Asked by JL
Hi! So, I thought I was very comfortable with distinguishing among Arrhenius, Bronsted-Lowry, and Lewis acids and bases. But then, I answered this question wrong and am looking for help because I'm not sure if my "second choice" answer is correct.
The question asks: In the gas phase reacion below, NH3 is acting as a(n) _______ base but not as a(n) _______ base.
The reaction is illustrated in Lewis structures: NH3 (including the lone electron pair on the N) plus a single hydrogen, H+. Of course, the reaction produces NH4.
At this point, I want to say that NH3 is acting as an Arrhenius base but not a Lewis base. (My initial choice was the exact opposite - I guess not including water threw me off.)
ANY help or clarification would be greatly appreciated!
The question asks: In the gas phase reacion below, NH3 is acting as a(n) _______ base but not as a(n) _______ base.
The reaction is illustrated in Lewis structures: NH3 (including the lone electron pair on the N) plus a single hydrogen, H+. Of course, the reaction produces NH4.
At this point, I want to say that NH3 is acting as an Arrhenius base but not a Lewis base. (My initial choice was the exact opposite - I guess not including water threw me off.)
ANY help or clarification would be greatly appreciated!
Answers
Answered by
DrBob222
Arrhenius:
in water solution,
acid = provides H^+
base = provides OH^-
Bronsted-Lowry:
acid = proton donor
base = proton acceptor
Lewis:
acid = electron acceptor
base = electron donor
:NH3 + H^+ ==>NH4^+.
NH3 accepts a proton; therefore, it is acting as a Bronsted-Lowry base.
It is donating a pair of electrons, also, therefore, it is acting as a Lewis base.
No OH^- are involved; therefore, it is not acting as an Arrhenius base.
in water solution,
acid = provides H^+
base = provides OH^-
Bronsted-Lowry:
acid = proton donor
base = proton acceptor
Lewis:
acid = electron acceptor
base = electron donor
:NH3 + H^+ ==>NH4^+.
NH3 accepts a proton; therefore, it is acting as a Bronsted-Lowry base.
It is donating a pair of electrons, also, therefore, it is acting as a Lewis base.
No OH^- are involved; therefore, it is not acting as an Arrhenius base.
Answered by
JL
Thanks, Dr. Bob! It seems my initial thinking was correct. Perhaps the question was marked wrong by mistake!
Answered by
Anonymous
kjbouho
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