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.

Short answer

The concentration of hydrofluoric acid (HF) in the solution is 0.0847 M.

Step-by-step solution

Write the dissociation of hydrofluoric acid and its equilibrium expression with Ka

Hydrofluoric acid (HF) dissociates in water as follows: HF ⇌ H+ + F- The equilibrium constant expression for this reaction is given by Ka: Ka = \(\frac{[H^+][F^-]}{[HF]}\) Since we are given the Ka value, 6.8 x 10^{-4}, we can use it in our calculations.

Write the definition of pH and find the H+ concentration

The pH of a solution is defined as the negative logarithm of the concentration of H+ ions in the solution: pH = -log[H+] We are given the pH value as 2.12. We can find the H+ concentration by taking the inverse logarithm of the pH value: [H+] = 10^{-2.12} Now, calculate the concentration of H+: [H+] = 7.59 x 10^{-3} M

Apply the ICE table (Initial, Change, Equilibrium) method

Let's set up an ICE table to find the concentration of HF at equilibrium. | HF | H+ | F- | | 1. Initial | 0 | 0 | | 2. Change | -x | +x | +x | | 3. Equilibrium | [HF]-x | [H+] +x | x | We know that [H+] = 7.59 x 10^{-3} M and that at equilibrium: [H+] + x = 7.59 x 10^{-3} x = [F-]

Substitute the equilibrium concentrations into the Ka expression

Now, we'll substitute the values from the ICE table into the Ka expression: \(Ka = \frac{([H^+]+x)(x)}{[HF]-x}\) Substitute the known values: \(6.8 \times 10^{-4} = \frac{(7.59 \times 10^{-3} + x)(x)}{[HF] - x}\)

Apply the approximation that x is very small

In weak acid/base equilibria, we can make an approximation that x is very small compared to the initial concentrations. This allows us to make a simplification in our calculations: Approximation: x << [H+] Now, we can rewrite the equation as: \(6.8 \times 10^{-4} = \frac{(7.59 \times 10^{-3})(x)}{[HF]}\)

Solve for the concentration of HF

Let's solve the equation for [HF]: [HF] = \(\frac{(7.59 \times 10^{-3})(x)}{6.8 \times 10^{-4}}\) Now, notice that in our ICE table, we found that x = [F-], so we can substitute [F-] for x: [HF] = \(\frac{(7.59 \times 10^{-3})([F^-])}{6.8 \times 10^{-4}}\) We also know that [F-] = [H+], so we can plug the value of [H+] in the equation: [HF] = \(\frac{(7.59 \times 10^{-3})(7.59 \times 10^{-3})}{6.8 \times 10^{-4}}\) Now, calculate the value of [HF]: [HF] = 0.0847 M The concentration of hydrofluoric acid in the solution is 0.0847 M.

Key concepts

Acid-Base Equilibrium

In chemistry, acid-base equilibrium describes the balance between the acidic and basic ions in a solution. It involves a reversible reaction where acid molecules split into their ions, such as in the case of hydrofluoric acid (HF), which dissociates into hydrogen ions (H+) and fluoride ions (F-). The equilibrium state is where the rate of the forward reaction (dissociation of HF) matches the reverse reaction (reformation of HF).
This concept is vital because it helps predict how much acid has dissociated and how much remains intact in solution. Understanding equilibrium helps us calculate the concentration of acid in a given solution, essential in fields like chemistry and biology.

• Forward reaction: Dissociation of HF into H+ and F-
• Reverse reaction: Reformation of HF from H+ and F-

Recognizing the concept of acid-base equilibrium allows us to understand how changes in concentration, temperature, and pressure can influence system behavior.

pH Calculations

The measurement of acidity or alkalinity in a solution is called pH. The pH scale is a logarithmic scale from 0 to 14, with lower values being more acidic, and higher values more basic. Neutral solutions, like pure water, have a pH of 7.
pH is calculated using the formula:
\[ \text{pH} = -\log{[H^+]} \]
This equation relates pH to the concentration of free hydrogen ions \([H^+]\) in the solution. A lower pH value indicates a higher concentration of \([H^+]\).

• Important in understanding reaction environment
• Helps assess potential reaction outcomes

For our problem, a known pH of 2.12 allows us to find \([H^+] = 7.59 \times 10^{-3} M\). This is used in equilibrium calculations to determine hydrofluoric acid concentration.

Equilibrium Constant Expression

The equilibrium constant expression quantifies the ratio of concentrations of products to reactants at equilibrium. For acid-base reactions, the expression is typically denoted as \(K_a\) for acids.
The general form of the equilibrium constant expression for an acid like HF is:
\[ K_{a} = \frac{[H^+][F^-]}{[HF]} \]
This expression is central to determining the extent of acid dissociation. It provides a way to relate the concentrations of all species involved in the reaction to the equilibrium constant value \(K_a = 6.8 \times 10^{-4}\) for HF in this example.
Studying this expression allows scientists to:

• Predict the direction of equilibrium shifts
• Calculate unknown concentrations once equilibrium is reached

Understanding \(K_a\) is crucial for calculating the initial concentration of HF given a specified \([H^+]\) and determining the extent of dissociation.

ICE Table

An ICE table (Initial, Change, Equilibrium) is a helpful tool to organize and visualize changes in concentrations of reactants and products as a reaction progresses to equilibrium. It stands for Initial concentration, Change in concentration, and Equilibrium concentration.
Here's how you use it in the context of our exercise:

• **Initial (I)**: Start with known initial concentrations.
• **Change (C)**: Determine how concentrations change, utilizing variables like \(x\) to represent unknown changes.
• **Equilibrium (E)**: Determine concentrations at equilibrium by applying these changes to initial values.

The ICE table helps solve for unknowns in chemical reactions, especially when combined with the equilibrium constant equation. In our solution, it helps find the concentration of HF by visualizing how the formation of \(H^+\) and \(F^-\) at equilibrium relates to the initial concentration of HF and the given \(K_a\). Understanding how to set up and interpret an ICE table is a critical skill in solving equilibrium problems efficiently.