Problem 82 - Chemistry The Central Science

Question

Benzenesulfonic acid is a monoprotic acid with \(\mathrm{p} K_{a}=2.25\). Calculate the \(\mathrm{pH}\) of a buffer composed of \(0.150 \mathrm{M}\) benzenesulfonic acid and \(0.125 M\) sodium benzenesulfonate.

Step-by-Step Solution

Answer

The pH of the buffer solution composed of \(0.150 M\) benzenesulfonic acid and \(0.125 M\) sodium benzenesulfonate can be calculated using the Henderson-Hasselbalch equation: \(pH = pK_a + \log \frac{[\text{base}]}{[\text{acid}]}\). Plugging in the given values, we get \(pH = 2.25 + \log \frac{0.125}{0.150} \approx 2.07\). Therefore, the pH of the buffer solution is approximately \(2.07\).

Step 1: Identify the components of the buffer solution

The weak acid and its conjugate base are identified as benzenesulfonic acid and sodium benzenesulfonate, respectively.

Step 2: Write the given concentrations and pKa value

The concentration of benzenesulfonic acid is 0.150 M, the concentration of sodium benzenesulfonate is 0.125 M, and the pKa of benzenesulfonic acid is 2.25.

Step 3: Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution

The Henderson-Hasselbalch equation is: \( pH=pK_{a}+\log \frac{[\text{base}]}{[\text{acid}]} \) Plug in the given values to obtain the pH of the buffer solution: \( pH = 2.25 + \log \frac{0.125}{0.150} \)

Step 4: Calculate the logarithm and find the pH

Calculate the logarithm term: \( \log \frac{0.125}{0.150} \approx -0.177 \) Now find the pH: \( pH = 2.25 - 0.177 \approx 2.07 \) The pH of the buffer solution composed of 0.150 M benzenesulfonic acid and 0.125 M sodium benzenesulfonate is approximately 2.07.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is a vital tool in chemistry, especially when dealing with buffer solutions. A buffer solution typically consists of a weak acid and its conjugate base. The equation is used to estimate the pH of such solutions. The equation is:\[\text{pH} = \text{p}K_a + \log \frac{[\text{base}]}{[\text{acid}]}\]Here:

• \( \text{pH} \) is the measure of the acidity or basicity of the solution.
• \( \text{p}K_a \) is the negative logarithm of the acid dissociation constant, \( K_a \), of the weak acid.
• \([\text{base}]\) is the molarity of the conjugate base.
• \([\text{acid}]\) is the molarity of the weak acid.

This equation is especially useful because it allows us to calculate the pH without directly measuring it, as long as we know the concentrations of the acid and its conjugate base, and the \( \text{p}K_a \) of the acid.

pH Calculation

Calculating the pH of a buffer solution involves plugging the known values into the Henderson-Hasselbalch Equation. For instance, if we're working with a buffer made from benzenesulfonic acid and sodium benzenesulfonate, we first need to know:

• The concentration of the benzenesulfonic acid, the weak acid, which is 0.150 M.
• The concentration of the sodium benzenesulfonate, the conjugate base, which is 0.125 M.
• The \( \text{p}K_a \) of benzenesulfonic acid, which is given as 2.25.

By inserting these values into the Henderson-Hasselbalch Equation:\[pH = 2.25 + \log \frac{0.125}{0.150}\]The logarithmic term needs to be calculated; in this instance, \( \log \frac{0.125}{0.150} \approx -0.177 \). Therefore, the pH of the solution becomes:\[pH = 2.25 - 0.177 = 2.07\]This result indicates that the solution is slightly acidic, which is expected for a solution primarily containing a weak acid.

Acid-Base Equilibrium

Understanding acid-base equilibrium is crucial when dealing with buffer solutions. In a buffer, equilibrium between a weak acid and its conjugate base resists changes in pH. ### Equilibrium in Buffer SolutionsA weak acid, like benzenesulfonic acid, partially dissociates in a solution:\[\text{C}_6\text{H}_5\text{SO}_3\text{H} \rightleftharpoons \text{C}_6\text{H}_5\text{SO}_3^- + \text{H}^+\]Similarly, the conjugate base, like sodium benzenesulfonate, interacts with hydrogen ions:\[\text{C}_6\text{H}_5\text{SO}_3^- + \text{H}^+ \rightleftharpoons \text{C}_6\text{H}_5\text{SO}_3\text{H}\]### Role of EquilibriumThese equilibrium reactions allow the solution to resist drastic changes in pH. Adding a small amount of acid shifts the equilibrium to favor the formation of more weak acid, while adding base shifts it the other way. The buffer's ability to maintain equilibrium makes it essential for many chemical and biological processes. Understanding this balance helps in designing solutions with desired pH characteristics and preparing for reactions where a controlled pH is necessary.