Question

Weak Acid	Ka	Weak Base	Kb
CH3COOH (Acetic Acid)	1.8 X 10-5	NH3 (Ammonia)	1.76 X 10-5
C6H5COOH (Benzoic Acid)	6.5 X 10-5	C6H5NH2 (Aniline)	3.9 X 10-10
CH3CH2CH2COOH (Butanoic Acid)	1.5 X 10-5	(CH3CH2)2NH (Diethyl amine)	6.9 X 10-4
HCOOH (Formic Acid)	1.8 X 10-4	C5H5N (Pyridine)	1.7 X 10-9
HBrO (Hypobromous Acid)	2.8 X 10-9	CH3CH2NH2 (Ethyl amine)	5.6 X 10-4
HNO2 (Nitrous Acid)	4.6 X 10-4	(CH3)3N (Trimethyl amine)	6.4 X 10-5
HClO (Hypochlorous Acid)	2.9 X 10-8	(CH3)2NH (Dimethyl amine)	5.4 X 10-4
CH3CH2COOH (Propanoic Acid)	1.3 X 10-5	CH3CH2CH2NH2 (Propyl amine	3.5 X 10-4
HCN (Hydrocyanic Acid)	4.9 X 10-10

When you titrate 17.2 mL of a 0.14 M solution of (CH₃)₃N (trimethylamine), it required 17.0 mL of hydrochloric acid solution. Assuming that you have reached the equivalence point, what is the pH of the resulting solution?

Answer 1

at equivalence point, all (CH₃)₃N is converted to (CH₃)₃NH⁺,

             (CH₃)₃N   +      H₃O⁺       --->    (CH₃)₃NH⁺    +    H₂O

Initial:          2.408E-3 mol             0                               0

add:                    0                    2.408E-3 mol                 0

change:    -2.408E-3 mol      -2.408E-3 mol          +2.408E-3 mol  

equilibrium:        0                           0                       2.408E-3 mol  

the final concentration of (CH₃)₃NH⁺ = 2.408E-3 mol/ (0.017.2 L + 0.017 L) = 0.0704 M

so the only reaction that occurs at equivalence point is the hydrolysis of (CH₃)₃NH⁺

          (CH₃)₃NH⁺        +   H₂O    ----> (CH₃)₃NH + H₃O⁺

I:      0.0704 M

C:            -X                                                  X                 X

E:   0.0704 M - X                                          X                  X

                   Ka_{protonated trimethyl amine} = Kw/kb = 1.0E-14 / 6.4E-5 = 1.56E-10

                               Ka = [(CH₃)₃NH][H₃O⁺] / [(CH₃)₃NH⁺]

                               1.56E-10 = x² / 0.0704 M - X

         since, 0.0704 M/ 1.56E-10 >>>1000, we can neglect x in the denominator

          1.56E-10 = x² / 0.0704 M

                                              x² = (1.56E-10)(0.0704 M)

                                               x = 3.313E-6 M

                                       pH = -log 3.313E-6 M = 5.48