Ch.15 - Acid and Base EquilibriumWorksheetSee all chapters
All Chapters
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
Problem

At 40 °C, the value of Kw = 2.92 × 10–14. An acidic solution at 40 °C will therefore have:

A. [H3O+] > [OH] and [H3O+] = 1.70 × 10–7 M

B. [H3O+] > [OH] and [H3O+] > 1.70 × 10–7 M

C [H3O+] > [OH] and [H3O+] = 2.96 × 10–7 M

D. [H3O+] = [OH] and [OH] = 1.70 × 10–7 M

E. [H3O+] < [OH] and [OH] > 1.70 × 10–7 M

Solution

We’re being asked to determine [H3O+] and [OH-] of an acidic solution a40 °C if the value of Kw is 2.92 × 10–14.



Recall that Kw is the autoionization constant of water, and is given by:


Kw=H3O+OH-


Solution BlurView Complete Written Solution