The Preparation of Porous TiO[sub 2] by Immersing Ti in NaOH Solution

Abstract

The formation of a porous TiO 2 layer by immersing Ti in 5 M NaOH at 303 K was investigated using electrochemical measurements, a scanning electron microscope, and grazing incident X-ray diffraction. The porous layer was readily obtained in the case of the presence of titanium hydride (TiH 2 ) on the surface before immersion. The crystalline structure of the porous layer is composed of a mixture of the rutile and anatase type TiO 2 . The porous layer is hard to produce without the presence of TiH 2 . The TiH 2 is directly changed to TiO 2 by a dissolution reaction in alkaline solution. The presence of TiH 2 on the surface is an important factor for the preparation of the porous TiO 2 layer.

Large-Area Sodium Titanate Nanorods Formed On Titanium Surface Via NaOH Alkali Treatment

“…Fig. 2 (a), (b), and (c) show the pore morphology that formed on the surface of the Ti disk, and many researchers have reported that such a morphology can only be obtained when Ti is immersed into an NaOH solution [12,13]. The increase in temperature during heat treatment densifies the pores and changes them into rod-like features.…”

Effect of electrical-discharging on formation of nanoporous biocompatible layer on titanium

“…Furthermore, a nanoporous and thick titanium oxide surface can induce bony ingrowth into the porous structure, resulting in morphological fixation of the implants to the bone [8][9][10]. Previously, numerous procedures for forming titanium oxide films have been examined [8][9][10][11][12][13][14]. Among them, anodization is very popular.…”

Effect of electro-discharging on formation of biocompatible layer on implant surface

“…Hence, metal-based implants are recommended for use as coatings on bloodcontacting implants. The adhesion of blood cells on various implants with smooth and rough surfaces following surface treatment has been discussed elsewhere [22,23]. Cell adhesion was higher on rougher surfaces, with a Ra in the range 0.2-0.5 mm [24].…”