Textbook Question
For each of the following pairs of reactions, indicate which one has the more favorable equilibrium constant (that is, which one most favors products):
1.
2.
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Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
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Table of contents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 1)31
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 2)65
- Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry84
- Ch. 3 - An Introduction to Organic Compounds:Nomenclature, Physical Properties, and Structure183
- Ch. 4 - Isomers: The Arrangement of Atoms in Space121
- Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics83
- Ch. 6 - The Reactions of Alkenes • The Stereochemistry of Addition Reactions175
- Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis119
- Ch. 8 - Delocalized Electrons: Their Effect on Stability, pKa, and the Products of a Reaction • Aromaticity and Electronic Effects: An Introduction to the Reactions of Benzene148
- Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides212
- Ch. 10 - Reactions of Alcohols, Ethers, Epoxides, Amines, and Sulfur-Containing Compounds131
- Ch. 11 - Organometallic Compounds60
- Ch. 12 - Radicals90
- Ch. 13 - Mass Spectrometry; Infrared Spectroscopy; UV/Vis Spectroscopy62
- Ch. 14 - NMR Spectroscopy95
- Ch. 15 - Reactions of Carboxylic Acids and Carboxylic Acid Derivatives123
- Ch. 16 - Reactions of Aldehydes and Ketones • More Reactions of Carboxylic Acid Derivatives141
- Ch. 17 - Reactions at the Alpha-Carbon101
- Ch. 18 - Reactions of Benzene and Substituted Benzenes144
- Ch. 19 - More About Amines • Reactions of Heterocyclic Compounds49
- Ch. 20 - The Organic Chemistry of Carbohydrates80
- Ch. 21 - Amino Acids, Peptides, and Proteins57
- Ch. 22 - Catalysis in Organic Reactions and in Enzymatic Reactions35
- Ch. 23 - The Organic Chemistry of the Coenzymes—Compounds Derived from Vitamins1
- Ch. 28 - Pericyclic Reactions58
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Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 71
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 71Chapter 3, Problem 71
a. Without using a calculator, estimate the pH of each of the following solutions:
1. [HO−] = 3.2 × 10−5
2. [H3O+] = 8.3 × 10−1
3. [H3O+] = 1.7 × 10−3
b. Determine the exact pH, using a calculator.
Verified step by step guidance1
Step 1: Recall the formula for calculating pH, which is defined as \( \text{pH} = -\log_{10}[\text{H}_3\text{O}^+] \). This formula applies to all solutions where the concentration of \( \text{H}_3\text{O}^+ \) (or \( \text{HO}^- \) indirectly) is known.
Step 2: For the first solution, \( [\text{HO}^-] = 3.2 \times 10^{-5} \), use the relationship between \( [\text{HO}^-] \) and \( [\text{H}_3\text{O}^+] \): \( [\text{H}_3\text{O}^+] \cdot [\text{HO}^-] = 10^{-14} \). Solve for \( [\text{H}_3\text{O}^+] \) by dividing \( 10^{-14} \) by \( 3.2 \times 10^{-5} \). Then, estimate the pH using \( \text{pH} = -\log_{10}[\text{H}_3\text{O}^+] \).
Step 3: For the second solution, \( [\text{H}_3\text{O}^+] = 8.3 \times 10^{-1} \), directly apply the formula \( \text{pH} = -\log_{10}[\text{H}_3\text{O}^+] \). Since the concentration is close to 1, the pH will be slightly less than 0. Estimate the value by considering the logarithmic scale.
Step 4: For the third solution, \( [\text{H}_3\text{O}^+] = 1.7 \times 10^{-3} \), again use \( \text{pH} = -\log_{10}[\text{H}_3\text{O}^+] \). Recognize that \( \log_{10}(1.7) \) is slightly greater than 0, and \( \log_{10}(10^{-3}) \) is -3. Combine these to estimate the pH as slightly above 3.
Step 5: To determine the exact pH values using a calculator, input the concentrations into the formula \( \text{pH} = -\log_{10}[\text{H}_3\text{O}^+] \) for each solution. For the first solution, calculate \( [\text{H}_3\text{O}^+] \) first using \( [\text{H}_3\text{O}^+] = \frac{10^{-14}}{[\text{HO}^-]} \), then find the pH. For the second and third solutions, directly compute the logarithm of the given \( [\text{H}_3\text{O}^+] \) values.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
pH Scale
The pH scale measures the acidity or basicity of a solution, ranging from 0 to 14. A pH less than 7 indicates an acidic solution, while a pH greater than 7 indicates a basic solution. The pH is calculated as the negative logarithm of the hydronium ion concentration: pH = -log[H3O+]. Understanding this scale is crucial for estimating and calculating pH values.
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The pH scale vs. the pKa scale.
Hydronium Ion Concentration
The concentration of hydronium ions ([H3O+]) in a solution directly influences its pH. A higher concentration of H3O+ corresponds to a lower pH, indicating a more acidic solution. Conversely, a lower concentration results in a higher pH, indicating a more basic solution. This relationship is fundamental for both estimating and calculating pH values in various solutions.
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Logarithmic Relationships
The pH scale is logarithmic, meaning each whole number change on the scale represents a tenfold change in H3O+ concentration. For example, a solution with a pH of 3 has ten times more hydronium ions than a solution with a pH of 4. This concept is essential for understanding how small changes in ion concentration can significantly affect pH, which is critical for both estimation and precise calculation.
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Related Practice
Textbook Question
Given that pH + pOH = 14 and that the concentration of water in a solution of water is 55.5 M, show that the pKa of water is 15.7. (Hint: pOH=−log [HO−])
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Textbook Question
Citrus fruits are rich in citric acid, a compound with three COOH groups. Explain the following:
a. The first pKa (for the COOH group in the center of the molecule) is lower than the pKa of acetic acid.
2
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Textbook Question
You are planning to carry out a reaction that produces protons. The reaction will be buffered at pH = 10.5. Would it be better to use a protonated methylamine/methylamine buffer or a protonated ethylamine/ethylamine buffer? (pKa of protonated methylamine = 10.7; pKa of protonated ethylamine = 11.0)
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Textbook Question
Which of the four reactions has the most favorable equilibrium constant?
1.
1
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Textbook Question
Citrus fruits are rich in citric acid, a compound with three COOH groups. Explain the following:
b. The third pKa is greater than the pKa of acetic acid.
1
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