Problem 98 - Chemistry An Atoms First Approach

Question

Calculate \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}^{+}\right],\) and the \(\mathrm{pH}\) of \(0.40 M\) solutions of each of the following amines (the \(K_{\mathrm{b}}\) values are found in Table 13-3). a. aniline b. methylamine

Step-by-Step Solution

Answer

For aniline, \(\left[\mathrm{OH}^{-}\right] = x ≈ 1.3 \times 10^{-5} M\), \(\left[\mathrm{H}^{+}\right] ≈ 7.7 \times 10^{-10} M\), and \(\mathrm{pH} ≈ 9.11\). For methylamine, \(\left[\mathrm{OH}^{-}\right] = x ≈ 6.63 \times 10^{-3} M\), \(\left[\mathrm{H}^{+}\right] ≈ 1.5 \times 10^{-12} M\), and \(\mathrm{pH} ≈ 11.82\).

Step 1: Write the equilibrium expressions for both amines

We will write the equilibrium expressions for both amines as they react with water to form hydroxide ions and their respective conjugate acids. The general reaction is: Amine + H2O \(\rightleftharpoons\) Conjugate Acid + OH- For aniline (C6H5NH2): C6H5NH2 + H2O \(\rightleftharpoons\) C6H5NH3+ + OH- For methylamine (CH3NH2): CH3NH2 + H2O \(\rightleftharpoons\) CH3NH3+ + OH-

Step 2: Set up the ICE table for the equilibrium expressions

We will set up an ICE (Initial, Change, Equilibrium) table for each amine to determine how the concentrations change during the reaction. For aniline: \[ \begin{array}{cccc} &{} & C6 H5 N H2 & & + & & H2 O & \rightleftharpoons & C6 H5 N H3^{+} & + & O H^{-} \\ I & & 0.40 & M & - & - & - & & 0 & M & & 0 & M \\ C & & - x & & - & - & - & & + x & & + x \\ E & & 0.40 - x & & - & - & - & & x & & x \end{array} \] For methylamine: \[ \begin{array}{cccc} &{} & C H3 N H2 & & + & & H2 O & \rightleftharpoons & C H3 N H3^{+} & + & O H^{-} \\ I & & 0.40 & M & - & - & - & & 0 & M & & 0 & M \\ C & & - x & & - & - & - & & + x & & + x \\ E & & 0.40 - x & & - & - & - & & x & & x \end{array} \]

Step 3: Use the Kb values to solve for x, and then find [OH-]

For aniline, given \(K_b = 4.2 \times 10^{-10}\): \[K_b = \frac{[C6H5NH3^+][OH⁻]}{[C6H5NH2]} = \frac{x^2}{0.40-x}\] For methylamine, given \(K_b = 4.4 \times 10^{-4}\): \[K_b = \frac{[CH3NH3^+][OH⁻]}{[CH3NH2]} = \frac{x^2}{0.40-x}\] For each amine, we will solve for x, which represents the equilibrium concentration of OH⁻. Since \(K_b\) values are small, we can assume that \(x << 0.40\) and simplify the equations. For aniline: \[4.2 \times 10^{-10} \approx \frac{x^2}{0.40}\] Solve for x to find \(\left[OH^{-}\right]\). For methylamine: \[4.4 \times 10^{-4} \approx \frac{x^2}{0.40}\] Solve for x to find \(\left[OH^{-}\right]\).

Step 4: Calculate [H+] and pH

Using the obtained OH⁻ concentration, we can find the H⁺ concentration using the relationship: \(K_w = \left[H^+\right]\left[OH^{-}\right]\), where \(K_w = 1.0 \times 10^{-14}\). Solve for the [H+] for both amines and use the formula to find the pH: \(pH = -\log\left([H^+]\right)\) Calculate the pH of both the aniline and methylamine solutions.

Key Concepts

ICE table

The ICE table is a handy tool in chemistry for tracking the concentrations of reactants and products in a chemical reaction as it reaches equilibrium. Let's break it down:

An ICE table consists of three main parts:

• Initial (I): The starting concentrations of the reactants and products.
• Change (C): The change in concentrations as the reaction moves towards equilibrium.
• Equilibrium (E): The concentrations of the reactants and products when the system is at equilibrium.

For aniline and methylamine, we begin with 0.40 M concentrations of each amine and 0 M of their respective conjugate acids or hydroxide ions. As the reaction proceeds, an amount 'x' of the reactive species is changed, resulting in the formation of hydroxide ions and the conjugate acid. The equilibrium expressions add clarity to how concentrations adjust as equilibrium is achieved. Understanding each part of the ICE framework is vital for solving equilibrium problems efficiently.

Equilibrium expressions

Equilibrium expressions are equations derived from the balanced chemical reactions. They help us understand how the concentrations of reactants and products are related at equilibrium. For a base dissociation reaction, the general expression can be written from the equilibrium law:

For amines, the equilibrium reactions are:

• Aniline: \(C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH_3^+ + OH^-\)
• Methylamine: \(CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-\)

The equilibrium constant expression for the dissociation of these weak bases (denoted by the base dissociation constant, \(K_b\)) is crucial. It compares the concentration of the products to the concentration of the reactants at equilibrium. For example, for aniline, the expression is:\[ K_b = \frac{[C_6H_5NH_3^+][OH^-]}{[C_6H_5NH_2]} \]This equation shows us the relationship between the concentration of aniline, its conjugate acid, and hydroxide ions once equilibrium is reached.

Base dissociation constant (Kb)

The base dissociation constant, \(K_b\), quantifies the strength of a base in solution. It indicates the extent to which a base dissociates to form hydroxide ions in water. A larger \(K_b\) value means the base is stronger and will dissociate more to produce OH⁻ ions.

Both aniline and methylamine have their own \(K_b\) values, reflecting their base strength:

• Aniline: \(4.2 \times 10^{-10}\)
• Methylamine: \(4.4 \times 10^{-4}\)

Given these \(K_b\) values, we calculate the concentration of hydroxide ions, \([OH^-]\), in the solution using the formula:\[ K_b = \frac{x^2}{[Initial \, Concentration] - x} \]For simplicity, since \(K_b\) is quite small relative to the initial concentration (0.40 M), we often approximate by assuming \(x\) is much smaller than 0.40 M. This approximation enables us to simplify the calculation and solve for \(x\), which represents the \([OH^-]\) concentration.

pH and pOH relationship

Understanding the relationship between pH and pOH is crucial in calculating the acidity or basicity of a solution. It stems from the ion product constant of water, \(K_w\), which is always \(1.0 \times 10^{-14}\) at 25°C. This constant maintains a balance:\[ [H^+][OH^-] = K_w \]Knowing \([OH^-]\), we can find \([H^+]\) using the formula:\[ [H^+] = \frac{K_w}{[OH^-]} \]Once \([H^+]\) is calculated, the pH is determined using:\[ pH = -\log([H^+]) \]These steps showcase how hydroxide ion concentrations lead to pH calculations and extend into real-world applications, like understanding soil acidity, water treatment, and more. This method of calculation allows us to bridge the gap between the theoretical expressions and practical assessments of acidity or basicity in a given solution.