Problem 54 - Chemistry

Question

The \(K_{\mathrm{a}}\) for hydrofluoric acid is \(7.1 \times 10^{-4}\). Calculate the \(\mathrm{pH}\) of a \(0.15-M\) aqueous solution of hydrofluoric acid at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The pH of the 0.15-M HF solution is approximately 2.99.

Step 1: Write the Dissociation Reaction

The dissociation reaction for hydrofluoric acid (HF) in water is given by:\[ \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \]

Step 2: Write the Expression for Ka

The acid dissociation constant \(K_a\) for HF is expressed as:\[ K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} \]

Step 3: Determine the Initial Concentrations

The initial concentration of HF is 0.15 M, and initially, the concentration of \(\text{H}^+\) and \(\text{F}^-\) ions are 0 M.

Step 4: Set Up the ICE Table

The ICE (Initial, Change, Equilibrium) table helps track the concentration changes:\[\begin{array}{c|c|c|c} & [\text{HF}] & [\text{H}^+] & [\text{F}^-] \\hline\text{Initial (I)} & 0.15 & 0 & 0 \\text{Change (C)} & -x & +x & +x \\text{Equilibrium (E)} & 0.15-x & x & x \\end{array}\]

Step 5: Substitute Equilibrium Concentrations into Ka

Substitute the equilibrium concentrations from the ICE table into the \(K_a\) expression:\[ K_a = \frac{x^2}{0.15 - x} = 7.1 \times 10^{-4} \]

Step 6: Approximating for Small x

Assume \(x\) is small compared to 0.15, so \(0.15 - x \approx 0.15\). Thus,\[ \frac{x^2}{0.15} = 7.1 \times 10^{-4} \]

Step 7: Solve for x

Solve the equation from Step 6 to find \(x\):\[ x^2 = (7.1 \times 10^{-4}) \times 0.15 \]\[ x^2 = 1.065 \times 10^{-4} \]\[ x = \sqrt{1.065 \times 10^{-4}} \approx 0.0103 \]

Step 8: Calculate the pH

The \(\text{pH}\) is calculated using the concentration of \(\text{H}^+\):\[ \text{pH} = -\log[\text{H}^+] = -\log(0.0103) \approx 2.99 \]

Key Concepts

Acid Dissociation Constant

The acid dissociation constant, often denoted as \(K_a\), is a critical concept when discussing weak acids, such as hydrofluoric acid (HF). It provides insight into the extent of ionization of the acid in solution.

In simple terms, it measures how readily an acid releases its hydrogen ions (\(H^+\)) into solution. For HF, the dissociation reaction is:

• \( \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \)

The value of \(K_a\) is calculated based on the concentrations of the ions at equilibrium:

• \( K_a = \frac{[\text{H}^+][\text{F}^-]}{[\text{HF}]} \)

Here, a smaller \(K_a\) value indicates a weaker acid, signifying that HF does not dissociate completely in water.

This dissociation is key to understanding the pH of the solution. The provided value of \(K_a\) for HF is \(7.1 \times 10^{-4}\), which reflects its moderate acidity compared to strong and highly dissociating acids.

ICE Table

The ICE table is a pivotal tool for solving equilibrium problems, particularly those involving the dissociation of weak acids. ICE stands for Initial, Change, and Equilibrium, spelling out each stage of concentration during the reaction.

Here's a simple breakdown:

• Initial: Concentrations at the start before any dissociation occurs. For HF, you begin with 0.15 M of HF, and 0 M of \(H^+\) and \(F^-\).
• Change: The amount by which concentrations change as the reaction proceeds to equilibrium. This is represented as \(x\).
• Equilibrium: The final concentrations when equilibrium is established. For HF, these become \(0.15 - x\) for HF and \(x\) for both \(H^+\) and \(F^-\).

The ICE table organizes these values systematically, making it easier to visualize and solve the equilibrium expression for \(K_a\). It's very handy for identifying and simplifying steps in equilibrium problems, particularly when assumptions about small changes \((x)\) can be made to streamline calculations.

Hydrofluoric Acid

Hydrofluoric acid is an interesting and unique acid because, unlike most common acids, it only partially dissociates in water, hence being classified as a weak acid. This means not all hydrofluoric acid molecules disassociate to release hydrogen ions (\(H^+\)) when in solution.

Hydrofluoric acid's molecular formula is HF, and it's known for forming an equilibrium between its molecular form and its constituent ions (\(H^+\) and \(F^-\)). This dissociation is described with the reaction:

• \( \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \)

Since HF is a weak acid, it has a relatively low \(K_a\) value of \(7.1 \times 10^{-4}\), indicating its limited ionization.

The weak nature of hydrofluoric acid is due to the strength of the hydrogen-fluorine bond, which is stronger compared to bonds in other hydrogen halides, thus less efficiently breaking apart into ions. As a result, this bond strength directly impacts the solution's pH, making calculations required to understand HF's acidity level more intricate than those for strong acids where \(pH\) can be more directly related to initial concentration.