Textbook Question
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.
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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 73
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 73Chapter 3, Problem 73
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−])
Verified step by step guidance1
Step 1: Start by recalling the relationship between pH, pOH, and the ionization constant of water (Kw). The equation pH + pOH = 14 is derived from the fact that Kw = [H3O+] × [HO−] = 1 × 10^−14 at 25°C.
Step 2: Use the hint provided, which states that pOH = −log [HO−]. This means the concentration of hydroxide ions ([HO−]) can be expressed as 10^(−pOH).
Step 3: Recognize that the ionization constant of water (Kw) can also be expressed in terms of pKa. The relationship is pKa = −log Ka, where Ka is the acid dissociation constant. For water, Ka = Kw / [H2O].
Step 4: Substitute the known values into the equation for Ka. Kw = 1 × 10^−14 and the concentration of water ([H2O]) is given as 55.5 M. Therefore, Ka = (1 × 10^−14) / 55.5.
Step 5: Calculate the pKa using the formula pKa = −log Ka. Substitute the value of Ka obtained in Step 4 into this formula to find the pKa of water, which is approximately 15.7.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
pH and pOH
pH is a measure of the hydrogen ion concentration in a solution, while pOH measures the hydroxide ion concentration. The relationship pH + pOH = 14 is fundamental in aqueous solutions at 25°C, indicating that as the concentration of one ion increases, the other decreases. This relationship is crucial for understanding acid-base equilibria and calculating the acidity or basicity of a solution.
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The pH scale vs. the pKa scale.
Ion Product of Water (Kw)
The ion product of water, Kw, is the equilibrium constant for the self-ionization of water, defined as Kw = [H+][OH-]. At 25°C, Kw is approximately 1.0 x 10^-14. This value is essential for calculating pKa and pKb values, as it relates the concentrations of hydrogen and hydroxide ions in pure water and helps determine the acidity of solutions.
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pKa and its Relation to Water
pKa is the negative logarithm of the acid dissociation constant (Ka) and indicates the strength of an acid in solution. For water, the pKa can be derived from the relationship between pH, pOH, and Kw. Since the pKa of water is derived from the dissociation of water into H+ and OH-, it reflects the intrinsic acidity of water, which is relevant for understanding acid-base reactions in aqueous solutions.
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Related Practice
Textbook Question
At what pH does 80% of the acid exist in its acidic form?
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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.
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Textbook Question
How could you separate a mixture of the following compounds? The reagents available to you are water, ether, 1.0 M HCl, and 1.0 M NaOH.
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Textbook Question
If an acid with a pKa of 5.3 is in an aqueous solution of pH 5.7, what percentage of the acid is present in its acidic form?
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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.
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