Answer:
9.95 pH
Step-by-step explanation:
To determine the pH of a sodium fluoride solution, we need to calculate the concentration of the fluoride ion, F⁻, using the dissociation constant expression for sodium fluoride:
Kb = [HF][OH⁻] / [F⁻]
where Kb is the base dissociation constant of sodium fluoride, [HF] is the concentration of hydrogen fluoride, [OH⁻] is the concentration of hydroxide ions, and [F⁻] is the concentration of fluoride ions.
Sodium fluoride is a salt of a weak acid (hydrogen fluoride, HF) and a strong base (sodium hydroxide, NaOH), and so it undergoes hydrolysis in water to produce hydroxide ions and the weak acid HF:
NaF + H₂O → NaOH + HF
The Kb expression for sodium fluoride then becomes:
Kb = [HF][OH⁻] / [NaF]
Since the solution is 0.43 M in sodium fluoride, the concentration of fluoride ions is also 0.43 M. Assuming complete hydrolysis, the concentration of hydroxide ions can be calculated from the Kb expression:
Kb = [HF][OH⁻] / [F⁻]
1.8 x 10⁻¹¹ = x² / 0.43
x = 1.12 x 10⁻⁶ M
The pH of the solution can then be calculated from the concentration of hydroxide ions:
pH = 14 - pOH = 14 - (-log[OH⁻]) = 14 - (-log(1.12 x 10⁻⁶)) = 9.95
Therefore, the pH of the sodium fluoride solution is approximately 9.95.