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Science
Chemistry
The pOH of an aqueous solution of 0.532 M caffeine (a weak base with the formula C8 H10 N4O2) is (Assume that Kb (C8H10N4O2) = 4.10 × 10-4.)
The pOH of an aqueous solution of 0.532 M caffeine (a weak base with the formula C8 H10 N4O2) is (Assume that Kb (C8H10N4O2) = 4.10 × 10-4.)
BUY
Chemistry
10th Edition
ISBN:
9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
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1 Chemical Foundations
2 Atoms, Molecules, And Ions
3 Stoichiometry
4 Types Of Chemical Reactions And Solution Stoichiometry
5 Gases
6 Thermochemistry
7 Atomic Structure And Periodicity
8 Bonding: General Concepts
9 Covalent Bonding: Orbitals
10 Liquids And Solids
11 Properties Of Solutions
12 Chemical Kinetics
13 Chemical Equilibrium
14 Acids And Bases
15 Acid-base Equilibria
16 Solubility And Complex Ion Equilibria
17 Spontaneity, Entropy, And Free Energy
18 Electrochemistry
19 The Nucleus: A Chemist's View
20 The Representative Elements
21 Transition Metals And Coordination Chemistry
22 Organic And Biological Molecules
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Transcribed Image Text:
The pOH of an aqueous solution of 0.532 M caffeine (a weak base with the formula \( \text{C}_8\text{H}_{10}\text{N}_4\text{O}_2 \)) is \(\_\_\_\_\_\_\_\_\_\_\_\_\_\). (Assume that \( K_b (\text{C}_8\text{H}_{10}\text{N}_4\text{O}_2) = 4.10 \times 10^{-4} \).)
Transcribed Image Text:
The text reads: "The pH of an aqueous solution of 0.0959 M ammonium perchlorate, \( \text{NH}_4\text{ClO}_4 \, (\text{aq}), \) is [_____]. (Assume that \( K_b(\text{NH}_3) = 1.80 \times 10^{-5}. \)) This solution is [dropdown menu]." Explanation: - The question asks for the pH calculation of a 0.0959 M ammonium perchlorate solution. - It provides the base dissociation constant (\( K_b \)) for ammonia (\( \text{NH}_3 \)) as \( 1.80 \times 10^{-5} \). - There is a blank space to fill in the pH value. - There is a dropdown menu indicating the nature of the solution (likely options such as acidic, neutral, or basic).
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