How is NaOH a base? [closed]

Question

Firstly, how is $NaOH$ created? My teacher said that before it is $NaOH$, it is $NaO^-$.

$$NaO^- + H_2O = NaOH + OH^-$$

If so, that means that $NaO^-$ is a base, which would make $NaOH$ it's conjugate acid. However, $NaOH$ is a base. So, perhaps my teacher said wrong, and the fact is, the "prelude" to $NaOH$ is $NaOH_2^+$, i.e.:

$$NaOH_2^+ + H_2O = NaOH + H_3O^+$$

Meaning that $NaOH_2^+$ is the acid, and $NaOH$ is the conjugate base.

Okay, so assuming the latter is correct, I still don't see how NaOH is a base.

Bases, in their most general definition, are substances that easily accept proton(s). Now, this would make sense for $NaOH$ if the equation looked like this:

$NaOH + H_2O = NaOH_2^+ + OH^-$

This would make the solution more basic, due to the higher concentration of $OH-$, though considering how strong of a base $NaOH$ is, that means that $NaOH_2^+$ is a weak acid, and the solution would therefore revert back a lot (double arrow thingy).

However, I have never seen the equation like that. Instead, I've seen it like this:

$$NaOH + H_2O = Na^+ (aq) + OH^-(aq)$$

Now, that would increase the concentration of $OH-$, would would make the solution more basic, but there is no proton being accepted here. The so-called base that is $NaOH$ does not accept a proton, but rather releases an $OH^-$ group, which is quite the opposite of accepting a proton. Now, this would make sense if $NaOH$ was classified as "a basic salt", as in, a salt that makes a solution more basic. So, with such an explanation, $NaOH$ is just a salt, not a base, but it is a salt that makes the solution more basic.

Now, since it is a substance that makes a water solution more basic, it would by definition become an Arrhenius base. I can accept that, because the Arrhenius base definition only requires the substance to make the water solution more basic. $NaOH$ does that. However, I cannot see in any way how $NaOH$ fits the Brønsted-Lowry definiton of a base.

Please explain to me simply, but thoroughly, if you can. For some reason, the Acid-Base part of chemistry has evaded my comprehension for a while now.

Answer 1

Let me analyze your sentences gradually, and try to correct the numerous mistakes.

1. Your teacher says that "before it is $\ce{NaOH}$, it was $\ce{NaO^-}$". No. First $\ce{NaO-}$ does not exist and cannot exist. Second NaOH has not been created out of an ion. Sodium hydroxide is usually produced from the sodium metal after reaction with water according to : $$\ce{2 Na + 2 H2O -> 2 NaOH + H2}$$ This equation may also be written in order to show the ions : $$\ce{2 Na + 2 H2O -> 2 Na^+ + 2 OH^- + H2 }$$

2. Your teacher said that $\ce{NaOH}$ is the conjugate acid of the base $\ce{NaO^-}$. No ! it is impossible to find a structure for $\ce{NaO^-}$, as $\ce{Na}$ cannot be covalently bound to an oxygen atom. $\ce{Na}$ can only be bound to an oxygen atom by an ionic bond, and this bond will be immediately broken when dissolving it into water. Atoms included inside a polyatomic ion are always bound by covalent bonds, never by ionic bonds.

3. Your teacher said that "the prelude of NaOH is $\ce{NaOH2^+}$". First there are no "preludes" to $\ce{NaOH}$. Second $\ce{NaOH2^+}$ does not exist and cannot exist. If such an ion was imagined in any solution, it would immediately get decomposed into $\ce{Na+ + H2O}$

4. Your teacher believes that a reaction can produce simultaneously $\ce{NaOH}$ and $\ce{H3O+}$. This is wrong ! As soon as the smallest amount of $\ce{NaOH}$ and $\ce{H3O+}$ are in contact one another they would react producing $\ce{NaOH + H3O+ -> Na+ + 2 H2O}$

5. Your teacher said that $\ce{NaOH2^+}$ is a weak acid. No ! It is not a weak acid, as it doe snot exist.

6. You said that if NaOH is a base, it must have $\ce{NaOH2^+}$ as conjugated acid. No ! $\ce{NaOH}$ is not a base. It only contains the ion $\ce{OH^-}$ which is a base. $\ce{NaOH}$ is not a molecule. It is a juxtaposition of the two ions $\ce{Na+}$ and $\ce{OH-}$- When dissolved into water, this two ions are separated and can react indepentedly from one another.

7. It is right to say that $\ce{NaOH}$ behaves like a basic salt.

8. Apparently your teacher has never mentioned the fundamental property of $\ce{NaOH}$ which is its dissociation in water according to $$\ce{NaOH -> Na+ + OH-}$$ and this is why $\ce{NaOH}$ is basic in solution

Answer 2

However, I cannot see in any way how NaOH fits the Brønsted-Lowry definiton of a base.

Here are the two chemical equations that show you how the hydroxide anion present in an aqueous NaOH solution acts as a Brønsted-Lowry base:

$$\ce{OH-(aq) + H3O+(aq) <=> H2O(l)}\tag{1}$$

This shows that adding NaOH to water lowers the hydrogen ion concentration (i.e. raising the pH) by hydroxide reacting with hydronium ions to form water.

$$\ce{OH-(aq) + H2O(aq) <=> H2O + OH-(aq)}\tag{2}$$

This one is a bit strange, and the reason why NaOH is not your typical base (like ammonia or sodium acetate etc.). The hydroxide ion reacts with water to abstract a proton, turning into water. At the same time, water turns into hydroxide. The proton transfer is difficult to see because reactants and products are the same unless you somehow tag the two oxygen atoms involved.

My teacher said ...

As for your teacher, see Maurice's answer and the comments (which already contain the content of my answer but I think it helps to have it as an answer specifically about the question of Brønsted-Lowry bases).