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$$\ce{NH3_{(aq)} + H2O_{(l)} <=> NH4^{+}_{(aq)} + \fbox{$\ce{OH^{-}_{(aq)}}$}}$$

Now one might say, due to the presence of $\ce{OH^-}$, the aqueous solution is alkaline. But I'd like to make the opposite claim:

$$\ce{NH3_{(aq)} + H2O_{(l)}<=>\fbox{$\ce{NH_4^{+}_{(aq)}}$} + OH^{-}_{(aq)}}$$

Due to the presence of ammonium ions, can't we say that the solution is acidic?

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  • $\begingroup$ Solutions of NH3 and NH4Cl have very different ratio of NH3/NH4+ concentrations. Involve respective equilibrium dissociation constants for water and for ammonia and you will see. $\endgroup$
    – Poutnik
    Nov 23 at 7:25
  • $\begingroup$ It is not only about the possibility to donate a proton at all, it is about the relative strength of an acid in comparison of what could act like a proton acceptor/base and (e.g., in the case of water) a limit imposed by the solvents (e.g., in water, there can't be an acid stronger than $\ce{H3O+}$). So, what is the pKa of $\ce{NH^+4}$ in water? $\endgroup$
    – Buttonwood
    Nov 23 at 7:46
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    $\begingroup$ @Buttonwood Wow! - \fbox - something new. $\endgroup$
    – Poutnik
    Nov 23 at 7:59
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    $\begingroup$ Let's distinguish facts from interpretations. Aqueous solution of ammonia is considered alkaline because it is alkaline. Then one may ask why it is like that, and find the answer in the previous comments. $\endgroup$ Nov 23 at 8:25
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    $\begingroup$ @Poutnik Though I don't remember the day I started using LaTeX directly or/and as an intermediate format and (assume to) know the bits and bolts relevant for me, I know there are unknown unknowns (Rumsfeld matrix). No, not packages which sound like programming (e.g., ifthen), but finds like this recent post on tex.se. I found joining a user group an appealing avenue to go further (based on your information, cstug.cz may be interesting for you.) $\endgroup$
    – Buttonwood
    Nov 23 at 15:02