Dimethylamine, (CH3)2NH, is an organic base with pKb= 3.23at 298 K. In an experi
ID: 1066118 • Letter: D
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
Explanation / Answer
(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