Question

Hydrogen fluoride behaves as a weak acid in solution. Two equilibria influence which fluorine containing species are present in solution.

HF (g) + H₂O (l) --> H₃O⁺ (aq) + F- (aq) (Ka= 1.1*10⁻³)
F⁻(aq) +HF (g) --> HF₂⁻ (aq) (Ka= 2.6*10⁻¹)


a. Is fluoride in pH 7.00 drinking water more likely to be present as F⁻ or HF₂⁻

Answer 1

In pH 7.00 drinking water, fluoride is more likely to be present as fluoride ions (F-) rather than bifluoride ions (HF2-) due to the neutral conditions which do not favor excess hydrogen ions that would lead to greater amounts of HF2-.

Step-by-step explanation:

Hydrogen fluoride (HF) behaves as a weak acid in aqueous solution, establishing an equilibrium between its molecules and the ions produced upon dissolution. When dissolved in water, HF interacts with water molecules to form hydronium ions (H3O+) and fluoride ions (F-). Additionally, fluoride ions can react with undissociated HF to form bifluoride ions (HF2-). The relevant equilibria are: HF (aq) ⇌ H+ (aq) + F- (aq) with a Ka = 1.1 x 10-3, and F- (aq) + HF (aq) ⇌ HF2- (aq) with a Ka = 2.6 x 10-1.

At pH 7.00, the concentration of hydrogen ions ([H+]) is equal to 1 x 10-7 M, which means that the solution is neutral and there is not an excess of H+ that would favor the formation of HF2-. Therefore, in pH 7.00 drinking water, fluoride is more likely to be present as F- rather than as HF2-.

This is supported by considering the ionization of acids in solution and the stability of the ionic species in neutral conditions. The presence of F- ions acting as weak Brønsted-Lowry bases in the reaction with water to form hydroxide ions (OH-) can also cause the pH to rise slightly above 7, indicating a basic tendency, but under neutral conditions, the formation of OH- is minimal, and therefore, F- remains the dominant fluorine-containing species.