Reaction between Zn2+ and excess NaOH?

Question

What is the product between Zn2+ and excess NaOH? I observed that the addition of Zn2+ and NaOH yields a white precipitate, and that I can explain it. However, what I don't get is the equation between the precipitate with excess NaOH. What happened? Can anyone show me the equation? (no need to be balanced) I would also appreciate it if you would explain why that happened.

Thanks to whoever answered the question seriously!

Answer 1

zn+2NAOH  gives NA2zno2

sodium zincate    and liberates hydrogen

Answer 2

Zn2+ forms an insoluble hydroxide upon initial addition of sodium hydroxide.

Zn2+(aq) + 2NaOH(aq) → Zn(OH)2(s) + 2Na+(aq)

However, upon further addition of the hydroxide, a complex zinc ion is formed, one which is water soluble. It is called a zincate ion.

Zn(OH)2(s) + 2NaOH(aq) → 2Na+(aq) + Zn(OH)4 ²¯(aq)

This is sometimes used as a test for zinc, but aluminium and lead can behave similarly and would give false positives.