Question

A chemist titrates 80.0mL of a 0.6050M trimethylamine CH33N solution with 0.2965M HNO3 solution at 25°C . Calculate the pH at equivalence. The pKb of trimethylamine is 4.19 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO3 solution added.

Answer 1

use:

pKb = -log Kb

4.19= -log Kb

Kb = 6.457*10^-5

find the volume of HNO3 used to reach equivalence point

M((CH3)3N)*V((CH3)3N) =M(HNO3)*V(HNO3)

0.605 M *80.0 mL = 0.2965M *V(HNO3)

V(HNO3) = 163.2378 mL

Given:

M(HNO3) = 0.2965 M

V(HNO3) = 163.2378 mL

M((CH3)3N) = 0.605 M

V((CH3)3N) = 80 mL

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

mol(HNO3) = 0.2965 M * 163.2378 mL = 48.4 mmol

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

mol((CH3)3N) = 0.605 M * 80 mL = 48.4 mmol

We have:

mol(HNO3) = 48.4 mmol

mol((CH3)3N) = 48.4 mmol

48.4 mmol of both will react to form (CH3)3NH+ and H2O

(CH3)3NH+ here is strong acid

(CH3)3NH+ formed = 48.4 mmol

Volume of Solution = 163.2378 + 80 = 243.2378 mL

Ka of (CH3)3NH+ = Kw/Kb = 1.0E-14/6.456542290346549E-5 = 1.549*10^-10

concentration of(CH3)3NH+,c = 48.4 mmol/243.2378 mL = 0.199 M

(CH3)3NH+ + H2O -----> (CH3)3N + H+

0.199 0 0

0.199-x x x

Ka = [H+][(CH3)3N]/[(CH3)3NH+]

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 ((1.549*10^-10)*0.199) = 5.551*10^-6

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

so, x = 5.551*10^-6 M

[H+] = x = 5.551*10^-6 M

use:

pH = -log [H+]

= -log (5.551*10^-6)

= 5.26

Answer: 5.26