Ch.12 - SolutionsWorksheetSee all chapters
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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: The Henry’s law constant for oxygen gas in water at 25 °C, kH is 1.3×10-3M/atm. What is the partial pressure of O2 above a solution at 25 °C with an O2 concentration of 2.3×10-4 M at equilibrium?A. 0.18 atm B. 5.7 atmC. 1.3 x 10-3 atmD. 2.3 x 10-4 atmE. 3.0 x 10-7 atm

Problem

The Henry’s law constant for oxygen gas in water at 25 °C, kH is 1.3×10-3M/atm. What is the partial pressure of O2 above a solution at 25 °C with an O2 concentration of 2.3×10-4 M at equilibrium?

A. 0.18 atm 

B. 5.7 atm

C. 1.3 x 10-3 atm

D. 2.3 x 10-4 atm

E. 3.0 x 10-7 atm

Solution

We’re being asked to calculate the partial pressure of O2 above a solution at 25 °C


Recall that the solubility of a gas is given by Henry’s law:


Sgas=kH·Pgas


where:

 Sgas = solubility of the gas (in mol/L or M)

kH = Henry’s law constant for the gas

Pgas = partial pressure of the gas


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