Ch.15 - Acid and Base EquilibriumWorksheetSee 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: At body temperature of 37.0°C, K w = 2.5 x 10-14. Calculate the [H3O+] if the [OH -] = 3.9 x 10-3 M.A. 2.5 x 10-12 MB. 6.4 x 10-12 MC. 3.9 x 10-3 MD. 1.0 x 10-7 ME. 2.5 x 10-14 M

Solution: At body temperature of 37.0°C, K w = 2.5 x 10-14. Calculate the [H3O+] if the [OH -] = 3.9 x 10-3 M.A. 2.5 x 10-12 MB. 6.4 x 10-12 MC. 3.9 x 10-3 MD. 1.0 x 10-7 ME. 2.5 x 10-14 M

Problem

At body temperature of 37.0°C, K w = 2.5 x 10-14. Calculate the [H3O+] if the [OH -] = 3.9 x 10-3 M.

A. 2.5 x 10-12 M

B. 6.4 x 10-12 M

C. 3.9 x 10-3 M

D. 1.0 x 10-7 M

E. 2.5 x 10-14 M

Solution

We’re being asked to determine the H3O+ concentration if [OH] = 3.9 × 10–3 M. We’re given Kw = 2.5 × 10–14 at 37.0 ˚C.


Recall that Kw is the auto-ionization product of water. For the following reaction:


2 H2O(l)  H3O+(aq) + OH(aq)


the auto-ionization product constant is given by:



Note that each concentration is raised by the stoichiometric coefficient: [H3O+] and [OH] are raised to 1.


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