I. The Nature of Mixtures

Mixtures Heterogeneous Mixture – a non-uniform or unequal blend of two or more pure substances Suspension – mixture containing particles that will settle out Colloid – mixture containing particles with a size of 1nm – 1000nm, and do not separate – stay suspended

Mixtures Tyndall Effect – Caused by dilute colloids, which appear to be homogeneous. Is the scattering of light as it passes though a dilute colloid

Mixtures Solutions - homogeneous mixtures Solute - substance being dissolved Solvent - present in greater amount

Homogeneous Mixtures A substance that dissolves in a solvent is soluble. Two liquids that are soluble in each other in any proportion are miscible. A substance that does not dissolve in a solvent is insoluble. Two liquids that can be mixed but separate shortly after are immiscible.

Solutions Solvation – the process of dissolving – sugar dissolves in to water solute particles are surrounded by solvent particles First... solute particles are separated and pulled into solution http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html Then...

Solvation Non- Electrolyte Weak Electrolyte Strong Electrolyte + sugar - + acetic acid - + salt Non- Electrolyte Weak Electrolyte Strong Electrolyte solute exists as molecules only solute exists as ions and molecules solute exists as ions only

Solvation C6H12O6(s)  C6H12O6(aq) Molecular Solvation molecules stay intact C6H12O6(s)  C6H12O6(aq)

NaCl(s)  Na+(aq) + Cl–(aq) Solvation Dissociation separation of an ionic solid into aqueous ions http://chemmovies.unl.edu/ChemAnime/NACL1D/NACL1D.html NaCl(s)  Na+(aq) + Cl–(aq)

Solvation “Like Dissolves Like” NONPOLAR POLAR

Solvation Heat of Solution – energy change that occurs during the formation of the solution. Exothermic – solvation produces a warm solution Endothermic – solvation produces a cold solution

Factors Affecting Solvation (solubility) Agitation – stirring or shaking – allows for more collisions (mixing) between solute and solvent Surface Area – smaller pieces – more collisions Temperature – in general solids dissolve faster in higher temps. – more collisions Opposite for gases – colder the better

Solubility Solubility maximum grams of solute that will dissolve in 100 g of solvent at a given temperature varies with temp based on a saturated soln Book Reference: p 969 R-3 and p 974 R-8 http://chemmovies.unl.edu/ChemAnime/SLBIL1D/SLBIL1D.html

Solubility Rules In general: Group I ions and Ammonium are soluble Acetates and Nitrates are soluble Cl, Br, I are soluble, except with Pb, Ag, Hg2+2 Sulfates are soluble, except with Ba, Sr, Pb, Ca, Ag, Hg2+2 Carbonates, Hydroxides, oxides, sulfides, phosphates are INsoluble

Solubility UNSATURATED SOLUTION more solute dissolves no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

Solubility Solubility Curve shows the dependence of solubility on temperature

Solubility Solids are more soluble at... high temperatures. Gases are more soluble at... low temperatures & high pressures (Henry’s Law).

Solubility Types of Reactions in Aqueous Solutions When two solutions that contain ions as solutes are combined, the ions might react. If they react, it is always a double replacement reaction. Three products can form: precipitates, water, or gases.

Solubility Replacement Reactions Double Replacement Reactions occur when ions exchange between two compounds. This figure shows a generic double replacement equation.

Solubility Aqueous solutions of sodium hydroxide and copper(II) chloride react to form the precipitate copper(II) hydroxide. 2NaOH(aq) + CuCl2(aq) → 2NaCl(aq) + Cu(OH)2(s) Ionic equations that show all of the particles in a solution as they actually exist are called complete ionic equations. 2Na+(aq) + 2OH–(aq) + Cu2+ (aq)+ 2Cl–(aq) → 2Na+(aq) + 2Cl–(aq) + Cu(OH)2(s)

Types of Reactions in Aqueous Solutions (cont.) Ions that do not participate in a reaction are called spectator ions and are not usually written in ionic equations. Formulas that include only the particles that participate in reactions are called net ionic equations. 2OH–(aq) + Cu2+(aq) → Cu(OH)2(s)

Solutions II. Concentration

A. Concentration is... a measure of the amount of solute dissolved in a given quantity of solvent A concentrated solution has a large amount of solute A dilute solution has a small amount of solute thus, only qualitative descriptions

