# Reaction between sodium hydroxide and aluminium oxide [closed]

I am a 10th grade student.

Is it possible to know the reason behind the formation of sodium aluminate when sodium hydroxide and aluminum oxide reacts. Is there an intuitive reason for this other than mere observation.

I would like to know an easy explanation as I am a 10th grade student

The strange property of $$\ce{Al2O3}$$ is due to the position of aluminium in the periodic table. As it is not obvious, I have to start a rather long explanation. Sorry !

Let's start with the definition of acids used in the beginning of the $$20$$th century. At this time an acid was a substance containing a Hydrogen atom and able to produce $$\ce{H^+}$$ (or $$\ce{H3O^+}$$) in water. Their formula were beginning with $$\ce{H}$$, and hydroxide had formula ending in $$\ce{H}$$. The main property of the acids was that they could react with Hydroxides like NaOH, to produce salts, according to the general equation :

Acid + Hydroxide -> Salt + Water.

With this definition, an acid is made of Hydrogen, maybe some Oxygen and another element which is electronegative. All non-metals are in this group. But there are a couple of metallic elements which are nearly at the border of the limit metals - non-metals in the periodic table. These elements deserve to be discussed.

Aluminium is in this border group. It makes an hydroxide $$\ce{Al(OH)3}$$ which usually works as a hydroxide and sometimes as an acid. Usually it works as a hydroxide in reaction with strong acids, like $$\ce{Al(OH)3 + 3HCl -> AlCl3 + 3 H2O}$$ $$\ce{2 Al(OH)3 + 3 H2SO4 -> Al2(SO4)3 + 6 H2O}$$

But $$\ce{Al(OH)3}$$ is a rather weak hydroxide. It does produce extremely few $$\ce{OH-}$$ ions in water. And such weak hydroxides can sometimes behave as weak acids, if they are in contact with a strong acid. Matter of fact, $$\ce{Al(OH)3}$$ reacts with NaOH, which is a strong hydroxide, as if it was a weak acid $$\ce{H3AlO3}$$. The equation can be : $$\ce{H3AlO3 + 3 NaOH -> Na3AlO3 + 3 H2O }$$ or, depending on the concentrations : $$\ce{H3AlO3 + NaOH -> NaAlO2 + 2 H2O}$$ Today we know that the aluminate ion is either $$\ce{[Al(OH)4]^{-}}$$ or $$\ce{[Al(OH)6]^{3-}}$$. But it does not matter for the present development.

Now you should understand that the strange reaction of aluminium hydroxide is due to the position of aluminium at the border of the separation metals - non-metals in the periodic table.

Your question was not related to aluminium hydroxide, but to aluminium oxide. All that has been said about aluminium hydroxide can also be said about oxide, because in water, the following equilibrium can occur : $$\ce{2 Al(OH)3 <=> Al2O3 + 3 H2O}$$ so that $$\ce{Al2O3}$$ may react with acids and hydroxides exactly like $$\ce{Al(OH)3}$$

• Strange? Arguably, most metals are amphoteric to some extent. Aug 29 '21 at 18:00
• Elements from columns $1$ and $2$ are not amphoteric Aug 29 '21 at 18:17
• That doesn't contradict what I said. Then again, Be is amphoteric. Aug 29 '21 at 18:34
• Be is at the border, as its neighbor the boron B is already a non-metal. Aug 29 '21 at 18:48
• Elements are metals or non-metals, For me, Boron is not a metal. As a consequence, it is a non-metal. In my dictionary, metalloid is an old name for non-metal Aug 29 '21 at 20:30