Problem 71 - Chemistry: The Central Science

Question

Calculate the molar concentration of \(\mathrm{OH}^{-}\)in a \(0.075 \mathrm{M}\) solution of ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} ; K_{b}=6.4 \times 10^{-4}\right)\). Calculate the \(\mathrm{pH}\) of this solution.

Step-by-Step Solution

Answer

The molar concentration of OH⁻ ions in the 0.075 M solution of ethylamine is approximately \(6.93 \times 10^{-3}\, M\). The pH of the solution is approximately 11.84.

Step 1: Write the base ionization equation for ethylamine

C₂H₅NH₂ + H₂O ⇌ C₂H₅NH₃⁺ + OH⁻ In this step, we write the equilibrium equation for the reaction of ethylamine with water. Ethylamine is a weak base, so it reacts with water to form the corresponding conjugate acid (ethylammonium ion) and hydroxide ions.

Step 2: Write the Kb expression for the equilibrium

\(K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]}\) Using the ionization equation from Step 1, we write the base ionization constant expression for the given \(K_b\). The concentration of the ethylammonium ions and hydroxide ions are in the numerator, and the concentration of the ethylamine is in the denominator.

Step 3: Set up an ICE table

Initial Concentrations: [C₂H₅NH₂] = 0.075 M [C₂H₅NH₃⁺] = 0 M [OH⁻] = 0 M Change in Concentrations: [C₂H₅NH₂] = -x M [C₂H₅NH₃⁺] = +x M [OH⁻] = +x M Equilibrium Concentrations: [C₂H₅NH₂] = 0.075 - x M [C₂H₅NH₃⁺] = x M [OH⁻] = x M In this step, we construct an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations of each species involved in the reaction.

Step 4: Calculate base ionization constant (Kb) and find [OH⁻]

\(6.4 \times 10^{-4} = \frac{x^2}{0.075-x}\) Next, substitute the equilibrium concentrations from the ICE table into the Kb expression to solve for x, which represents the concentration of OH⁻ ions at equilibrium. We can ignore the -x in the denominator as x is small compared to 0.075. \(6.4 \times 10^{-4} = \frac{x^2}{0.075}\) Now, solve for x: \(x^2 = 0.075(6.4 \times 10^{-4})\) \(x^2 = 4.8 \times 10^{-5}\) \(x = \sqrt{4.8 \times 10^{-5}}\) \[x = 6.93 \times 10^{-3}\] Thus, the molar concentration of OH⁻ ions at equilibrium is approximately \(6.93 \times 10^{-3}\, M\).

Step 5: Calculate the pOH and pH of the solution

To find the pOH of the solution, use the formula: pOH = -log[OH⁻] pOH = -log(\(6.93 \times 10^{-3}\)) pOH ≈ 2.16 Now, to find the pH of the solution, use the relationship between pH and pOH: pH + pOH = 14 pH = 14 - pOH pH = 14 - 2.16 pH ≈ 11.84 The pH of the 0.075 M solution of ethylamine is approximately 11.84.

Key Concepts

Molar Concentration

Molar concentration is essential in understanding how substances behave in a solution. It is the measure of the number of moles of a solute per liter of solution. For instance, if you dissolve a known amount of ethylamine in water, you determine how concentrated the solution is by calculating its molar concentration.
This value tells us about the availability of reactive molecules in the solution, which directly influences the reaction rates and equilibria.
To calculate molar concentration, we typically use the formula:

• Molarity (M) = moles of solute / liters of solution

This is a straightforward yet powerful way to quantify the concentration of any solute in a solution.

pH Calculation

Calculating the pH of a solution is crucial for determining its acidity or basicity. pH is a scale used to specify the acidity or basicity of an aqueous solution.
Here's the fundamental relationship we use:

• pH = 14 - pOH

The pOH value is derived from the concentration of hydroxide ions ( [OH^-]), which we can find using the equation:

• pOH = -log[OH^-]

In cases like ethylamine solutions, which are basic, the focus is on the generation of OH⁻ ions. Knowing their concentration lets us transition to finding the pH, using the intrinsic relationship between pH and pOH. This process allows us to complete the picture of the solution's characteristics.

Base Ionization

Base ionization involves a base reacting with water to donate hydroxide ions (OH⁻) into the solution. For weak bases like ethylamine, this reaction is not complete, and instead, an equilibrium is set up.
The ionization equation for ethylamine shows how it reacts with water: C₂H₅NH₂ + H₂O ⇌ C₂H₅NH₃⁺ + OH⁻ The base ionization constant ( K_b) is a measure of the strength of a base in solution. It’s represented by the expression:

• K_b = [C₂H₅NH₃⁺][OH⁻] / [C₂H₅NH₂]

This formula quantifies the extent to which the base dissociates into its ions.
Understanding this concept helps us predict how additional solutes or changes in conditions might affect the balance of ions, thus giving insight into the solution's behavior.

Chemical Equilibrium

Chemical equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction in a chemical process. At this point, the concentrations of all reactants and products remain constant over time.
For the ionization of ethylamine, setting up an ICE table helps visualize this equilibrium:

• Initial, Change, Equilibrium concentrations (ICE)

This approach provides a systematic way to track changes in concentration as the reaction progresses to equilibrium.
By applying the equilibrium expression, we calculate the concentrations at equilibrium, revealing how much of each ion is present when the equilibrium is established. Grasping chemical equilibria is crucial for making sense of reactions in real-world scenarios, as it underscores the dynamic balance that reactions strive to maintain.