When you titrate 17.2 mL of a 0.14 M solution of (CH 3 ) 3 N (trimethylamine), i
ID: 907470 • Letter: W
Question
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?
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 10-10Explanation / Answer
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
Kaprotonated 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
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