Pages 466-486 Notes One Unit Six– Chapter 13 Solutions Definitions Types of Mixtures Example Solutions Factors Affecting Solubility Like Dissolves Like Solubility of Solids Changes with Temperature Solubility of Gases Changes with Temperature Pressure Factor Molar Concentration Finding Molarity From Mass and Volume Finding Mass from Molarity and Volume Finding Volume from Molarity and Mass
Definitions • Solutions are homogeneous mixtures. • Uniform throughout. • Solvent. • Determines the state of solution • Largest component • Solute. • Dissolved in solvent
Common Mixtures SOLUTE EXAMPLE SOLVENT Type liquid mayonnaise liquid emulsion gas whipped cream liquid liquid foam solid dust in air gas aerosol liquid hair spray gas aerosol solid ruby glass solid solid liquid pearl solid emulsion gas Styrofoam solid solid foam
Solution Types SOLUTE EXAMPLE SOLVENT PHASE gas a i r gas gas gas soda pop liquid liquid liquid antifreeze liquid liquid liquid filling solid solid solid seawater liquid liquid solid brass solid solid
Factors Affecting Solubility • 1. Nature of Solute / Solvent. • 2. Temperature Increase • i) Solid/Liquid • ii) gas • 3. Pressure Factor - • i) Solids/Liquids - Very little • ii) gas • iii) squeezes gas into solution.
Like Dissolves Like • Non-polar in Non-polar • Butter in Oil • Non-polar in polar • Oil in H2O • Polar in Polar • C2H5OH in H2O • Ionic compounds in polar solvents • NaCl in H2O
Solubility of solids Changes with Temperature • How does the solubility Δ with temperature Increase? • How many grams of potassium chromate will dissolve in100g water at 70oC? • 70g • How many grams of lead(II) nitrate will precipitate from 250g water cooling from 70oC to 50oC? solids gases 101g 82g 250g _____ 19gx =48g 100g
Solubility of Gases Changes with Temperature • a) Why are fish stressed, if the temperature of the water increases? • How much does the solubility of oxygen change, for a 20oC to 60oC change? • 0.90-0.60=0.30mg 0.90mg 0.60mg
Pressure Factor Greater pressure… more dissolved gas
Molar Concentration • M=n/V • n=MxV • V=n/M
Finding Molarity From Mass and Volume • Calculate molarity for 25.5 g of NH3 in 600. mL solution. • 1) Calculate Formula Mass: • 2) Calculate the moles of solute: • 3) Calculate the Moles/Liters Ratio E Mass # N 14.0 1x 14.0 = 3.0 H 3x 1.0 = 17.0g/m 25.5g ÷ 17.0g/m= 1.50m M = 1.50m 0.600L / 2.52 mol/L M =
Finding Volume from Molarity and Mass • How many milliliters of 2.50M solution can be made using 25.5grams of NH3? • 1)Calculate formula mass: • 2)Calculate the moles of solute: • 3)Calculate Volume: • V= Mass E # 14.0 N 1x 14.0 = 3.0 H 3x 1.0 = 17.0g/m 25.5g ÷ 17.0g/m= 1.50m V=n/M V= (1.50m) / (2.50M) 0.600L solution 600.mL
Finding Mass from Molarity and Volume • How many grams of NH3 are in 600. mL solution at 2.50M? • 1) Calculate formula mass: • 2) Calculate moles • n=1.50m • 3) Calculate mass • g=25.5g NH3 Mass E # 14.0 N 1x 14.0 = 3.0 H 3x 1.0 = 17.0g/m g ÷ fm= mol n = x L M x n= 2.50M 0.600L x g = n fm x g= (17.0g/m) (1.50m)
Notes Two Unit Six– Chapter 13 Solutions Pages 487-501 • Saturated versus Unsaturated • Colligative properties of water • Forming a Saturated Solution • How Does a Solution Form? • Colligative Properties • Vapor Pressure • Boiling and Freezing Point • BP Elevation and Freezing FP Depression • Calculating Freezing Point Depression Mass
Characteristics of Saturated Solutions water precipitate precipitate dissolve dissolve dissolve Solid Unsaturated Saturated Unsaturated Dynamic Equilibrium Cooling causes precipitation. Warming causes dissolving.
Solvation • As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.
Colligative Properties • Colligative properties depend on moles dissolved particles. • Vapor pressure lowering • Boiling point elevation • Melting point depression • Osmotic pressure
Vapor Pressure Lowering • The particles of solute are surrounded by and attracted to particles of solvent. • Now the solvent particles have less kinetic energy and tend less to escape into the space above the liquid. • So the vapor pressure is less.
