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: Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving 156 g of the salt in 100 mL of water at 100 oC and then slowly cooling it to 20 oC. Becau

Solution: Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving 156 g of the salt in 100 mL of water at 100 oC and then slowly cooling it to 20 oC. Becau

Problem

(1) A seed crystal of sodium acetate is added to supersaturated solution. The amount of sodium acetate dissolved is greater than its solubility at this temperature, and a white substance is visible in a beaker holding a clear liquid in a photograph. (2) Excess sodium acetate starts to crystallize from solution. A photograph shows a substance with many white threadlike extensions at the bottom of the liquid and extending down from the surface of the liquid. (3) Solution is saturated and excess solute has crystallized. A photograph shows a crystal filling most of the beaker.

Precipitation from a supersaturated sodium acetate solution. The solution on the left was formed by dissolving 156 g of the salt in 100 mL of water at 100 oC and then slowly cooling it to 20 oC. Because the solubility of sodium acetate in water at 20 oC is 46 g per 100 mL of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess solute to crystallize from solution.

How many grams of sodium acetate are in solution in the third beaker?

Solution

In this problem, we are asked to find the grams of sodium acetate that are in solution in the third beaker in the figure above.

A solution is a mixture of two or more substances.

  • It can be classified as saturated, unsaturated or supersaturated
    • In a Saturated solution, the solvent has reached its maximum amount of dissolved solute.
    • In an Unsaturated solution, additional amounts of solute can be further dissolved in the solvent.
    • In a Supersaturated solution, the solvent has dissolved beyond its maximum solubility.

Solubility is the ability of a given substance, a solute, to be dissolved in a solvent.


In solutions, there are only two possible reactions of solute to the solvent, that is crystallization and dilution. 

  • Crystallization is the process of formation of solid crystals from solution.
  • Dilution is the process of dissolving (solid) solute into the solvent. 

The rate of which is different depending on the type of solution.

  • In a saturated solution, the rate of crystallization = the rate of dilution.
  • In an unsaturated solution, the rate of crystallization < the rate of dilution.
  • In a supersaturated solution, the rate of crystallization > the rate of dilution.

To solve this problem, we must analyze the process of the crystallization of the salt in the solution.


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