Question

A \(1.000 \mathrm{~L}\) sample of \(\mathrm{HF}\) gas at \(20.0^{\circ} \mathrm{C}\) and \(0.601 \mathrm{~atm}\) pressure was dissolved in enough water to make \(50.0 \mathrm{~mL}\) of hydrofluoric acid (a) What is the \(\mathrm{pH}\) of the solution? (b) To what volume must you dilute the solution to triple the percent dissociation. Why can't you solve this problem by using the result obtained in Problem 15.144?

Short answer

(a) pH ≈ 1.74; (b) Dilute to 9221.4 mL; can't use 15.144 due to differing conditions.

Step-by-step solution

Calculate Number of Moles of HF

First, we need to calculate the number of moles of HF gas using the ideal gas law: \[ PV = nRT \]Given:- \( P = 0.601 \ \text{atm} \)- \( V = 1.000 \ \text{L} \)- \( R = 0.0821 \ \text{L atm/mol K} \)- \( T = 273.15 + 20 = 293.15 \ \text{K} \)Rearranging the ideal gas law, we find:\[ n = \frac{PV}{RT} = \frac{(0.601)(1.000)}{(0.0821)(293.15)} \approx 0.0249 \ \text{mol} \]

Calculate Initial Concentration of HF

Now, determine the initial molarity \((M)\) of the HF solution, which is formed by dissolving the moles in \(50.0 \ \text{mL} = 0.0500 \ \text{L}\):\[ M = \frac{0.0249 \ \text{mol}}{0.0500 \ \text{L}} = 0.498 \ \text{M}\]

Setup Equilibrium Reaction for HF

For the weak acid HF, the dissociation in water is:\[ \text{HF} \leftrightarrow \text{H}^+ + \text{F}^- \]The initial concentration of HF is \(0.498 \ \text{M}\). Let the change in concentration at equilibrium be \(x\), giving:\[ \begin{align*}\text{Initial:} & \ 0.498 \, M, \, 0, \, 0 \\text{Change:} & \ -x, \, +x, \, +x \\text{Equilibrium:} & \ 0.498 - x, \, x, \, x \\end{align*} \]

Apply Ka Expression

The equilibrium constant expression (\( K_a \)) for HF is:\[ K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} = \frac{x^2}{0.498 - x} \]Given \( K_a \approx 6.6 \times 10^{-4} \) for HF, we solve:\[ x^2 \approx K_a \, (0.498 - x) \]Assume \( x \ll 0.498\), simplify to:\[ x^2 \approx 6.6 \times 10^{-4} \times 0.498 \]\[ x \approx \sqrt{3.2868 \times 10^{-4}} \approx 0.0181 \]

Calculate pH

The concentration of \( \text{H}^+ \) ions is \( x = 0.0181 \ \text{M} \). Calculate the pH:\[ \text{pH} = -\log(0.0181) \approx 1.74 \]

Determine New Volume for Triple Percent Dissociation

Percent dissociation initial is:\[ \frac{0.0181}{0.498} \times 100 \approx 3.63\% \]To triple the percent dissociation, we need approximately \(10.89\%\) dissociation, which implies:\[ 3x \approx \frac{0.498 \times 10.89}{100} \approx 0.0542 \]This means \( [\text{HF}] = 0.498 - 0.0542 \approx 0.4438 \ \text{M} \), and using a new \([\text{H}^+]=0.0542\):\[ K_a = \frac{(0.0542)^2}{0.4438} \approx 6.6 \times 10^{-4} \]Solving for needed concentration, since dilution decreases initial concentration to \(\approx 0.0542/3\). Hence, new volume ::\[ V_{2} = \frac{0.498}{0.0542/3} \times 50.0 \ \text{mL} \approx 9221.4 \ \text{mL} \]

Explain Why Previous Solution Can't Be Used

Problem 15.144 likely involves a different initial setup or reaction conditions that do not apply to the special conditions of this HF solution, such as differing initial concentrations, reaction products or dilute states that affect dissociation.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in Chemistry used to describe the behavior of gases under various conditions. It combines several more specific gas laws into a single formula:

• Pressure (P)
• Volume (V)
• Temperature (T)
• Amount in moles (n)

These factors are related by the equation \[ PV = nRT \]where R is the ideal gas constant, valued typically at 0.0821 L atm/mol K.
This law helps in predicting how a gas will behave when subjected to changes in the mentioned conditions. To solve the current problem, we first used the Ideal Gas Law to determine the number of moles of HF gas present in a given volume, at a specific temperature and pressure. The calculated number of moles was crucial for further calculations, such as determining the molarity of the resulting HF solution after dissolving it in water.

Chemical Equilibrium

In chemistry, equilibrium is the stage where the forward and reverse reactions occur at the same rate. As a result, the concentration of reactants and products remains constant over time. For weak acids like HF, establishing equilibrium is fundamental as they do not completely dissociate in water. The equilibrium reaction for HF in an aqueous solution is:\[ \text{HF} \leftrightarrow \text{H}^+ + \text{F}^- \]In the case of HF, the initial concentration (0.498 M) of the acid begins to dissociate into ions (H^+ and F^-), but not completely. By setting up an ICE (Initial, Change, Equilibrium) table, we can track the changes in concentrations of species involved in the equilibrium. This information is essential for calculating the pH and assessing the degree of dissociation.

Acid Dissociation Constant

The Acid Dissociation Constant, known as K_a, quantifies the strength of a weak acid in solution. It considers the equilibrium concentration of all species involved when the acid partially dissociates in water. For HF, the dissociation can be described by the equation:\[ K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} \]The smaller the K_a, the weaker the acid. For HF, K_a \approx 6.6 \times 10^{-4}, signifying a relatively weak acid compared to stronger acids.
In the exercise, knowing K_a allowed us to calculate the equilibrium concentration of H^+ ions, which in turn led to determining the pH of the solution.

Hydrofluoric Acid

Hydrofluoric Acid ( HF ) is a weak acid that partially dissociates in water. It's widely used in industrial chemistry, albeit with safety precautions due to its corrosive nature. Although it doesn't dissociate fully like stronger acids, its equilibrium dynamics are critical for understanding reactions involving HF .
Being aware of its physical and chemical properties helps in appropriately handling it in lab scenarios. In this specific exercise:

• We found that the provided specific conditions allow HF to reach a particular level of dissociation.
• The calculated pH showed how much H^+ was present in a relatively dilute solution.
• Adjustments to the solution, like dilution, can significantly affect dissociation levels, which is crucial in various applications.