Ionic vs Molecular Solutes • Ionic solutes produce two or more ion particles in solution. • They affect the colligative properties proportionately more than molecular solutes (that do not ionize). • The effect is proportional to the number of particles of the solute in the solution.
How many particles do each of the following give upon solvation? • NaCl • CaCl2 • Glucose
Example • Salt is added to melt ice by reducing the freezing point of water.
Example • Addition of ethylene glycol C2H6O2 (antifreeze) to car radiators.
Freezing Point Depression and Boiling Point Elevation Boiling Point Elevation • ∆Tb =imkb (for water kb=0.51 oC/m) • Freezing Point Depression • ∆Tf=imkf (for water kf=1.86 oC/m) • Note: m is the molality of the particles, so if the solute is ionic, multiply by the #of particles it dissociates to. (i is # of particles called van hoff factor)
Which is more effective for lowering the freezing point of water? • NaCl or CaCl2
Example 1: • Find the new freezing point of 3m NaCl in water.
Example 2: • Find the new boiling point of 3m NaCl in water.
Molarity versus Molality moles of solute ________________ Molality (m) = kilograms solvent moles of solute ________________ Molarity (M) = liters of solution
Calculating Tf andTb • Calculate the freezing and boiling points of a solution made using 1000.g antifreeze (C2H6O2) in 4450g water. • 1) Calculate Moles • 2) Calculate molality • 3) Calculate Temperature Change • Δt=Kxm • ΔTf = • Tf = • ΔTb = • Tb = Mass E # 24.0 C 2x 12.0 = 1000.g ÷ 62.0g/mol = 16.1 moles 32.0 O 2x 16.0 = 6.0 H 6x 1.0 = 16.1 mole ÷ 4.45 Kg water = 62.0g/m 3.62m (1.858oC/m) (3.62 m) = 6.73oC 0.000oC- 6.73oC= -6.73oC (0.512oC/m) (3.62 m) = 1.96oC 100.000oC + 1.96oC = 101.96oC
Calculating Boiling Point Elevation Mass • A solution containing 18.00 g of glucose in 150.0 g of water boils at 100.34oC. Calculate the molecular weight of glucose. • 1.)Calculate Temperature Change • ΔTb = • 2.)Calculate moles per Kilograms • ΔTb = Kb x m m = ΔTb /Kb • m =0.67m/kg • 3.)Calculate grams / kilograms • g = • g =120 g/kg • MW=120 g/0.67m • 180g/m 100.34oC- 100.00oC= 0.34oC 0.34÷ 0.512oC/m= m 18.00 g ÷ 0.1500kg
One Molal Solution of Water solid 1 atm Liquid Pressure gas Kb Kf Temperature 0.512oC 1.858oC
Notes Three Unit Six • Ice-cream Lab A Calculating Freezing Point • Depression Mass • Colligative Properties of Electrolytes • Distillation • Osmotic Pressure • Dialysis Pages 487-501
Calculating Freezing Point Depression Mass • 1.)Calculate Temperature Change • ΔTf = • 2.)Calculate moles per Kilograms • ΔTf = Kfx m m = ΔTf /Kf • m =1.83mol/kg • 3.)Calculate grams / kilograms • g = • g =36.4 g/kg • MW=36.4 g/1.83m • 19.9g/m 1.1oC- (-2.3oC)= 3.4oC 3.4oC÷ 1.858oC/m = m 0.05196kg 1.89 g ÷
Colligative Properties of Electrolytes • Colligative properties depend on the number of particles dissolved. • NaClNa+1+Cl-1 CH3OHAl2(SO4)32Al+3 + 3SO4-2 C6H12O6
Osmotic Pressure • Hypertonic • > 0.92% (9.g/L) • Crenation • Isotonic Saline • = 0.92% (9.g/L) • Hypotonic • < 0.92% (9.g/L) • Rupture
Finding Molarity From Mass and Volume • Calculate molarity for 14.0 g of sodium peroxide(Na2O2) in 615 mL solution. • 1) Calculate Formula Mass: • 2) Calculate the moles of solute: • 3) Calculate the Moles/Liters Ratio Mass E # 46.0 Na 2x 23.0 = 32.0 O 2x 16.0 = 78.0g/m 14.0g ÷ 78.0g/m= 0.179m 0.289M 0. 179moles ÷ 0.615L = M = n/ v M =