The title says it all.
I had this question when helping someone with their high-school chemistry homework. What are the reasons that someone would use an (from my point of view) obfuscated notation?
Answer
The lone pair at the nitrogen is very well suited to form hydrogen bonds. This bond is especially strong, since ammonia has a quite high proton affinity, it is a medium strong base.
Now consider the autoprotolysis reactions in pure water and (liquid) ammonia: \begin{align} \ce{2H2O &<=> H3O+ + OH-}\tag1\\ \ce{2NH3 &<=> NH4+ + NH2-}\tag2 \end{align}
Mixing these systems, the reaction of ammonia with water will yield \begin{align} && \ce{2H2O + 2NH3 &<=> H3O+ + OH- + NH4+ + NH2-}\tag3 \\ & (~-~)& \ce{H3O+ + NH2- &-> H2O + NH3}\\\hline && \ce{H2O + NH3 &<=> NH4+ + OH-}\tag4\label{eq:ammonia+water} \end{align} \eqref{eq:ammonia+water} is incorporated into a huge and strong network of hydrogen bonds. In this case the equilibrium will be very much pushed to the right side, meaning, that almost all ammonia molecules will be protonated.
Therefore it is more appropriate to write: \begin{align} \ce{NH4OH (aq) &<=> NH4+ (aq) + OH- (aq)}\tag{4a} \end{align}
However, it is always advisable to look at the conditions of the whole reaction system.
Note that $\ce{NH4OH}$ is a pretty outdated notation, if you would want to consider one water entity explicitly, $\ce{NH3.H2O}$ would be more appropriate. See the answer by Maurice for some more information: Can acid-base reactions occur in a non-aqueous medium?
No comments:
Post a Comment