Say if you add NH3 to a solution, and it accepts a proton, it still becomes NH4+ in a solution, which is also acidic. A base should decrease the hydrogen ion concentration in a solution, but how does a solution become more basic if it now has an increased concentration of NH4+?

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    $\begingroup$ When $NH_3$ accepts a proton, what molecule does the proton come from, and what does this molecule look like after losing a proton? Are there any other acids or bases in the solution, and how will their concentrations change? $\endgroup$ – Blaise Feb 8 at 18:05
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    $\begingroup$ Indeed, H+ is acidic, and so is NH4+, but there is a catch. Indeed, Mike Tyson is strong, and so is an elephant, and so is a nuclear blast, and so is the black hole's gravity. One may be tempted to say they are all the same, but there is a catch. $\endgroup$ – Ivan Neretin Feb 8 at 18:57

Severe misconception. Hydrolysis reaction is $\ce{NH3 + H2O ⇌ NH4+ + OH-}$

Hence, dissolving $\ce{NH3}$ in water produces $\ce{OH-}$ ions, making the solution more basic, as $\ce{[OH^{-}] > [H+]}$


Just to elaborate more on what Mr Organic has said.

Basically what made a solution acidic is when $\ce{[H3O+]>[OH^{-}]}$ and what made a solution alkaline is when $\ce{[H3O+]<[OH^{-}]}$.

You are right about $\ce{NH4^{+}}$ is acidic, but why it is acidic? Because you need to remember the following happen when $\ce{NH4^{+}}$ is dissolved water and caused $\ce{[H3O+]}$ to increase $$\ce{NH4^{+} + H2O ⇌ NH3 + H3O+}$$

But then you would say, the $\ce{NH3}$ will accept a proton again and decrease the $\ce{[H3O+]}$. The thing is the reaction is a reversible reaction and will eventually become equilibrium which means the concentration of all reactants and products will not changed. And at that moment, the increase of $\ce{[H3O+]}$ caused by $\ce{NH4^{+}}$ is far more than the decrease of $\ce{[H3O+]}$ caused by $\ce{NH3}$. Thus $\ce{[H3O+]>[OH^{-}]}$ and make the solution acidic.

Now back to your question. First the $\ce{NH3}$ will increase the $\ce{[OH^{-}]}$ by the following reaction $$\ce{NH3 + H2O ⇌ NH4^{+} + OH^{-}}$$.

But the $\ce{NH4^{+}}$ will release proton and increase $\ce{[H3O+]}$. However, the increase of $\ce{[OH^{-}]}$ caused by $\ce{NH3}$ is far more than the increase of $\ce{[H3O+]}$ caused by $\ce{NH4^{+}}$ . Thus $\ce{[H3O+]<[OH^{-}]}$ and make the solution alkaline


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