# How to work out equations with period 3 oxides?

I know the general equation of reacting an acid with a base to be the salt and water, but I am not able to apply this to the following equations. So in what way can I derive the end product if I have never seen the reactants before?

For example, I can see the water is liberated out in the reaction between silicon dioxide and sodium hydroxide, producing this equation: $$\ce{SiO2 + 2NaOH -> Na2SiO3 + H2O}.$$

However if the reactants were aluminium oxide and sodium hydroxide, how could I work out the equation: $$\ce{Al2O3 + 2NaOH + 3H2O-> 2NaAl(OH)4}$$

I would have thought the product formed would be: $$\ce{Al2O3 + 3NaOH ->2Al(OH)3 + 2Na+}$$

• It is actually possible to precipitate $\ce{Al(OH)3}$ from $\ce{Na[Al(OH)]4}$ solutions. This is a key step of the Bayer process. May 25 '16 at 15:41

You can write a reaction that forms $\ce{Al(OH)3}$: $$\ce{Al2O3 + 3H2O -> 2Al(OH)3}$$
However, beyond about pH 6 the $\ce{Al(OH)3}$ will start to dissolve to form $\ce{[Al(OH)4]-}$. Once you get to around pH 7.5 all of the solid has essentially dissolved and you are left with solution of $\ce{Na+}$ and $\ce{[Al(OH)4]-}$ (see footnote). So unless you are very sparing with your addition on $\ce{NaOH}$ you will get the tetrahydroxo product in solution.
$$\ce{Al(OH)3 + OH- -> [Al(OH)4]-}$$
Footnote: The aluminium species present in aqueous solution are strongly dependent on the pH, particularly around the neutral point. There is a nice graph here showing the different species. The notation is $\ce{Al}$ corresponds to $\ce{[Al(H2O)6]^{3+}}$ and then $\ce{AlH_{-n}}$ denotes the loss of n protons from the hexaaqua complex. This was originally found by @Jan for this question.