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1) A mixture of 5.1 x 10-3 moles of HSO4 (hydrogen sulfate) is dissolved in 25 mL of water. At equilibrium the vessel contains SO4² (sulfate) and H3O+ (hydronium). What is the pH at equilibrium? HSO4 (aq) + H2O(l) = SO4²¯(aq) + H3O+(aq) Ka = 1.2 x 10-2

1) A mixture of 5.1 x 10-3 moles of HSO4 (hydrogen sulfate) is dissolved in 25 mL of water. At equilibrium the vessel contains SO4² (sulfate) and H3O+ (hydronium). What is the pH at equilibrium? HSO4 (aq) + H2O(l) = SO4²¯(aq) + H3O+(aq) Ka = 1.2 x 10-2

Principles of Modern Chemistry
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter15: Acid–base Equilibria
Section: Chapter Questions
Problem 82AP
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Give the explanation formula which using for the calculation.

1) A mixture of 5.1 x 10-3 moles of HSO4 (hydrogen sulfate) is dissolved in 25 mL of water. At equilibrium the vessel contains SO4² (sulfate) and H3O+ (hydronium). What is the pH at equilibrium? HSO4 (aq) + H2O(l) = SO4²¯(aq) + H3O+(aq) Ka = 1.2 x 10-2

1) A mixture of 5.1 x 10-3 moles of HSO4 (hydrogen sulfate) is dissolved in 25 mL of water. At equilibrium the vessel contains SO4² (sulfate) and H3O+ (hydronium). What is the pH at equilibrium? HSO4 (aq) + H2O(l) = SO4²¯(aq) + H3O+(aq) Ka = 1.2 x 10-2
Transcribed Image Text:1) A mixture of 5.1 x 10-3 moles of HSO4 (hydrogen sulfate) is dissolved in 25 mL of water. At equilibrium the vessel contains SO4² (sulfate) and H3O+ (hydronium). What is the pH at equilibrium? HSO4 (aq) + H2O(l) = SO4²¯(aq) + H3O+(aq) Ka = 1.2 x 10-2
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