A. Concentration Describing Concentration Quantitatively % by mass % by volume Molarity Molality Mole Fraction

B. Percent

B. Percent What is the percent solution if 25 mL of CH3OH is diluted to 150 mL with water? 17% 4.8 g of NaCl are dissolved in 82.0 mL of solution. What is the percent of the solution? 5.53% How many grams of salt are there in 52 mL of a 6.3 % solution? 3.3g

C. Molarity Volume of solvent only 1 kg water = 1 L water

C. Molarity Find the molarity of a solution containing 75 g of MgCl2 in 250 mL of water. 75 g MgCl2 1 mol MgCl2 95.21 g MgCl2 0.25 L water = 3.2 M MgCl2

C. Molarity How many grams of NaCl are req’d to make a 1.54 M solution using 0.500 L of water? 0.500 L water 1.54 mol NaCl 1 L water 58.44 g NaCl 1 mol NaCl = 45.0 g NaCl

D. Molality 1 L water = 1 Kg water so … Molality (m) = moles of solute / Kg of solvent Basically the same calculations, but different units

E. Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

E. Dilution What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M) V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

F. Preparing Solutions 1.54m NaCl in 0.500 kg of water 500 mL of 1.54M NaCl mass 45.0 g of NaCl add water until total volume is 500 mL mass 45.0 g of NaCl add 0.500 kg of water 500 mL water 45.0 g NaCl 500 mL mark volumetric flask

F. Preparing Solutions 250 mL of 6.0M HNO3 by dilution measure 95 mL of 15.8M HNO3 95 mL of 15.8M HNO3 combine with water until total volume is 250 mL 250 mL mark Safety: “Do as you oughtta, add the acid to the watta!” water for safety

Solution Preparation Lab Turn in one paper per team. Complete the following steps: A) Show the necessary calculations. B) Write out directions for preparing the solution. C) Prepare the solution. For each of the following solutions: 1) 100.0 mL of 0.500M NaCl 2) 0.250m NaCl in 100.0 mL of water 3) 100.0 mL of 1.00M Red Solution from 12.1M concentrate.

III. Colligative Properties Solutions III. Colligative Properties

A. Definition Colligative Property property that depends on the concentration of solute particles, not their identity

A. Electrolytes Dissolved Solute particles disrupt the normal Intermolecular Forces of the Solvent Molecules – Count as one dissolved solute particle Example: CH3OH = 1 solute particle Ionic Compounds – number of solute particles is equal to the total number of ions in the neutral formula Example: AlCl3 = 4 solute particles 1 Al and 3 Cl ions

B. Types Vapor Pressure Lowering Solutions have a lower vapor pressure than the original pure solvent Because the solute particles are attracted to solvent particles cause more IMF So more IMF = less solvent particles becoming vapors

B. Types Freezing Point Depression (tf) f.p. of a solution is lower than f.p. of the pure solvent Boiling Point Elevation (tb) b.p. of a solution is higher than b.p. of the pure solvent

Freezing Point Depression B. Types Freezing Point Depression http://chemmovies.unl.edu/ChemAnime/SOLND/SOLND.html http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html

Boiling Point Elevation B. Types Boiling Point Elevation Solute particles increase IMF of the solvent.

C. Phase Diagram

D. Applications salting icy roads making ice cream Antifreeze - cars (-64°C to 136°C)

E. Calculations  t: change in temperature (°C) t = k · m · n  t: change in temperature (°C) k: constant based on the solvent (°C·kg/mol) m: molality (m) n or i: # of particles

E. Calculations # of Particles Nonelectrolytes (covalent) remain intact when dissolved 1 particle Electrolytes (ionic) dissociate into ions when dissolved 2 or more particles

E. Calculations At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil? GIVEN: b.p. = ? tb = ? kb = 3.56°C·kg/mol WORK: m = 0.73mol ÷ 0.225kg tb = (3.56°C·kg/mol)(3.2m)(1) tb = 11°C b.p. = 181.8°C + 11°C b.p. = 193°C m = 3.2m n = 1 tb = kb · m · n

E. Calculations Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water. GIVEN: f.p. = ? tf = ? kf = 1.86°C·kg/mol WORK: m = 0.48mol ÷ 0.100kg tf = (1.86°C·kg/mol)(4.8m)(2) tf = 18°C f.p. = 0.00°C - 18°C f.p. = -18°C m = 4.8m n = 2 tf = kf · m · n