Ch.14 - Chemical 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

Solution: Chromium(VI) forms two different oxyanions, the orange dichromate ion, Cr 2O72-, and the yellow chromate ion, CrO42-. (See the following photos.) The equilibrium reaction between the two ions isCr2O7 2-(aq) + H2O(l) ⇌ 2CrO4 2-(aq) + 2H+(aq)Explain why orange dichromate solutions turn yellow when sodium hydroxide is added.

Problem

Chromium(VI) forms two different oxyanions, the orange dichromate ion, Cr 2O72-, and the yellow chromate ion, CrO42-. (See the following photos.) The equilibrium reaction between the two ions is

Cr2O7 2-(aq) + H2O(l) ⇌ 2CrO4 2-(aq) + 2H+(aq)

Explain why orange dichromate solutions turn yellow when sodium hydroxide is added.