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What is the pH of a solution that is 0.43 M in sodium fluoride?

User Shakthi
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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.

User Undefitied
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