Problem 56 - Chemistry: The Central Science

Question

A \(0.100 \mathrm{M}\) solution of bromoacetic acid \(\left(\mathrm{BrCH}_{2} \mathrm{COOH}\right)\) is \(13.2 \%\) ionized. Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{BrCH}_{2} \mathrm{COO}^{-}\right]\), and \(\left[\mathrm{BrCH}_{2} \mathrm{COOH}\right]\).

Step-by-Step Solution

Answer

At equilibrium, the concentrations of the species are: \([H^+] = 0.0132\ \mathrm{M}\), \([BrCH_2COO^-] = 0.0132\ \mathrm{M}\), and \([BrCH_2COOH] = 0.0868\ \mathrm{M}\).

Step 1: Determine the equation for the ionization of bromoacetic acid

The ionization of bromoacetic acid can be represented as: BrCH₂COOH (aq) ⇌ H⁺ (aq) + BrCH₂COO⁻ (aq)

Step 2: Determine the initial concentrations of the species

The initial concentration of bromoacetic acid is given as 0.100 M: - [BrCH₂COOH]₀ = 0.100 M - [H⁺]₀ = 0 - [BrCH₂COO⁻]₀ = 0

Step 3: Calculate the change in concentrations at equilibrium

Since the percent ionization of bromoacetic acid is given as 13.2%, we can determine the change in concentrations at equilibrium using the initial concentration of BrCH₂COOH: - Δ[H⁺] = Δ[BrCH₂COO⁻] = 13.2% × 0.100 M = 0.132 × 0.100 M = 0.0132 M - Δ[BrCH₂COOH] = -0.0132 M

Step 4: Determine the concentrations at equilibrium

We can now find the concentrations of the species at equilibrium by adding or subtracting the changes from the initial concentrations: - [H⁺] = [H⁺]₀ + Δ[H⁺] = 0 + 0.0132 M = 0.0132 M - [BrCH₂COO⁻] = [BrCH₂COO⁻]₀ + Δ[BrCH₂COO⁻] = 0 + 0.0132 M = 0.0132 M - [BrCH₂COOH] = [BrCH₂COOH]₀ + Δ[BrCH₂COOH] = 0.100 M - 0.0132 M = 0.0868 M So, at equilibrium we have: - [H⁺] = 0.0132 M - [BrCH₂COO⁻] = 0.0132 M - [BrCH₂COOH] = 0.0868 M

Key Concepts

Bromoacetic Acid

Bromoacetic acid, with the chemical formula \(\mathrm{BrCH}_{2}\mathrm{COOH}\), is a weak organic acid. Its significance lies primarily in its structure because it contains a bromine atom near the acidic carboxyl group. This bromine atom increases the acid's ionization potential by weakening the \(\mathrm{O-H}\) bond in the carboxyl group, making it easier for the acid to donate a hydrogen ion \(\left(\mathrm{H}^{+}\right)\). This is an important aspect of understanding how bromoacetic acid behaves in solution.

When dissolved in water, bromoacetic acid partially dissociates, as depicted in the equation:

BrCH₂COOH (aq) ⇌ H⁺ (aq) + BrCH₂COO⁻ (aq).

This equilibrium represents the acid in its ionized and non-ionized forms, which helps determine concentrations at equilibrium states in solutions.

Percent Ionization

The percent ionization of an acid is a key concept in understanding acid strength and behavior in solution. It tells us what percentage of the original acid molecules have ionized into ions in a solution. For bromoacetic acid in our exercise, the percent ionization is 13.2%.

This means that out of all the bromoacetic acid molecules originally in the solution, 13.2% have dissociated to form hydrogen ions \(\left(\mathrm{H}^{+}\right)\) and conjugate base ions \(\left(\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right)\).

To calculate the percent ionization, you can use the formula:

\[\text{Percent Ionization} = \left(\frac{\left[\mathrm{H}^{+}\right]_{eq}}{\left[\mathrm{HA}\right]_{\text{initial}}}\right) \times 100\]%

Knowing the percent ionization helps in determining both the degree to which a weak acid dissociates in water and the concentrations of ions at equilibrium.

Equilibrium Concentrations

Equilibrium concentrations are pivotal in chemistry, especially in acid-base reactions, to understand the outcome of the reaction's progress. For bromoacetic acid, equilibrium concentrations are calculated based on initial concentrations and changes due to ionization.

Initially, the concentration of bromoacetic acid was 0.100 M, with no \(\left[\mathrm{H}^{+}\right]\) or \(\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right]\). As ionization occurs, these concentrations change.

Given a 13.2% ionization rate, the change in concentration for each ion and the molecule is calculated as follows:

- Change in \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{BrCH}_{2}\mathrm{COO}^{-}\right]\) = 13.2% of 0.100 M = 0.0132 M
- Change in \(\left[\mathrm{BrCH}_{2}\mathrm{COOH}\right]\) is negative due to consumption and equals -0.0132 M.

So, the equilibrium concentrations are:

• [H⁺] = 0.0132 M
• [BrCH₂COO⁻] = 0.0132 M
• [BrCH₂COOH] = 0.0868 M

These values indicate the balance achieved in the reaction, reflecting both ionized and non-ionized forms of bromoacetic acid.