How Fluoride Protects Dental Enamel from Demineralization

INTRODUCTION: How fluoride (F(–)) protects dental enamel from caries is here conveyed to dental health-care providers by making simplifying approximations that accurately convey the essential principles, without obscuring them in a myriad of qualifications. MATERIALS AND METHODS: We approximate that dental enamel is composed of calcium hydroxyapatite (HAP), a sparingly soluble ionic solid with the chemical formula Ca(10)(PO(4))(6)(OH)(2). RESULTS: The electrostatic forces binding ionic solids together are described by Coulomb’s law, which shows that attractions between opposite charges increase greatly as their separation decreases. Relatively large phosphate ions (PO(4)(3–)) dominate the structure of HAP, which approximates a hexagonal close-packed structure. The smaller Ca(2+) and OH(–) ions fit into the small spaces (interstices) between phosphates, slightly expanding the close-packed structure. F(–) ions are smaller than OH(–) ions, so substituting F(–) for OH(–) allows packing the same number of ions into a smaller volume, increasing their forces of attraction. Dental decay results from tipping the solubility equilibrium Ca(10)(PO(4))(6)(OH)(2) (s) ⇔ 10Ca(2+) (aq) + 6PO(4)(2–) (aq) + 2OH(–) (aq) toward dissolution. HAP dissolves when the product of its ion concentrations, [Ca(2+)](10)×[PO(4)(3–)](6)×[OH(–)](2), falls below the solubility product constant (Ksp) for HAP. CONCLUSION: Because of its more compact crystal structure, the Ksp for fluorapatite (FAP) is lower than the Ksp for HAP, so its ion product, [Ca(2+)](10)×[PO(4)(3–)](6)×[F(–)](2), must fall further before demineralization can occur. Lowering the pH of the fluid surrounding enamel greatly reduces [PO(4)(3–)] (lowering the ion products of HAP and FAP equally), but [OH(–)] falls much more rapidly than [F(–)], so FAP better resists acid attack.