Problem 56 - Chemistry and Chemical Reactivity

Question

A hypothetical weak base has \(K_{\mathrm{b}}=5.0 \times 10^{-4}\) Calculate the equilibrium concentrations of the base, its conjugate acid, and \(\mathrm{OH}^{-}\) in a \(0.15 \mathrm{M}\) solution of the base.

Step-by-Step Solution

Answer

[B] = 0.1413 M, [BH⁺] = 8.7 × 10⁻³ M, [OH⁻] = 8.7 × 10⁻³ M

Step 1: Write the Equilibrium Expression

For a weak base, we start by writing the dissociation in water:- Base (B) + Water (H₂O) ⇌ Conjugate Acid (BH⁺) + Hydroxide (OH⁻)The expression for the base dissociation constant is: \[ K_{\mathrm{b}} = \frac{[\mathrm{BH}^+][\mathrm{OH}^-]}{[\mathrm{B}]} \]

Step 2: Set Up Initial Conditions

Assume the initial concentration of the base (B) is given as 0.15 M, and initially, there are no products, BH⁺ or OH⁻. This provides us with these initial concentrations: - [B] = 0.15 M - [BH⁺] = 0 M - [OH⁻] = 0 M.

Step 3: Define Change in Concentrations

As the reaction reaches equilibrium, let the change in concentration of B be \(-x\), then the changes for BH⁺ and OH⁻ will be \(+x\) each because they are produced in a 1:1 ratio.This leads to equilibrium concentrations as follows:- [B] = 0.15 - x- [BH⁺] = x- [OH⁻] = x.

Step 4: Write the Expression for Kb at Equilibrium

Substitute the equilibrium concentrations into the expression for the base dissociation constant:\[ 5.0 \times 10^{-4} = \frac{x \cdot x}{0.15 - x} = \frac{x^2}{0.15 - x} \]

Step 5: Make an Assumption for Simplification

Assume that \(x\) is much smaller than 0.15, so \(0.15 - x \approx 0.15\). This allows us to simplify the equation to:\[ 5.0 \times 10^{-4} = \frac{x^2}{0.15} \]

Step 6: Solve the Simplified Equation

Solve for \(x\) by rearranging the equation:\[ x^2 = 5.0 \times 10^{-4} \times 0.15 \]\[ x^2 = 7.5 \times 10^{-5} \]\[ x = \sqrt{7.5 \times 10^{-5}} \]Determining \(x\) gives:\[ x \approx 8.7 \times 10^{-3} \]

Step 7: Determine Equilibrium Concentrations

Substitute \(x\) back to find the equilibrium concentrations:- \([\mathrm{B}] = 0.15 - 8.7 \times 10^{-3} \approx 0.1413\) M- \([\mathrm{BH}^+] = x = 8.7 \times 10^{-3}\) M- \([\mathrm{OH}^-] = x = 8.7 \times 10^{-3}\) M.

Key Concepts

Weak Base

A weak base is a base that does not completely dissociate into its ions in solution. Unlike strong bases that completely ionize, weak bases only partially break down into ions. This means that only a fraction of the base molecules donate their lone electron pairs to water molecules to produce hydroxide ions (OH⁻).
Examples of weak bases include ammonia (NH₃) and organic bases like methylamine (CH₃NH₂).
In any aqueous solution of a weak base, an equilibrium is established between the dissociated ions and the intact base molecules. This equilibrium depends on the strength of the base, which is expressed in terms of the base dissociation constant, denoted as \(K_b\). Understanding how weak bases interact in water helps predict the pH of the solution and the behavior of the solution when mixed with acids or other substances.

Base Dissociation Constant

The base dissociation constant, or \(K_b\), is a critical value that indicates the strength of a weak base. It defines the equilibrium concentration of the products and reactants when the base is dissolved in water.
Expressed mathematically, \(K_b\) is given by the equation:

• \[ K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} \]

This formula shows that \(K_b\) is the ratio of the concentration of the conjugate acid (BH⁺) and hydroxide ions (OH⁻) to the undissociated base (B). A larger \(K_b\) value indicates a stronger weak base, meaning more of the base dissociates to release OH⁻ ions in solution. Conversely, a smaller \(K_b\) signifies a weaker base with less dissociation.
Understanding \(K_b\) helps to predict and calculate the pH of the weak base solution and determine the concentrations of various species at equilibrium.

Equilibrium Constant Calculation

The calculation of equilibrium concentrations in a weak base solution is crucial for many chemistry problems. The equilibrium constant expression, based on \(K_b\), allows us to find the concentrations of various chemical species once equilibrium has been reached.
Formulating the Equilibrium Problem
We start by setting up an equation that involves the initial concentrations and the changes after dissociation. For a weak base B dissociating in water:

• Initial concentrations are given, e.g., [B]=0.15 M, [BH⁺]=0, [OH⁻]=0.
• At equilibrium, [BH⁺] and [OH⁻] = \(x\), and [B] = \(0.15 - x\).
• The equilibrium expression is \(K_b = \frac{x^2}{0.15 - x}\).

Solving the Equation
Because weak bases have low dissociation, \(x\) is typically much smaller than initial concentrations, allowing simplifications in the equation and easier determination of \(x\). Calculating \(x\) gives us the hydroxide ion concentration, as well as the concentrations of the base and its conjugate acid at equilibrium.

OH- Concentration

The concentration of hydroxide ions, \([\text{OH}^-]\), is a vital factor in determining the basicity of a solution. Higher \([\text{OH}^-]\) indicates more basic or alkaline conditions.
In a weak base solution, \([\text{OH}^-]\) arises from the partial dissociation of the base. To find \([\text{OH}^-]\), we calculate it directly through the dissociation process using the equilibrium concentration given by \(K_b\).
Calculating \([\text{OH}^-]\)
Based on the equilibrium equation, \([\text{OH}^-]\) is equal to \(x\), where \(x\) was determined from the solved equation for \(K_b\):

• Calculate \(x\), given the simplified equation: \(x = \sqrt{5.0 \times 10^{-4} \times 0.15}\).
• The result, \(x \approx 8.7 \times 10^{-3}\), is the \([\text{OH}^-]\) at equilibrium.

Understanding \([\text{OH}^-]\) helps determine the pH of the solution and is essential for chemical reactions involving bases.