Question

Dimethylamine, (CH3)2NH, is an organic base with pKb= 3.23at 298 K. In an experiment, a 40.0 mL sample of 0.105mol/L-1(CH3)2NH(aq)is titrated with 0.150mol/L-11HI(aq)at 298 K.

(a) Write a balanced chemical equation for the neutralization reaction upon which this titration is based, and indicate clearly which atom is being protonated. Then, calculate the equilibrium constant for the neutralization reaction. (Hint: To calculate the equilibrium constant, you may find it helpful to represent the neutralization reaction as the sum of twoor morereactions.)

(b) Calculate the pH, [(CH3)2NH], and[(CH3)2NH2+] at the following stages of the titration. Justify any approximations.

i) before the addition of any HI solution

ii) after the addition of 20.0 mL of HI solution

iii) at the equivalence point

iv) after the addition of 60.0 mL of HI solution

Answer 1

(CH3)2NH(aq) + HI(aq) ------> (CH3)2NH2+(aq) + I-(aq)

kb of (CH3)2NH = 10^-kb

              = 10^-3.23

             kb = 5.9*10^-4

b)

i) before the addition of any HI solution

       pH of (CH3)2NH = 14 - 1/2(pkb-logC)

                      = 14 - 1/2(3.23-log0.105)

                      = 11.9

ii) after the addition of 20.0 mL of HI solution

   no of mole of (CH3)2NH = M*V = 40*0.105 = 4.2 mmole

   no of mole of HI = M*V = 20*0.15 = 3 mmole

   pH = 14 - (pkb+log(salt(or)acid/base)

      = 14 - (3.23+log(3/(4.2-3))

      = 10.37

iii) at the equivalence point

no of mole of (CH3)2NH = M*V = 40*0.105 = 4.2 mmole

no of mole of HI = M*V = 20*0.15 = 4.2 mmole
  
volume of HI added =n/M = 4.2/0.15 = 28 ml

concentration of salt = 4.2/(28+40) = 0.0617 M

pH = 7-1/2(pkb+logC)

      = 7-1/2(3.23+log0.0617)

      = 6

iv)

   concentration of excess HI = (60 - 28)*0.15/(60+40) = 0.048 M

   pH = -logH+

       = -log0.048

       = 1.32