Ch.17 - Additional Aspects of Aqueous Equilibria

Problem 26a

Chapter 17, Problem 26a

You are asked to prepare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid 1C6H5COOH2 and any amount you need of sodium benzoate 1C6H5COONa2. (a) What is the pH of the benzoic acid solution prior to adding sodium benzoate?

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Identify the chemical species involved: benzoic acid (C6H5COOH) is a weak acid, and its dissociation in water can be represented as C6H5COOH ⇌ C6H5COO⁻ + H⁺.

Use the expression for the acid dissociation constant (Ka) for benzoic acid: Ka = [C6H5COO⁻][H⁺] / [C6H5COOH].

Look up the Ka value for benzoic acid, which is typically around 6.3 x 10⁻⁵.

Assume that the initial concentration of benzoic acid is 0.0200 M and that the change in concentration due to dissociation is small compared to the initial concentration.

Set up an ICE (Initial, Change, Equilibrium) table to solve for the concentration of H⁺ ions at equilibrium, and then calculate the pH using the formula pH = -log[H⁺].

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Key Concepts

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

pH and pKa

pH is a measure of the acidity or basicity of a solution, defined as the negative logarithm of the hydrogen ion concentration. The pKa is the negative logarithm of the acid dissociation constant (Ka) and indicates the strength of an acid. For weak acids like benzoic acid, the pH can be calculated using the formula pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid.

Benzoic Acid as a Weak Acid

Benzoic acid (C6H5COOH) is a weak acid that partially dissociates in solution. Its dissociation can be represented by the equation C6H5COOH ⇌ C6H5COO- + H+. The extent of this dissociation is characterized by its Ka value, which is relatively low, indicating that at equilibrium, the concentration of undissociated benzoic acid remains significant compared to its dissociated ions.

Buffer Solutions

A buffer solution is a system that resists changes in pH upon the addition of small amounts of acid or base. It typically consists of a weak acid and its conjugate base, which can neutralize added acids or bases. In this case, the benzoic acid and sodium benzoate will form a buffer system, allowing the solution to maintain a stable pH when sodium benzoate is added.

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Buffer Solutions

Related Practice

Open Question

A buffer is prepared by adding 10.0 g of ammonium chloride (NH4Cl) to 250 mL of 1.00 M NH3 solution. (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

Textbook Question

You are asked to prepare a pH = 3.00 buffer solution starting from 1.25 L of a 1.00 M solution of hydrofluoric acid (HF) and any amount you need of sodium fluoride (NaF). (a) What is the pH of the hydrofluoric acid solution prior to adding sodium fluoride?

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Textbook Question

You are asked to prepare a pH = 3.00 buffer solution starting from 1.25 L of a 1.00 M solution of hydrofluoric acid (HF) and any amount you need of sodium fluoride (NaF). (b) How many grams of sodium fluoride should be added to prepare the buffer solution? Neglect the small volume change that occurs when the sodium fluoride is added.

1235

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Textbook Question

You are asked to prepare a pH = 4.00 buffer starting from 1.50 L of 0.0200 M solution of benzoic acid 1C6H5COOH2 and any amount you need of sodium benzoate 1C6H5COONa2. (b) How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

1095

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Textbook Question

A buffer contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. (a) What is the pH of this buffer?

678

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Textbook Question

A buffer contains 0.10 mol of acetic acid and 0.13 mol of sodium acetate in 1.00 L. (b) What is the pH of the buffer after the addition of 0.02 mol of KOH?

1041

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