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Ch.16 - Aqueous Equilibria: Acids & Bases
Chapter 16, Problem 15

For a solution of two weak acids with comparable values of Ka, there is no single principal reaction. The two acid dissociation equilibrium equations must therefore be solved simultaneously. Calculate the pH in a solution that is 0.10 M in acetic acid (CH3CO2H, Ka = 1.8 * 10^-5) and 0.10 M in benzoic acid (C6H5CO2H, Ka = 6.5 * 10^-5). (Hint: Let x = [CH3CO2H] that dissociates and y = [C6H5CO2H] that dissociates; then [H3O+] = x + y.)

Verified step by step guidance
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Step 1: Write the dissociation equations for both acetic acid and benzoic acid. For acetic acid: \( \text{CH}_3\text{CO}_2\text{H} \rightleftharpoons \text{CH}_3\text{CO}_2^- + \text{H}_3\text{O}^+ \) and for benzoic acid: \( \text{C}_6\text{H}_5\text{CO}_2\text{H} \rightleftharpoons \text{C}_6\text{H}_5\text{CO}_2^- + \text{H}_3\text{O}^+ \).
Step 2: Set up the equilibrium expressions for each acid using their respective \( K_a \) values. For acetic acid: \( K_{a1} = \frac{[\text{CH}_3\text{CO}_2^-][\text{H}_3\text{O}^+]}{[\text{CH}_3\text{CO}_2\text{H}]} = 1.8 \times 10^{-5} \) and for benzoic acid: \( K_{a2} = \frac{[\text{C}_6\text{H}_5\text{CO}_2^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{CO}_2\text{H}]} = 6.5 \times 10^{-5} \).
Step 3: Assume \( x \) moles per liter of acetic acid dissociate and \( y \) moles per liter of benzoic acid dissociate. Therefore, \([\text{H}_3\text{O}^+] = x + y\).
Step 4: Substitute \([\text{H}_3\text{O}^+] = x + y\) into the equilibrium expressions: \( K_{a1} = \frac{x(x + y)}{0.10 - x} \) and \( K_{a2} = \frac{y(x + y)}{0.10 - y} \).
Step 5: Solve the system of equations simultaneously to find the values of \( x \) and \( y \), which will allow you to calculate \([\text{H}_3\text{O}^+] = x + y\) and subsequently the pH using \( \text{pH} = -\log([\text{H}_3\text{O}^+]) \).

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Weak Acids and Acid Dissociation Constant (Ka)

Weak acids are substances that partially dissociate in solution to produce hydrogen ions (H3O+). The strength of a weak acid is quantified by its acid dissociation constant (Ka), which measures the extent of dissociation. A higher Ka value indicates a stronger weak acid, while a lower Ka suggests a weaker acid. In this problem, both acetic acid and benzoic acid are weak acids with given Ka values, which are essential for calculating their contributions to the pH.
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Simultaneous Equilibrium Calculations

When multiple weak acids are present in a solution, their dissociation reactions can affect each other, leading to a situation where no single principal reaction dominates. To find the pH, one must set up and solve the equilibrium expressions for each acid simultaneously. This involves defining variables for the concentrations of dissociated species and applying the equilibrium constants (Ka) to create a system of equations that can be solved together.
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pH Calculation from Hydronium Ion Concentration

The pH of a solution is a measure of its acidity, defined as the negative logarithm of the hydronium ion concentration: pH = -log[H3O+]. In this scenario, the total concentration of hydronium ions is the sum of the contributions from both weak acids, represented as [H3O+] = x + y, where x and y are the amounts of acetic and benzoic acids that dissociate, respectively. Calculating the pH requires determining this total concentration from the equilibrium expressions.
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Hydronium Ion Concentration Example