Dimethylamine, (CH3)2NH, is an organic base with pKb= 3.23at 298 K. In an experi

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 (CH3)2NH(aq)is titrated with 0.150mol/HI(aq)at 298 K

2. Dimethylamine, (CH32NH, is an organic base with pKb 3.23 at 298 K. In an experiment, a 40.0 mL sample of 0.105 mol L (CH32NH(aq) is titrated with 0.150 mol L HI(aq at 298 K. (a) (5 marks) 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 two or more reactions.)

Explanation / Answer

Titration

(a) Neutralization equation

(CH3)2NH + HI ---> (CH3)2NH2I

(CH3)2NH2+ + H2O <==> (CH3)2NH + H3O+

Keq = [(CH3)2NH][H3O+]/[(CH3)2NH2+]

(a) Addition of HI volumes

(i) before any HI was added

(CH3)2NH + H2O <==> (CH3)2NH2+ + OH-

Kb = [(CH3)2NH2+][OH-]/[(CH3)2NH]

6 x 10^-4 = x^2/0.105

x = [OH-] = 7.94 x 10^-3 M

[H3O+] = 1 x 10^-14/7.94 x 10^-3 = 1.26 x 10^-12 M

pH = -log[H3O+] = 11.90

(ii) 20 ml HI added

moles of (CH3)2NH = 0.105 M x 40 ml = 4.2 mmol

moles of HI added = 0.150 M x 20 ml = 3 mmol

[(CH3)2NH2+] formed = 3 mmol/60 ml = 0.05 M

[(CH3)2NH] remained = 1.2 mmol/60 ml = 0.02 M

pH = pKa + log(base/acid)

     = 10.77 + log(0.02/0.05)

     = 10.37

(iii) equivalence point

[(CH3)2NH2+] formed = 4.1 mmol/68 ml = 0.0603 M

(CH3)2NH2+ + H2O <==> (CH3)2NH + H3O+

Ka = 1.7 x 10^-11 = x^2/0.0603

x = [H3O+] = 1.01 x 10^-6 M

pH = -log[H3O+] = 5.995

(iv) 60 ml HI added

excess [H+] = 0.150 M x 32 ml/100 ml = 0.048 M

pH = -log[H+] = 1.32

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