Question

A 20.8 mL sample of 0.304 M triethylamine, (C2H5)3N, is titrated with 0.341 M hydroiodic acid.

(1) Before the addition of any hydroiodic acid, the pH is (2) After adding 7.53 mL of hydroiodic acid, the pH is

(3) At the titration midpoint, the pH is

(4) At the equivalence point, the pH is

(5) After adding 26.5 mL of hydroiodic acid, the pH is

Please show step by step instructions,

many thanks!

Answer 1

trimethyl amine = Kb = 6.3 x 10-5

pKb = 4.2

millimoles of trimethyl amine = 20.8 x 0.304 = 6.32

(a) initial pH

pOH = 1/2 [pKb- logC]

pOH = 1/2 [pKb - log 0.125]

          = 1/2 (4.2 - log 0.304)

          = 2.36

pH + pOH = 14

pH = 14 - pOH

      = 11.64

pH = 11.64

b) 7.53 mL titrant added :

millimoles of HCl = 7.53 x 0.341 = 2.57

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

   6.32      2.57                                     0

   3.75      0                                        2.57

here salt and base remained . so it forms basic buffer

pOH = pKb + log [salt / base]

         = 4.2 + log [2.57 / 3.75]

          = 4.04

pH = 9.96

c) At the titration midpoint, the pH is

here pOH = pKb

pOH = 4.2

pH + pOH = 14

pH = 9.80

(4) At the equivalence point, the pH is

at this point volume of HCl needed = 18.5 mL

here only salt remains = salt concentration = C = 20.8 x 0.304 / (20.8 + 18.5) = 0.161 M

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

pH = 7- 1/2 [4.20 + log 0.161]

pH = 5.30