Problem: (b) What is the pH of 0.40 M triethylammonium chloride, (CH3CH2)3NHCl?

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FREE Expert Solution

The dissociation of (CH3CH2)3NHCl is as follows:


(CH3CH2)3NHCl(aq)  Cl(aq)(CH3CH2)3NH+(aq)

[(CH3CH2)3NHCl] = [(CH3CH2)3NH+] = 0.40 M


The hydrolysis of (CH3CH2)3NH+:

(CH3CH2)3NH+(aq) + H2O(l) H3O+(aq) + (CH3CH2)3N(aq)


Kb of (CH3CH2)3NH = 5.2 × 10–4


Kw=KaKbKwKb=KaKbKbKa=KwKbKa=1.0×10-145.2×10-4

Ka = 1.92 x 10–11 

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Problem Details

(b) What is the pH of 0.40 M triethylammonium chloride, (CH3CH2)3NHCl?

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Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.