ABSTRACT: The Solubility-Reducing Effect on Teeth by SnF

It is well established that stannous fluoride (SnF₂) reduces the solubility of teeth, whereas stannous chloride or sodium fluoride salts have no such effect. Objective: To examine whether fluoride at the concentration and pH at which it is present in 0.4% SnF₂, can reduce solubility of teeth without the presence of Sn²⁺. As the pH of an aqueous solution of 0.4 % SnF₂ is 3, due to an immediate hydrolysis reaction (Sn²⁺ +2H₂O > Sn(OH)₂ +2H⁺), it appears that the fluoride part of the SnF₂-molecule is always exposed to acidic conditions. This fluoride is thus present as partly non-dissociated hydrofluoric acid (HF). Methods: Whole intact human premolars were pre-treated with 0.4% SnF₂ or 0.1% HF (10 min) and then exposed to 0.1M citric acid or 0.01M HCl (30 min). Tooth solubility after acid exposure was estimated by measuring the [Ca²⁺] in the acids, by atomic absorption spectroscopy. Teeth were also examined by EDXA and SEM (Phillips type XL30 ESEM). Results: Solubility reduction by both 0.4% SnF₂ and 0.1% HF were similar in both acids, and not statistically different. This indicated that the fluoride part of SnF₂ alone accounts for its solubility reduction and that Sn²⁺ appears only to provide a suitable pH for the fluoride. The teeth treated with HF were covered with dense layers of CaF₂, which probably caused the solubility reduction in acids. Conclusion: It is thus possible to separate the solubility reduction by SnF₂ from stannous ions. As Sn²⁺ may cause discoloration of teeth, and tend to be both oxidized and hydrolyzed during storage, the aqueous solution of HF may have advantages and be of potential interest in the prevention of dental erosions.