 M * 30 mL = 12.75 mmol

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

mol((CH3)3N) = 0.4 M * 25 mL = 10 mmol

We have:

mol(HNO3) = 12.75 mmol

mol((CH3)3N) = 10 mmol

10 mmol of both will react

excess HNO3 remaining = 2.75 mmol

Volume of Solution = 30 + 25 = 55 mL

[H+] = 2.75 mmol/55 mL = 0.05 M

use:

pH = -log [H+]

= -log (5*10^-2)

= 1.301

Answer: 1.30