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Passivating titanium - chemical equation

When passivating Ti6-4 as per ASTM B600 (HF &: nitric acid), can someone supply a balanced chemical equation for the basic chemical reaction which takes place?

Cheers

Peta Wingrove
medical equipment - Sydney, NSW, Australia
2003

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2003

The basic equations look like this:

Ti --> Ti4+ + 4e-

Ti4+ + 2H2O --> TiO2 + 4H+

Now, this is highly simplified. Due to the presence of HF, another reaction that takes place is the following:

Ti + mHF + nHNO3 => oTiF4 + pH2O + qNO

Sorry, I don't have the balanced equation for above. The F from HF is slowly consumed by the formation of fluoro and oxofluoro complexes with the Ti4+ from the first reaction. Also, TiO2 is not the only oxide species that is formed. The oxide layer is actually composed of TiO near the Ti substrate, then Ti2O3, and then TiO2 at the surface.

HF is a reducing acid, and therefore promotes more aggressive attack of the titanium surface. HNO3 is an oxidizing acid, and therefore promotes the formation of a stable, passive film. The ratio needs to be a minimum 7:1 (HNO3:HF) in order to limit the uptake of hydrogen into the titanium.

More information on this subject can be obtain from the following references:

"The behaviour of titanium in nitric-hydrofluoric acid solutions", Corrosion Science Volume 30, Issues 4-5 , 1990 , Pages 461-476. (Available from Elsevier/Science Direct) http://www.nsc.co.jp/gikai/en/contenthtml/n85/8512.pdf http://www.timet.com/pdfs/corrosion.pdf

Toby Padfield
Automotive supplier - Michigan

adv.