Question

Consider a \(0.67-M\) solution of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) a. Which of the following are major species in the solution? i. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) iii. OH \(^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) b. Calculate the \(p\) H of this solution.

Short answer

a. The major species in the solution are: i. C2H5NH2 iii. OH⁻ iv. H2O v. C2H5NH3⁺ b. The pH of the solution is approximately 10.48.

Step-by-step solution

Identify the Major Species.

First, we need to identify which of the listed species can be created from a solution of C2H5NH2 (ethylamine), using the given Kb value. Let's think about the potential reactions that can occur with C2H5NH2: - C2H5NH2 can act as a base since it has a nitrogen atom that can accept a proton (H⁺). When it accepts a proton, it forms C2H5NH3⁺. - Water is from the solvent and will be present in the solution. - Since C2H5NH2 is a base, it will react with water (H2O) by accepting a proton, releasing OH⁻ ions in the process. So, we can conclude the following: i. C2H5NH2 is present as it is the given compound. ii. H⁺ is not present, as no acid has been added. iii. OH⁻ is present due to the basic nature of the ethylamine reacting with water. iv. H2O will be present as it is the solvent. v. C2H5NH3⁺ will be present as it is formed when C2H5NH2 reacts with a proton.

Set up an Equilibrium Expression for the Base Ionization Reaction.

Now, we need to determine the concentration of H⁺ ions in the solution. We can set up an equilibrium expression for the reaction between C2H5NH2 and water: C2H5NH2 + H2O ↔ C2H5NH3⁺ + OH⁻ The given Kb for C2H5NH2 is 5.6 X 10⁻⁴, so we can write the expression: Kb = \(\frac{[C2H5NH3^+][OH^-]}{[C2H5NH2]}\) We're given that the initial concentration of C2H5NH2 is 0.67 M. Let x represent the change in concentrations due to the reaction, so the concentration of C2H5NH3⁺ and OH⁻ are both x at equilibrium. The concentration of C2H5NH2 at equilibrium will be 0.67 M – x. We can now rewrite the Kb expression: 5.6 × 10⁻⁴ = \(\frac{x^2}{(0.67-x)}\)

Solve for the Concentration of OH⁻ Ions.

As the concentration of C2H5NH2 is significantly higher than the given Kb, we can make an assumption that the change in concentration, x, is very small, which means 0.67 - x is approximately equal to 0.67. Now, we can simplify the expression: 5.6 × 10⁻⁴ ≈ \(\frac{x^2}{0.67}\) Now, we'll solve for x: x^2 = 5.6 × 10⁻⁴ × 0.67 x ≈ 3.02 × 10⁻⁴ Since x represents the concentration of the OH⁻ ions at equilibrium, [OH⁻] = 3.02 × 10⁻⁴ M.

Calculate the pH of the Solution.

First, we need to find the pOH of the solution, using the concentration of OH⁻ ions: pOH = -log10([OH⁻]) pOH = -log10(3.02 × 10⁻⁴) ≈ 3.52 From the relationship between pH and pOH (pH + pOH = 14), we can now calculate the pH of the solution: pH = 14 - pOH pH = 14 - 3.52 ≈ 10.48 So, the pH of the 0.67 M ethylamine solution is approximately 10.48.

Key concepts

Base Ionization Reaction

In chemistry, a base ionization reaction involves a base, which is a substance capable of accepting protons (H⁺ ions), reacting with water to produce hydroxide ions (OH⁻) and the corresponding conjugate acid. For example, ethylamine ((C₂H₅NH₂) accepts a proton from a water molecule, forming ethylammonium ion ((C₂H₅NH₃⁺) and a hydroxide ion ((OH⁻), as seen in the reaction: (C₂H₅NH₂) + H₂O ↔ (C₂H₅NH₃⁺) + OH⁻.
This reaction is crucial as it dictates the behavior of bases in aqueous solutions and influences the pH level of the solution. The degree to which a base ionizes is represented by its base ionization constant (K_b), which is specific to each base. A higher K_b value indicates a stronger base, meaning it ionizes more completely in water.

Equilibrium Expression

An equilibrium expression quantifies the balance between the reactants and products in a reversible chemical reaction at equilibrium. For any base ionization reaction, such as that of ethylamine in water, the expression takes the form: K_b = \(\frac{[C₂H₅NH₃⁺][OH⁻]}{[C₂H₅NH₂]}\).
In this formula, K_b is the base ionization constant, [C₂H₅NH₃⁺] and [OH⁻] represent the concentrations of the products, and [C₂H₅NH₂] denotes the concentration of the initial base. The square brackets signify molar concentrations in moles per liter (M). To predict the pH of a solution involving a weak base, one can use this expression to find the concentrations of ions at equilibrium. It's important to note that water, the solvent, does not appear in the expression because its concentration is much larger and changes insignificantly.

Concentration of Hydroxide Ions

The concentration of hydroxide ions ((OH⁻) is a direct measure of the basicity of a solution. In the context of a base ionization reaction, calculating this concentration can help us understand how basic a solution is, and consequently determine its pH. The concentration of (OH⁻ ions is established at equilibrium and is calculated by assuming the concentration of the base does not change substantially, particularly when K_b is much smaller than the initial concentration of the base.
For example: If the initial concentration of ethylamine is 0.67 M and the ionization constant is K_b = 5.6 × 10⁻⁴, we determine the equilibrium concentration of OH⁻ ions by the simplification x²/(0.67 - x) ≈ x²/0.67, where x represents the change in the concentration of OH⁻. This simplification is valid due to the small K_b value, thus x is negligible when compared to 0.67 M. After solving the equation for x, we find the concentration of OH⁻, which is critical in calculating the pOH and ultimately the pH of the solution.