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Question

What is the pH of a 0.095 M solution of sodium fluoride (NaF). The Ka(HF) = 3.2 x 10-7.

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Blake

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Solution By Steps

Step 1: Identify the Reaction

Sodium fluoride (NaF) dissociates in water to form Na⁺ and F⁻ ions. The fluoride ion (F⁻) reacts with water (H₂O) to form HF and OH⁻, which is a basic reaction. The reaction is: F⁻ + H₂O ⇌ HF + OH⁻.

Step 2: Write the Kb Expression

Given Ka(HF) = 3.2 x 10⁻⁷, we find Kb for F⁻ using the relation Kw = Ka * Kb, where Kw = 1.0 x 10⁻¹⁴ at 25°C. Thus, Kb = Kw / Ka = 1.0 x 10⁻¹⁴ / 3.2 x 10⁻⁷.

Step 3: Calculate Kb

Kb = 1.0 x 10⁻¹⁴ / 3.2 x 10⁻⁷ = 3.125 x 10⁻⁸.

Step 4: Set Up the ICE Table

For the reaction F⁻ + H₂O ⇌ HF + OH⁻, initially, [F⁻] = 0.095 M, and [OH⁻] and [HF] are 0. We assume x M of F⁻ reacts to form x M of OH⁻ and HF.

Step 5: Write the Kb Expression with Concentrations

Kb = [OH⁻][HF] / [F⁻] = x² / (0.095 - x). Since x is small, approximate to Kb = x² / 0.095.

Step 6: Solve for x (Concentration of OH⁻)

3.125 x 10⁻⁸ = x² / 0.095. Solving for x gives x = √(3.125 x 10⁻⁸ * 0.095) = √(2.96875 x 10⁻⁹) = 5.45 x 10⁻⁵ M.

Step 7: Calculate pOH and pH

pOH = -log[OH⁻] = -log(5.45 x 10⁻⁵) = 4.26.

pH = 14 - pOH = 14 - 4.26 = 9.74.

Final Answer

The pH of the 0.095 M solution of sodium fluoride (NaF) is 9.74.

Key Concept

Acid-Base Equilibrium

Key Concept Explanation

The pH calculation for a solution of sodium fluoride involves understanding the acid-base equilibrium. Sodium fluoride, when dissolved in water, creates a basic solution due to the hydrolysis of the fluoride ion, which increases the concentration of hydroxide ions (OH⁻) in the solution. The equilibrium constant (Kb) for this reaction helps in calculating the concentration of OH⁻, which is then used to find the pH of the solution. This process illustrates the dynamic nature of acid-base reactions and their impact on the pH of solutions.

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