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I'll try to approach the equilibrium from both sides, and see what happens. Stay with me for the ride guys!

Case 1:

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

Let the initial concentration of $\ce{NH_3}$ is 1M. Now, let $\pu{\alpha M}$ NH3 react with water to produce $\pu{\alpha M}$ NH4+ and $\pu{\alpha M}$ OH-. Now, the concentrations at equilibrium will be,

$[NH_3]=(1-\alpha)M$

$[NH_4^{+}]=\alpha M$

$[OH^{-}]=\alpha M$

$$K_b=\frac{[NH_4^{+}][OH^{-}]}{[NH_3]}$$

$$1.8\times 10^{-5}=\frac{\alpha^2}{(1-\alpha)}$$

$$\alpha^2+1.8\times 10^{5}\alpha-1.8\times 10^{5}=0$$

$$\alpha=-0.0042516502330501,0.0042336502330501$$

Ignoring the negative value, because degree of dissociation can't be negative here,

$$\alpha=0.0042(approx.)$$

Now the equilibrium concentrations,

$[NH_3]=0.9958M$

$[NH_4^{+}]=0.0042 M$

$[OH^{-}]=0.0042 M$

So, we can see that the equilibrium situates far to the left: the concentration of NH3 barely decreses.

Case 2:

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

Let the initial concentrations of NH4+ and OH- are 1M and 1M respectively. Let $\pu{\alpha M}$ NH4+ and $\pu{\alpha M}$ OH- react to produce $\pu{\alpha M}$ NH3. Now, the concentrations at equilibrium will be,

$[NH_3]=(\alpha)M$

$[NH_4^{+}]=(1-\alpha)M$

$[OH^{-}]=(1-\alpha)M$

Now, here the equilibrium constant will be the reciprocal of Kb since we're going the other way around:

$$\frac{1}{K_b}=\frac{[NH_3]}{[NH_4^{+}][OH^{-}]}$$

$$\frac{1}{1.8\times 10^{-5}}=\frac{\alpha}{{(1-\alpha)}^2}$$

$$1-2\alpha+\alpha^2=1.8\times 10^{-5}\alpha$$

$$\alpha^2-\alpha(2+1.8\times10^{-5})+1=0$$

$$\alpha=1.004,0.995766$$

Neglecting the value that is greater than 1, the equilibrium concentrations are:

$[NH_3]=0.995766M$

$[NH_4^{+}]=0.004234M$

$[OH^{-}]=0.004234M$

These values are pretty close to the ones we got in case 1. So, I think we have done the math really nicely :)

My observations from case 2:

We can see that nearly all the NH4+ ions donate the proton to OH- and form H20 and NH3. So, I think NH4+ is an extremly strong acid like HCl and H2SO4, what do you guys think?

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    $\begingroup$ Alternatively, hydroxide is a very strong base. Consider adding ammonium chloride to neutral water. Do you still think ammonium is a strong acid? $\endgroup$
    – Andrew
    Nov 23 at 17:44
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    $\begingroup$ NH4+ is very weak acid with pKa about 9.25. HCl is neutralized at pH 7. At pH 9.25, a half of NH4+ still remains as an acid. $\endgroup$
    – Poutnik
    Nov 23 at 18:02
  • $\begingroup$ @Poutnik "At pH 9.25, a half of NH4+ still remains as an acid". Do you mean that 0.5M NH4+ remains in undissociated form? $\endgroup$
    – Tom Hardy
    Nov 23 at 18:26
  • $\begingroup$ Your NH4+ reacts with hydroxide(!) ions. Thing is, if you formulate your ammonia + water = ammonium + hydroxide equilibrium, the value of it is always valid. That is also true when you start with ammonium and hydroxide (for example, dissolving ammonium chloride in NaOH solution). The thermodynamic constant for the reaction is still valid. However, if you let Ammonium itself react with water to test its acidity in water, you will see that the equilibrium for hydronium ions is fairly small. $\endgroup$
    – 冰淇淋
    Nov 23 at 19:52
  • $\begingroup$ There is a stupid mistake in the sign. The $K_b$ constant is not $\pu{1.8 10^{5}}$ as you say. It is $\pu{1.8 10^{-5}}$ !! $\endgroup$
    – Maurice
    Nov 23 at 20:08