Question

Calculate the concentration of all species present and the \(\mathrm{pH}\) of a \(0.020 M \mathrm{HF}\) solution.

Short answer

In a 0.020 M HF solution, the equilibrium concentrations are approximately: [HF] ≈ 0.0162 M, [H+] ≈ 3.77 x 10⁻³ M, and [F-] ≈ 3.77 x 10⁻³ M. The pH of the solution is approximately 2.42.

Step-by-step solution

Write the dissociation equilibrium for HF and expression for Ka

The dissociation equilibrium for HF in water can be represented as follows: \( HF (aq) \rightleftharpoons H^+ (aq) + F^- (aq) \) Now, we can write the equilibrium constant expression (Ka) for the dissociation of HF: \(K_a = \frac{[H^+][F^-]}{[HF]}\)

Calculate the change in concentrations for reactants and products during the dissociation process

Let x be the change in concentrations during the equilibrium process, then: - The initial concentration of HF is 0.020 M. - At equilibrium, the concentration of HF has decreased by x M, so [HF] = 0.020 - x M. - At equilibrium, the concentration of H+ has increased by x M, so [H+] = x M. - At equilibrium, the concentration of F- has also increased by x M, so [F-] = x M.

Substitute equilibrium concentrations into the Ka expression and solve for x

The Ka value for HF is 7.1 x 10⁻⁴. Substituting the equilibrium concentrations into the Ka expression: \(7.1 \times 10^{-4} = \frac{x^2}{0.020 - x}\) To simplify the calculation, since x is very small compared to 0.020 M, we can assume (0.020 - x) ≈ 0.020: \(7.1 \times 10^{-4} ≈ \frac{x^2}{0.020}\) Now, we can solve for x: \(x \approx \sqrt{7.1 \times 10^{-4} \times 0.020} \approx \sqrt{1.42 \times 10^{-5}} \approx 3.77 \times 10^{-3}\)

Calculate the equilibrium concentrations of all species and the pH

Now, we can calculate the equilibrium concentrations of each species: - [HF] ≈ 0.020 - 3.77 x 10⁻³ ≈ 0.0162 M - [H+] ≈ 3.77 x 10⁻³ M - [F-] ≈ 3.77 x 10⁻³ M Finally, we can calculate the pH of the solution using the concentration of H+ ions: \(pH = -\log_{10}[H^+]\) \(pH \approx -\log_{10}(3.77 \times 10^{-3}) \approx 2.42\)

Present the solution

In the 0.020 M HF solution: - The equilibrium concentration of HF is ≈ 0.0162 M - The equilibrium concentration of H+ ions is ≈ 3.77 x 10⁻³ M - The equilibrium concentration of F- ions is ≈ 3.77 x 10⁻³ M - The pH of the solution is approximately 2.42.

Key concepts

Equilibrium Constant

The equilibrium constant, specifically the acid dissociation constant (Ka), plays a crucial role in understanding the behavior of weak acids in solution. In the context of the dissociation of hydrofluoric acid (HF) in water, the Ka value represents the ratio of the concentrations of the products of the dissociation reaction to the concentration of the undissociated acid.
The expression for the dissociation equilibrium of HF is:

• For HF dissociation: \( HF (aq) \rightleftharpoons H^+ (aq) + F^- (aq) \)
• The equilibrium constant expression: \( K_a = \frac{[H^+][F^-]}{[HF]} \)

Ka provides insight into how extensively an acid dissociates in solution. For HF, the relatively low Ka value of \(7.1 \times 10^{-4}\) indicates that it only partially dissociates, classifying it as a weak acid. This means at equilibrium, the concentrations of \(H^+\) and \(F^-\) are much lower compared to the original \(HF\) concentration. Understanding how to apply and interpret Kₐ helps predict the concentrations of species at equilibrium.

pH Calculation

Calculating the pH of a solution is essential to understand its acidity levels. The pH value provides a numerical scale to represent the concentration of hydrogen ions (\([H^+]\)) in solution, highlighting whether the environment is acidic or basic. In the case of our HF solution, we use the concentration of \(H^+\) ions found to determine pH.

• The formula for pH: \( pH = -\log_{10}[H^+] \)
• For the HF solution with \([H^+] \approx 3.77 \times 10^{-3} \ \text{M}\)
• The calculated pH: \( pH \approx -\log_{10}(3.77 \times 10^{-3}) \approx 2.42 \)

A pH of 2.42 confirms the acidic nature of the HF solution. The pH scale runs from 0 to 14, where lower values indicate higher acidity. Therefore, understanding the relationship between hydrogen ion concentration and pH is essential for interpreting the acidic or basic character of a solution.

Concentration Changes

In any equilibrium reaction, including the dissociation of HF, concentration changes for the reactants and products are key to understanding the system at equilibrium. As the reaction proceeds, the concentrations shift according to stoichiometry and the extent of dissociation dictated by Ka. Initial and equilibrium concentrations are typically compared to measure these changes:

• Initial concentration of \(HF\): \(0.020 \ \text{M}\).
• Change in concentration represented by \(x\): allows calculation of equilibrium concentrations.
• Equilibrium concentration calculations:
  • \([HF] \approx 0.020 - x\)
  • \([H^+] = x\)
  • \([F^-] = x\)

For our HF solution, \(x\) was calculated to be approximately \(3.77 \times 10^{-3}\). By substituting into the equations, we derive accurate figures for how concentrations have adjusted from initial to equilibrium states. Accounting for these changes is crucial to establishing the behavior of the system.

Dissociation Equilibrium

The concept of dissociation equilibrium involves examining how an acid like HF reaches a balance between its dissociated ions and its undissociated form in solution. Recognizing how these states shift and balance is key to understanding acid behavior in solution. At equilibrium, the rate at which the forward reaction (dissociation) occurs equals the rate of the reverse reaction (recombination), creating a steady state.

• State Representation: \(HF (aq) \rightleftharpoons H^+ (aq) + F^- (aq)\).
• Balance Evaluation: Equilibrium concentrations result from the stabilization of dissociation and recombination rates.
• Partial Dissociation: The low Ka value implies HF doesn't fully dissociate, maintaining significant undissociated HF at equilibrium.

Understanding these interactions helps in comprehensively analyzing and predicting the behavior of weak acids in varying conditions. This equilibrium perspective is foundational for solving many chemistry problems.