Problem 35 - Chemistry An Atoms First Approach

Question

Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.10 M\) acetic acid/0.25 \(M\) sodium acetate b. \(0.25 M\) acetic acid/0.10 \(M\) sodium acetate c. \(0.080 M\) acetic acid/0.20 \(M\) sodium acetate d. \(0.20 M\) acetic acid/0.080 \(M\) sodium acetate

Step-by-Step Solution

Answer

The pH values of the buffered solutions are as follows: a. \(pH = 4.74 + \log\frac{0.25}{0.10} = 5.57\) b. \(pH = 4.74 + \log\frac{0.10}{0.25} = 4.10\) c. \(pH = 4.74 + \log\frac{0.20}{0.080} = 5.40\) d. \(pH = 4.74 + \log\frac{0.080}{0.20} = 4.34\)

Step 1: Identify the weak acid and its conjugate base

In this solution, the weak acid (HA) is acetic acid with a concentration of 0.10 M and its conjugate base (A-) is the acetate ion coming from sodium acetate with a concentration of 0.25 M.

Step 2: Use the Henderson-Hasselbalch equation

Using the given concentrations and the pKa of acetic acid (4.74), we can calculate the pH of the solution as follows: \(pH = 4.74 + \log\frac{0.25}{0.10}\) b. \(0.25 M\) acetic acid/0.10 \(M\) sodium acetate

Step 1: Identify the weak acid and its conjugate base

In this solution, the weak acid (HA) is acetic acid with a concentration of 0.25 M and its conjugate base (A-) is the acetate ion coming from sodium acetate with a concentration of 0.10 M.

Step 2: Use the Henderson-Hasselbalch equation

Using the given concentrations and the pKa of acetic acid (4.74), we can calculate the pH of the solution as follows: \(pH = 4.74 + \log\frac{0.10}{0.25}\) c. \(0.080 M\) acetic acid/0.20 \(M\) sodium acetate

Step 1: Identify the weak acid and its conjugate base

In this solution, the weak acid (HA) is acetic acid with a concentration of 0.080 M and its conjugate base (A-) is the acetate ion coming from sodium acetate with a concentration of 0.20 M.

Step 2: Use the Henderson-Hasselbalch equation

Using the given concentrations and the pKa of acetic acid (4.74), we can calculate the pH of the solution as follows: \(pH = 4.74 + \log\frac{0.20}{0.080}\) d. \(0.20 M\) acetic acid/0.080 \(M\) sodium acetate

Step 1: Identify the weak acid and its conjugate base

In this solution, the weak acid (HA) is acetic acid with a concentration of 0.20 M and its conjugate base (A-) is the acetate ion coming from sodium acetate with a concentration of 0.080 M.

Step 2: Use the Henderson-Hasselbalch equation

Using the given concentrations and the pKa of acetic acid (4.74), we can calculate the pH of the solution as follows: \(pH = 4.74 + \log\frac{0.080}{0.20}\)

Key Concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is crucial for calculating the pH of buffer solutions. It is a mathematical representation that relates pH, pKa, and the ratio of the concentrations of a conjugate acid-base pair.

Here is the equation in its typical form: \[\begin{equation} pH = pKa + \text{log}\frac{[A^-]}{[HA]} \text{\end{equation}\]} In this equation, pH refers to the acidity or basicity of the solution, pKa is the acid dissociation constant, [A^-] is the concentration of the conjugate base, and [HA] is the concentration of the acid. By plugging in the known values into this formula, you can solve for the pH of the buffer. This relationship is pivotal, especially in solutions where the acid-base pair consist of a weak acid and its conjugate base. Moreover, knowing how to rearrange and solve this logarithmic equation is a fundamental skill in chemistry.

Acetic Acid and Acetate Buffer System

Buffer systems are designed to resist changes in pH when small amounts of acid or base are added. The acetic acid and acetate buffer system is a classic example involving a weak acid (acetic acid, CH3COOH) and its conjugate base (acetate, CH3COO^-).

This system works because the acetic acid partially dissociates in water, producing acetate ions and hydrogen ions. If extra hydrogen ions are added to the system, they will react with the acetate ions to form more acetic acid. Conversely, if hydroxide ions are added, they react with acetic acid to produce acetate ions and water. This reaction either way helps maintain the pH at a relatively constant value. The practical use of this buffer system is prevalent in biochemical processes and industrial applications where a stable pH is necessary.

Weak Acid and Conjugate Base

In the context of a buffer solution, a weak acid is one that does not fully dissociate in water. Its conjugate base is the species that is formed when the acid loses a proton (H+).

For the acetic acid and acetate buffer system, acetic acid (CH3COOH) is the weak acid, and acetate (CH3COO^-) is its conjugate base. Together, they create a equilibrium in solution that can absorb fluctuations in pH due to their ability to neutralize added acids or bases. This relationship is essential for understanding buffer chemistry, as the weak acid contributes to the acidic properties while the conjugate base counters by providing basic properties.

Logarithmic pH Calculations

pH is a logarithmic scale used to specify the acidity or basicity of an aqueous solution. The calculation of pH often involves logarithms due to the definition of pH as the negative logarithm of the hydrogen ion concentration.

Here is what the math looks like: \[\begin{equation} pH = -\text{log}[H^+] \text{\end{equation}\]} Using logarithms helps manage the broad range of hydrogen ion concentrations found in solutions. This scale compresses the range of ion concentrations, so pH 7 is neutral, values less than 7 are acidic, and values greater than 7 are basic or alkaline. An understanding of logarithms is crucial in pH calculations, especially when using the Henderson-Hasselbalch equation to determine the pH of buffer solutions.