Textbook Question
A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: (a) 20.0 mL. (b) 23.0 mL.
Ch.17 - Additional Aspects of Aqueous Equilibria
Chapter 17, Problem 45
A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (c) 34.5 mL, (d) 35.0 mL, (e) 35.5 mL, (f) 50.0 mL.
Verified step by step guidance1
Determine the initial moles of acetic acid (CH3COOH) using the formula: \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). For a 35.0 mL sample of 0.150 M acetic acid, convert the volume to liters and calculate the moles.
Calculate the moles of NaOH added for each volume given (34.5 mL, 35.0 mL, 35.5 mL, 50.0 mL) using the same formula: \( \text{moles} = \text{volume (L)} \times \text{molarity (M)} \). Convert each volume to liters before calculating.
For each volume of NaOH added, determine the moles of acetic acid and NaOH that react. Since the reaction is 1:1, the moles of NaOH will react with an equal number of moles of acetic acid.
Calculate the moles of acetic acid and acetate ion (CH3COO-) remaining after the reaction for each volume of NaOH added. Use the stoichiometry of the reaction: \( \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \).
Use the Henderson-Hasselbalch equation to calculate the pH for each scenario: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \right) \). The \( \text{pK}_a \) of acetic acid is approximately 4.76. Calculate the pH for each volume of NaOH added.
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Acid-Base Titration
An acid-base titration is a quantitative analytical method used to determine the concentration of an acid or base in a solution. In this process, a solution of known concentration (the titrant) is added to a solution of unknown concentration until the reaction reaches its equivalence point, where the amount of acid equals the amount of base. The pH changes during the titration can be monitored to identify this point.
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03:04
Acid-Base Titration
Buffer Solutions
Buffer solutions are mixtures that can resist changes in pH upon the addition of small amounts of acid or base. In the context of the titration of acetic acid with NaOH, the acetic acid and its conjugate base (acetate ion) form a buffer system that helps maintain a relatively stable pH until the equivalence point is reached. Understanding how buffers work is crucial for calculating pH at various stages of the titration.
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03:02Buffer Solutions
pH Calculation
The pH of a solution is a measure of its acidity or basicity, calculated as the negative logarithm of the hydrogen ion concentration. During titration, pH can be calculated using the Henderson-Hasselbalch equation for buffer regions or by using stoichiometry at the equivalence point. After the equivalence point, the pH can be determined based on the concentration of excess hydroxide ions from the added base.
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01:30pH Calculation Example
Related Practice
Textbook Question
A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: (c) 24.0 mL.
Textbook Question
A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: (d) 25.0 mL. (e) 30.0 mL.
Textbook Question
A 35.0-mL sample of 0.150 M acetic acid 1CH3COOH2 is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (b) 17.5 mL.
Textbook Question
Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: (b) 20.0 mL (c) 59.0 mL.
Textbook Question
Consider the titration of 30.0 mL of 0.050 M NH3 with 0.025 M HCl. Calculate the pH after the following volumes of titrant have been added: (e) 61.0 mL (f) 65.0 mL.