Question

O ACIDS AND BASES Calculating the pH of a weak ba.. An analytical chemist is titrating 79.4 ml of a 0.08300 M solution of trimethylamine (CH), N) with a 0.2900 M solution of HNO, The px, of trimethylamine is 4.19. Calculate the pH of the base solution after the chemist has added 24.1 mt of the HNO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO, solution added. Round your answer to 2 decimal places. x 6 ?

Answer 1

use:

pKb = -log Kb

4.19= -log Kb

Kb = 6.457*10^-5

Given:

M(HNO3) = 0.29 M

V(HNO3) = 24.1 mL

M((CH3)3N) = 0.083 M

V((CH3)3N) = 79.4 mL

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

mol(HNO3) = 0.29 M * 24.1 mL = 6.989 mmol

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

mol((CH3)3N) = 0.083 M * 79.4 mL = 6.5902 mmol

We have:

mol(HNO3) = 6.989 mmol

mol((CH3)3N) = 6.5902 mmol

6.5902 mmol of both will react

excess HNO3 remaining = 0.3988 mmol

Volume of Solution = 24.1 + 79.4 = 103.5 mL

[H+] = 0.3988 mmol/103.5 mL = 0.0039 M

use:

pH = -log [H+]

= -log (3.853*10^-3)

= 2.4142

Answer: 2.41