Problem 56 - Chemistry The Central Science In Si Units

Question

If a solution of hydrofluoric acid \(\left(\mathrm{HF} ; K_{a}=6.8 \times 10^{-4}\right)\) has a \(\mathrm{pH}\) of \(2.12,\) calculate the concentration of hydrofluoric acid.

Step-by-Step Solution

Answer

The concentration of hydrofluoric acid is approximately 0.085 M.

Step 1: Determine the \\( [H^+] \\\)

To find the hydrogen ion concentration, \( [H^+] \), from the given \( pH \), we use the formula \( [H^+] = 10^{-pH} \). Substituting the value of \( pH = 2.12 \) gives \[ [H^+] = 10^{-2.12} \approx 7.59 \times 10^{-3} \text{ M} \].

Step 2: Use the Ionization Equation

Hydrofluoric acid ionizes in water as follows: \( \text{HF} \rightleftharpoons \text{H}^+ + \text{F}^- \). At equilibrium, \( [H^+] = [F^-] \).

Step 3: Set Up the Expression for \\( K_a \\\)

The acid dissociation constant formula is \( K_a = \frac{[H^+][F^-]}{[HF]} \). Given \( K_a = 6.8 \times 10^{-4} \), substitute \( [H^+] \approx 7.59 \times 10^{-3} \) and \( [F^-] \approx 7.59 \times 10^{-3} \).

Step 4: Solve for \\( [HF] \\\)

Rearrange the \( K_a \) expression to solve for the initial concentration of \( \text{HF} \): \( [HF] = \frac{[H^+][F^-]}{K_a} \). Substitute the known values: \[ [HF] = \frac{(7.59 \times 10^{-3})^2}{6.8 \times 10^{-4}} \approx 0.08475 \text{ M} \].

Key Concepts

Understanding Hydrofluoric Acid

Hydrofluoric acid (HF) is a weak acid often utilized in industrial applications and laboratory settings. Unlike strong acids, which dissociate completely in water, hydrofluoric acid only partially ionizes in solution.
When HF is dissolved in water, it establishes an equilibrium:

• HF partially dissociates into hydrogen ions (H⁺) and fluoride ions (F^-).
• The reaction can reverse, recombining these ions into HF.

This behavior indicates that HF does not release all its hydrogen ions into the solution.
Its strength as an acid is represented by the acid dissociation constant, (\( K_a \)).
For HF, the \( K_a \) value of \(6.8 \times 10^{-4}\) is relatively low, emphasizing its nature as a weak acid.
A key implication of this property is that in any HF solution, both the parent acid and the ions exist in equilibrium.

Equilibrium Concentration Essentials

When chemical reactions reach a state where forward and reverse reactions occur at the same rate, the system is said to be in equilibrium.
For our hydrofluoric acid example, at equilibrium, the concentration of hydrogen ions equals that of fluoride ions:

• \([H^+] = [F^-]\)

Given that hydrofluoric acid partially dissociates, the balance between the dissociated and undissociated molecules forms the basis for calculating unknown concentrations.
The acid dissociation constant equation, \( K_a \), allows for determination of equilibrium concentrations when initial concentrations are known:\[K_a = \frac{[H^+][F^-]}{[HF]}\]Rearranging this formula can help discover unknown components by substituting known values.
The dynamic dance of molecules in reaching and maintaining this balance requires understanding of initial concentrations and total system viewpoint through analytical techniques.

Mastering pH Calculations

pH is a numerical scale used to measure the acidity of an aqueous solution, representing the negative base-10 logarithm of the hydrogen ion concentration.
It is an essential concept in understanding the behavior of acids like hydrofluoric acid. To deduce hydrogen ion concentration from a given pH:

• Use the formula \([H^+] = 10^{-pH}\)

For instance, with a pH of 2.12:
\[[H^+] = 10^{-2.12} \approx 7.59 \times 10^{-3} \text{ M}\]This step transforms a simple pH value into a more useful concentration figure linking to chemical equilibrium.
Once hydrogen ion concentration is assessed, calculations of both equilibrium conditions and initial concentrations can be refined.
Understanding pH's role in these processes enhances both theoretical chemistry understanding and practical abilities in analytical calculations.