Can you pls help me write a ionic equation for this question

The addition of excess sodium hydroxide solution to a suspension of aluminium hydroxide in water.

pls show me how you did this, thankyouu (:

This is the formation of a complex ion, and the simplest way to do it would be to double the charge on the metal, in this case Aluminum, and use that number of, in this case, hydroxides, as follows:

(Na+) + (OH-) + Al(OH)3 yields Al(OH)6 + (Na+)

but because it is a net ionic equation the sodium ions cancel out because they are only spectator ions and after balancing it, you are left with a net ionic equation of just

Al(OH)3 + 3(OH-) yields Al(OH)6