Problem 4 - Chemistry: The Central Science

Question

In which of the following cases is the approximation that the equilibrium concentration of \(\mathrm{H}^{+}(a q)\) is small relative to the initial concentration of HA likely to be most valid: (a) initial \([\mathrm{HA}]=0.100 \mathrm{M}\) and \(K_{a}=1.0 \times 10^{-6}\), (b) initial \([\mathrm{HA}]=0.100 \mathrm{M}\) and \(K_{a}=1.0 \times 10^{-4}\), (c) initial \([\mathrm{HA}]=0.100 \mathrm{M}\) and \(K_{a}=1.0 \times 10^{-3} ?[\) Section \(16.6]\)

Step-by-Step Solution

Answer

The approximation that the equilibrium concentration of H+ is small relative to the initial concentration of HA is most valid for case (a) with initial $[\mathrm{HA}]=0.100 \mathrm{M}$ and $K_{a}=1.0 \times 10^{-6}$. This is because it has the lowest Ka value, indicating a weaker acid, and therefore a smaller equilibrium concentration of H+ relative to the initial concentration of HA.

Step 1: Find the pH for each case

Use the given Ka values to find the pH of each solution. Since the initial concentration of A- is 0, the pH can be directly calculated from the Ka value. pH = -log(Ka) For case (a): Ka = 1.0 × 10^(-6) pH = -log(1.0 × 10^(-6)) = 6 For case (b): Ka = 1.0 × 10^(-4) pH = -log(1.0 × 10^(-4)) = 4 For case (c): Ka = 1.0 × 10^(-3) pH = -log(1.0 × 10^(-3)) = 3

Step 2: Calculate the [A-]/[HA] ratio for each case

Now, using the Henderson-Hasselbalch equation (pH = pKa + log[A-]/[HA]), calculate the ratio of dissociated HA to the undissociated HA concentration for each case. For case (a): pH = pKa + log[A-]/[HA] 6 = 6 + log[A-]/[0.1] [A-]/[HA] = 10^(6-6) = 1 For case (b): pH = pKa + log[A-]/[HA] 4 = 4 + log[A-]/[0.1] [A-]/[HA] = 10^(4-4) = 1 For case (c): pH = pKa + log[A-]/[HA] 3 = 3 + log[A-]/[0.1] [A-]/[HA] = 10^(3-3) = 1

Step 3: Compare the [A-]/[HA] ratios for each case, and determine which case is most valid

Since the [A-]/[HA] ratios are the same for all three cases, we should compare the Ka values. The lower the Ka value, the weaker the acid, which means it will dissociate less and have a smaller equilibrium concentration of H+ relative to the initial concentration of HA. Case (a) has the lowest Ka value, 1.0 × 10^(-6), which means that the approximation that the equilibrium concentration of H+ is small relative to the initial concentration of HA is most valid for this case.

Key Concepts

Equilibrium Concentration

In chemical equilibrium, particularly involving acids and bases, understanding the concept of equilibrium concentration is key. When a weak acid, such as HA, dissolves in water, it partially dissociates to form H⁺ ions and its conjugate base A⁻. The concentration of these ions at equilibrium determines the acidity of the solution. In equilibrium, the rate of the forward reaction (acid dissociation) equals the rate of the reverse reaction (reassociation of H⁺ and A⁻ to form HA), resulting in stable concentrations of all species involved.

**Key Points to Consider**:

• The initial concentration of the acid (HA) affects how much it dissociates at equilibrium.
• The acid dissociation constant, or Ka, reflects the acid's tendency to donate protons to the solution. A smaller Ka implies less dissociation and a smaller change in concentration.
• In scenarios where the equilibrium concentration of H⁺ is small compared to the initial concentration of HA, approximations can simplify calculations.

To effectively use approximations, the acid should have a small Ka, as seen in case (a) with Ka = 1.0 × 10⁻⁶, indicating that little dissociation occurs and equilibrium concentrations vary less dramatically from the initial values.

Acid-Base Equilibrium

The acid-base equilibrium involves the balance between the protonated form of a compound (HA) and its deprotonated form (A⁻) in an aqueous solution. Weak acids, such as those described in the exercise, partially dissociate. This balance can be influenced by factors including concentration and the strength of the acid, typically expressed via its Ka value.

**Important Aspects**:

• The position of the equilibrium is significantly influenced by the initial concentration of the acid and its Ka value.
• A larger Ka indicates a stronger acid, which dissociates more completely compared to a weak acid with a smaller Ka.
• The pH value of a solution provides insight into the equilibrium state, making it easier to evaluate the strengths of different acid solutions.

The approximation that the equilibrium concentration of H⁺ is small relative to HA is more valid when Ka is extremely low. This suggests that the acid doesn’t dissociate much, keeping the protons' concentration minimal in comparison to the original acid, as illustrated by case (a).

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a valuable tool for understanding the pH of a solution in relation to the acid dissociation constant (Ka) and the concentration ratio between the dissociated and undissociated forms of the acid. It's expressed as:
\[\text{pH} = \text{pKa} + \log \frac{[A^-]}{[HA]}\] This equation directly links the pH of a solution to the concentrations of its components at equilibrium.

**How It Works**:

• pKa is the negative logarithm of the Ka value and reflects the acid's strength.
• The ratio \([A^-]/[HA]\) indicates how much of the acid has dissociated.
• In scenarios where this ratio is significantly different across cases, it indicates varying extents of dissociation.
• The equation helps predict pH changes when an acid is in different concentration or strength states.

Using this equation, the step-by-step analysis of various cases in the exercise demonstrates that despite variations in pH values, understanding the ratio's implications is crucial. This highlights why case (a) is most suited for the approximation as minimal dissociation occurs, maintaining a closer initial ratio and reinforcing the simplified assumption.