# Colour of VO3^-?

To aquire the yellow colour of vanadium ion, it must be present in a 5+ oxdiation state.

This can be acheived (according to my textbook) by adding dilute sulfuric acid to $$\ce{NH4VO3}$$, then followed by zinc.

$$\ce{VO3- (aq) + 2H+ (aq) -> VO2^{+} (aq) + H2O (l)}$$ where $$\ce{VO2^{+}}$$ gives the yellow colour from the $$\ce{V^{5+}}$$ present in the compound

However, is not $$\ce{VO3-}$$(aq) yellow in colour?

If it is, then why is $$\ce{NH4VO3}$$ not yellow in colour but white?

Side question: what is the purpose of zinc in this reaction explained above - not including reduction to lower oxidation states?