Question

An analytical chemist is titrating 67.6 mL of a 0.1800M solution of trimethylamine ((CH3)3N) with a 0.7700M solution of HNO3. The pKb of trimethylamine is 4.19. calculate the pH of the base solution after the chemist has added 17.9 mL of the HNO3 solution to it.

Answer 1

use:
pKb = -log Kb
4.19= -log Kb
Kb = 6.457*10^-5
Given:
M(HNO3) = 0.77 M
V(HNO3) = 17.9 mL
M((CH3)3N) = 0.18 M
V((CH3)3N) = 67.6 mL


mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.77 M * 17.9 mL = 13.783 mmol

mol((CH3)3N) = M((CH3)3N) * V((CH3)3N)
mol((CH3)3N) = 0.18 M * 67.6 mL = 12.168 mmol



We have:
mol(HNO3) = 13.783 mmol
mol((CH3)3N) = 12.168 mmol

12.168 mmol of both will react
excess HNO3 remaining = 1.615 mmol
Volume of Solution = 17.9 + 67.6 = 85.5 mL
[H+] = 1.615 mmol/85.5 mL = 0.0189 M

use:
pH = -log [H+]
= -log (1.889*10^-2)
= 1.7238
Answer: 1.72