# Wouldn't adding a proton acceptor still make a solution just as acidic?

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+?

• 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? – Blaise Feb 8 at 18:05
• 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. – 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