# 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

The first thing to say is that you final equation doesn't balance. In fact it is impossible to balance your equation using hydroxide because you can never conserve charge with only neutral products.

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.

At first it'll form an Al(OH)3, but we should remember that Al(OH)3 is a Lewis acid since it'll need a pair of electron to complete the 'octet rule', and when it's formed in the solution (aqueous), the water (H2O)'s Oxygen will solvate the Al(OH)3 and it'll form an Al(OH)4-. Adding excess base will also resulting in the making of Al(OH)4-