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Solutions CH 13

Solutions CH 13. Two Types of Mixtures Homogeneous Same throughout, looks pure EX: Air Heterogeneous Different throughout EX: Sand

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Solutions

CH 13

Two Types of Mixtures Homogeneous

Same throughout, looks pure EX: Air

Heterogeneous Different throughout EX: Sand

A solution is any substance—solid, liquid, or gas—that is evenly distributed throughout another substance.

Solutions are homogeneous mixtures

EX: Brass which is a mixture of copper and zinc

A Solution contains… Solvent—the dissolving

medium Higher concentration EX: alcohol and water Water is the universal solvent!

Solute—being dissolved Lower concentration

Example

You are making Kool-Aid. What is the solute and what is the solvent?

The same substance can be a solute in one instance and a solvent in another.

EX: 15% alcohol solution: Water is the solvent and alcohol is the

solute EX: 85% alcohol solution:

Alcohol is the solvent and water is the solute

Types of Solutions

Suspension A mixture in which particles of

a material are more or less evenly dispersed throughout a liquid or gas.

EX: mud—will settle out over time

Large particles

Colloid A mixture consisting of tiny

particles that are intermediate in size between those in solution that are suspended in a liquid, solid, or gas.

EX: milk—particles do not settle out over time.

Alloys Two or more solid metals

mixed together EX: Steel, brass

Separating Mixtures

Decanting

Pouring off the liquid from a settled suspension

EX: mud

Centrifuge Separates substances by

their densities EX: milk—separates fat

out blood

Filtration Liquid goes through a filter and

the solid is left behind EX: ground coffee

Evaporation Liquid turns into a gas and the

solid is left behind EX: salt water

Chromatography Separates different dyes by

choosing the correct solvent

Distillation Separating liquids by

boiling points

Electrolytes Solutions that

conduct electricity Compounds EX: NaCl

Solvation The process of dissolving a

solute in a solvent Hydration—ions that are

surrounded by water molecules Dissociation—the

decomposition of a crystal into hydrated ions EX: NaCl Na+(aq) + Cl-(aq)

Polar molecules solvate! Attractions between polar

molecules may draw the solution components into a smaller volume

EX: 50 mL ethanol + 50 mL of water =

<100mL

Miscible—no apparent limit to the solubility of one substance in another

Immiscible—two liquids that don’t mix

Like dissolves like!

Solubility—the amount of substance needed to make a saturated solution at a given temperature.

EX: We use glass in lab because of its low solubility.

Factors that affect solubility

Liquids: T increases , solubility increases EX: rock candy

Supersaturated solution—add heat to make dissolve

Gases: T increases, solubility decreases EX: Coke and CO2

P increases, solubility increases Surface area increases, rate of

dissolving increases Agitating increases, rate of

dissolving increases T increases, rate of dissolving

increases

Henry’s Law States that at constant T, the

solubility of a gas in a liquid is directly proportional to the partial pressure of the gas on the surface of the liquid.

EX: Coke can S1 = S2

P1 P2

Saturated solution—contains the maximum amount of solute for a given amount of solvent at a constant T

Supersaturated solution—a solution contains more solute than it should theoretically at a given T

Concentrations

Concentration—how much of a substance a solution contains

ppm—parts per million Dilute solutions—contains low

concentration of solute Concentrated solutions—

contains high concentrations of solute

Molarity Molarity—the number of moles

solute dissolved in each liter of solution. (M)

M = mol / L**You have 500 mL of a 4M HCl

sol’n. You take out 100 mL, you still have a 4M sol’n.

Example 1 How many moles of

HCl are contained in 1.45 L of a 2.25M solution?

Solution 1 M = mol / L mol = M x L mol = 2.25 M x 1.45 L mol = 3.26 mol HCl

Example 2

Give the directions for the preparation of 2.50 L of a 1.34 M NaCl solution. (HINT: You need to find the number of g NaCl needed)

Solution 2 Step 1: Calculate the number of

moles: mol=1.34 M x 2.5 L = 3.35 mol NaCl

Step 2: Convert moles to grams 3.35 mol NaCl x 58.5 g NaCl = 196 g

1 mol NaCl

Dilutions Make solutions less concentrated by

diluting it with a solvent Since only the solvent changes, the

number of moles stays the same Moles solute before dilution = moles

solute after dilution Moles of solute = M x V (liters of

solution) So… M1V1 = M2V2