Solutions Properties of Water Solutions 1 Predict the % water in the following foods 2 Predict the % water in the following foods 88% water 94% water 85% water 86% water 3 Water in the Body water gain liquids 1000 mL food 1200 mL cells 300 mL water loss urine 1500 mL perspiring 300 mL exhaling 600 mL feces 100 mL Calculate the total water gain and water loss Total ______ mL _____ mL 4 Water Most common solvent A polar molecule O a hydrogen bond H + H + 5 Hydrogen Bonds Attract Polar Water Molecules 6 Explore: Surface Tension HW Fill a glass to the brim with water How many pennies can you add to the glass without causing any water to run over? Predict _________________ Actual _________________ Explain your results 7 Explore 1. Place some water on a waxy surface. Why do drops form? 2. Carefully place a needle on the surface of water. Why does it float? What happens if you push it through the water surface? 3. Sprinkle pepper on water. What does it do? Add a drop of soap. What happens? 8 Surface Tension Water molecules within water hydrogen bond in all directions Water molecules at surface cannot hydrogen bond above the surface, pulled inward Water surface behaves like a thin, elastic membrane or “skin” Surfactants (detergents) undo hydrogen bonding 9 Solute and Solvent Solutions are homogeneous mixtures of two or more substances Solute The substance in the lesser amount Solvent The substance in the greater amount 10 Nature of Solutes in Solutions Spread evenly throughout the solution Cannot be separated by filtration Can be separated by evaporation Not visible, solution appears transparent May give a color to the solution 11 Types of Solutions air O2 gas and N2 gas gas/gas soda CO2 gas in water gas/liquid seawater NaCl in water solid/liquid brass copper and zinc solid/solid 12 Discussion Give examples of some solutions and explain why they are solutions. 13 Learning Check SF1 (1) element (2) compound (3) solution A. water 1 2 3 B. sugar 1 2 3 C. salt water 1 2 3 D. air 1 2 3 E. tea 1 2 3 14 Solution SF1 (1) element (2) compound A. water 2 B. sugar 2 C. salt water 3 D. air 3 E. tea 3 (3) solution 15 Learning Check SF2 Identify the solute and the solvent. A. brass: 20 g zinc + 50 g copper solute = 1) zinc 2) copper solvent = 1) zinc 2) copper B. 100 g H2O + 5 g KCl solute = 1) KCl 2) H2O solvent = 1) KCl 2) H2O 16 Solution SF2 A. brass: 20 g zinc + 50 g copper solute solvent = = 1) zinc 2) copper B. 100 g H2O + 5 g KCl solute = 1) KCl solvent = 2) H2O 17 Learning Check SF3 Identify the solute in each of the following solutions: A. 2 g sugar (1) + 100 mL water (2) B. 60.0 mL ethyl alcohol(1) and 30.0 mL of methyl alcohol (2) C. 55.0 mL water (1) and 1.50 g NaCl (2) D. Air: 200 mL O2 (1) + 800 mL N2 (2) 18 Solution SF3 Identify the solute in each of the following solutions: A. 2 g sugar (1) B. 30.0 mL of methyl alcohol (2) C. 1.5 g NaCl (2) D. 200 mL O2 (1) 19 Like dissolves like A ____________ solvent such as water is needed to dissolve polar solutes such as sugar and ionic solutes such as NaCl. A ___________solvent such as hexane (C6H14) is needed to dissolve nonpolar solutes such as oil or grease. 20 Learning Check SF4 Which of the following solutes will dissolve in water? Why? 1) Na2SO4 2) gasoline 3) I2 4) HCl 21 Solution SF4 Which of the following solutes will dissolve in water? Why? 1) Na2SO4 Yes, polar (ionic) 2) gasoline No, nonnpolar 3) I2 No, nonpolar 4) HCl Yes, Polar 22 Formation of a Solution H2O Cl- Na+ Na+ Clsolute Na+ H2O Cl- Hydration Dissolved solute 23 Electrolyte and Non-electrolyte • Electrolyte: a substance that conducts electricity when dissolved in water. – Acids, bases and soluble ionic solutions are electrolytes. • Non-electrolyte: a substance that does not conduct electricity when dissolved in water. – Molecular compounds and insoluble ionic compounds are non-electrolytes. 24 Electrolytes • Some solutes can dissociate into ions. • Electric charge can be carried. 25 Types of solutes high conductivity Strong Electrolyte 100% dissociation, all ions in solution Na+ Cl26 Types of solutes slight conductivity Weak Electrolyte partial dissociation, molecules and ions in solution CH3COOH H+ CH3COO27 Types of solutes no conductivity Non-electrolyte No dissociation, all molecules in solution Sugar C6H12O6 28 Types of Electrolytes • Strong electrolyte dissociates completely. – Good electrical conduction. • Weak electrolyte partially dissociates. – Fair conductor of electricity. • Non-electrolyte does not dissociate. – Poor conductor of electricity. 29 Representation of Electrolytes using Chemical Equations A strong electrolyte: MgCl2(s) → Mg2+(aq) + 2 Cl- (aq) A weak electrolyte: -(aq) +H+(aq) CH3COOH(aq) → CH COO ← 3 A non-electrolyte: CH3OH(aq) 30 Electrolytes in Action 31 Strong Electrolytes Strong acids: HNO3, H2SO4, HCl, HClO4 Strong bases: MOH (M = Na, K, Cs, Rb etc) Salts: All salts dissolving in water are completely ionized. Stoichiometry & concentration relationship NaCl (s) Na+ (aq) + