Question

Determine the pH during the titration of 33.7 mL of 0.345 M triethylamine ((C₂H₅)₃N, K_b = 5.2×10^-4) by 0.345 M HI at the following points. (Assume the titration is done at 25 °C.)
Note that state symbols are not shown for species in this problem.
(a) Before the addition of any HI

(b) After the addition of 15.0 mL of HI

(c) At the titration midpoint

(d) At the equivalence point

(e) After adding 51.9 mL of HI

Answer 1

trimethyl amine = Kb = 5.2 x 10^-4

pKb = 3.28

millimoles of trimethyl amine = 33.7 x 0.345 = 11.6

(a) initial pH

pOH = 1/2 [pKb- logC]

pOH = 1/2 [pKb - log C]

          = 1/2 (3.28- log0.348)

          = 1.87

pH + pOH = 14

pH = 14 - pOH

      = 12.13

pH = 12.13

b) 15 mL titrant added :

millimoles of HCl = 15 x 0.345 = 5.175

(CH3)3N    + HCl --------------------> (CH3)3NHCl

11.6       5.175                                    0

6.425                 0                                       5.175

here salt and base remained . so it forms basic buffer

pOH = pKb + log [salt / base]

         = 3.28 + log [5.175 / 6.425]

          = 3.19

pH = 10.81

c) at mid point

pH = pKa

pH = 10.72

d) at equivalnece point

salt left = 11.6 / 33.7 + 33.7 = 0.172

pH = 7 - 1/2 [pKb + log C]

pH = 5.74

e) 51.9 ml added :

pH = 1.13