Problem 25 - Chemistry: The Central Science in SI Units, Global Edition

Question

You are asked to prepare a pH \(=2.50\) buffer solution starting from \(1.50 \mathrm{~L}\) of a \(0.75 \mathrm{M}\) solution of hydrofluoric acid (HF) and any amount you need of sodium fluoride (NaF). (a) What is the \(\mathrm{pH}\) of the hydrofluoric acid solution prior to adding sodium fluoride? (b) How many grams of sodium fluoride should be added to prepare the buffer solution? Neglect the small volume change that occurs when the sodium fluoride is added.

Step-by-Step Solution

Answer

The initial pH of the 0.75 M hydrofluoric acid solution is approximately 1.72. To prepare the desired pH 2.50 buffer solution, we need to add approximately 1.97 grams of sodium fluoride to the 1.50 L of 0.75 M hydrofluoric acid solution.

Step 1: 1. Ionization equation and ionization constant for hydrofluoric acid

Hydrofluoric acid (HF) is a weak acid and ionizes in water to form hydrogen ions (H+) and fluoride ions (F-): \[HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\] For weak acids, the equilibrium ionization constant (Ka) is given by: \[K_a = \frac{[H^+][F^-]}{[HF]}\]

Step 2: 2. Calculate the starting pH of the solution

To calculate the pH of the 0.75 M hydrofluoric acid solution, we use the Ka expression given below: \[ K_a = \frac{x^2}{0.75-x}\] The ionization constant of HF is approximately \(K_a = 6.6 \times 10^{-4}\). Since HF is a weak acid, we can assume \[ x \ll 0.75\] Therefore, \[ 6.6 \times 10^{-4} \approx \frac{x^2}{0.75}\] Solving for x: \[ x = \sqrt{6.6 \times 10^{-4} \times 0.75} = 0.01894\] Here, x represents the concentration of hydrogen ions [H+]. Thus, the initial pH of the HF solution can be determined using the formula: \[pH = -\log[H^+]\] \[ pH = -\log(0.01894) \approx 1.72\] The initial pH of the hydrofluoric acid solution is approximately 1.72.

Step 3: 3. Determine the required concentration of fluoride ions

We are asked to prepare a pH 2.50 buffer solution using HF and NaF. We will use the expression for Ka and the desired pH to determine the concentration of fluoride ions (F-) needed in the solution: \[ K_a = \frac{[H^+][F^-]}{[HF]}\] We know \([H^+] = 10^{-pH} = 10^{-2.5}\), and the final [HF] should remain constant at 0.75 M, as the volume change is negligible: \[ 6.6 \times 10^{-4} = \frac{(10^{-2.5})([F^-])}{0.75}\] Solve for [F-]: \[ [F^-] = \frac{6.6 \times 10^{-4} \times 0.75}{10^{-2.5}} \approx 0.0312\mathrm{M}\]

Step 4: 4. Calculate the amount of sodium fluoride needed

Now we know the required concentration of fluoride ions, we can calculate the amount of sodium fluoride, NaF, needed to be added: Moles of F- ions needed: \[moles = M_{required} \times V_{solution} = 0.0312 \times 1.50 = 0.0468 \: mol\] To find the mass of NaF required, we multiply the moles by the molar mass (42.0 g/mol for NaF): \[mass = moles \times molar \: mass = 0.0468 \times 42.0 = 1.97\mathrm{~g}\] To prepare the desired pH 2.50 buffer solution, we need to add approximately 1.97 grams of sodium fluoride to the 1.50 L of 0.75 M hydrofluoric acid solution.

Key Concepts

Hydrofluoric Acid

Hydrofluoric acid (HF) is a weak acid, which means it does not completely dissociate into ions in water like strong acids do. In its aqueous form, HF ionizes to produce hydrogen ions ( H^+ ) and fluoride ions ( F^- ). This relatively incomplete ionization is what characterizes it as a weak acid.
Hydrofluoric acid is quite different from other acids as it can etch glass by reacting with silicon dioxide, which makes it important in industries like glass and metal cleaning. However, due to its ionization properties and ability to penetrate tissue, HF is highly corrosive and must be handled with care in a lab environment.

Weak Acid Ionization

Weak acid ionization refers to the partial dissociation of weak acids in aqueous solutions. The ionization of hydrofluoric acid can be represented by the reaction:\[HF(aq) \rightleftharpoons H^+(aq) + F^-(aq)\]This reversible reaction illustrates that not all the HF molecules release hydrogen ions into the solution.
The degree of ionization for a weak acid is described by its ionization constant, K_a. For hydrofluoric acid, this is given by the equilibrium expression:\[K_a = \frac{[H^+][F^-]}{[HF]}\]Understanding weak acid ionization is crucial because it influences how equilibrium conditions are applied in chemical reactions, particularly when creating buffer solutions to resist changes in pH.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution based on the concentrations of an acid and its conjugate base. It is given by the formula:\[pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)\]Where [A^-] is the concentration of the conjugate base, [HA] is the concentration of the acid, and pK_a is the negative logarithm of the ionization constant K_a.
This equation is particularly helpful when determining how much of a substance, like sodium fluoride, must be added to a weak acid like hydrofluoric acid to create a buffer solution that maintains a specific pH level. It allows for a simplified view of how acid-base balance can be achieved in solution.

Sodium Fluoride

Sodium fluoride (NaF) is a salt which, when dissolved in water, dissociates completely into sodium ( Na^+ ) and fluoride ( F^- ) ions. In the context of a buffer solution, it plays a significant role by providing fluoride ions needed to form the conjugate base of hydrofluoric acid.
This helps to stabilize the hydrogen ion concentration, maintaining the desired pH of the solution effectively. In laboratory preparations, calculating the exact amount of sodium fluoride necessary is crucial to achieving a specific pH, as this depends on both the volume and concentration of the initial acid solution. Sodium fluoride is often used beyond laboratory settings in some dental products due to its ability to promote strong teeth.