I was introduced with an exercise which asks to find $K_\mathrm{b}$ of $\ce{NH3}$ when the $\mathrm{pH}$ of $\ce{NH3}$ solution with concentration $M_1$. Well, in order to solve it, I used the fact that $\mathrm{pOH} = 14 - \mathrm{pH}$ and the dissociation $\ce{NH3 + H2O <=> NH4+ + OH-}$ to compute $\ce{[OH-]}$ and then $K_\mathrm{b}$. However I don't fully understand what does $\mathrm{pH}$ mean, since in this dissociation there aren't any $\ce{H3O+}$ molecules. I think it relates to the conjugate dissociation of the $\ce{NH4+}$, i.e.: $\ce{NH4+ + H2O <=> NH3 + H3O+}$, so the $\mathrm{pH}$ of the $\ce{NH3}$ solution is in fact related to the concentration of $\ce{H3O+}$ which is formed from the $\ce{NH4+}$ dissociation. Is it indeed the case? I would really appreciate any explanation.

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    $\begingroup$ pH means the same as always. See, pure water also has pH, hence it must have some H3O+. How so? Well, just like that. $\endgroup$ – Ivan Neretin Jul 8 '19 at 15:57
  • $\begingroup$ Yes, but I try to understand where the $H_3O^+$ come from since the $NH_3$ forms $NH_4^+ + OH^-$ when reacting with water and not $H_3O^+$. I'll try to emphasize my question with another example - taking $NH_4Cl$ which dissociate to $NH_4^+ + Cl^- $ , how should I understand (or maybe view would be a better term) the $pH$ and the $pOH$ of $NH_4Cl$ solution? @IvanNeretin $\endgroup$ – user5721565 Jul 8 '19 at 16:16
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    $\begingroup$ For that you need to understand where the H3O+ comes from in pure H2O. $\endgroup$ – Ivan Neretin Jul 8 '19 at 16:20

As Ivan Neretin said in the comments, $\ce{H3O+}$ is already there, via the autodissociation of water:

$$\ce{2H2O <=> H3O+ + OH-}\tag{1}$$

Now water reacts with the ammonia that you add, according to the equilibrium reaction you provided:

$$\ce{NH3 + H2O <=> NH4+ + OH-}\tag{2}$$

This increases the hydroxide concentration further, so the first reaction is out of equilibrium, and consumes some hydronium and hydroxide by going in the net reverse direction, increasing the pH.

Alternatively, you can combine the two reactions to show that adding ammonia will lower the hydronium concentration:

$$\ce{NH3 + H3O+ <=> NH4+ + H2O}\tag{3}$$

All three reaction and the reverse reactions will occur simultaneously.

However I don't fully understand what does pH mean, since in this dissociation there aren't any H3O+ molecules.

For basic solutions, the hydronium ion is a minor species. For acidic solutions, the hydroxide ion is a minor species. Sometimes in calculations, we get away with ignoring a minor species for a while, but they are still there.


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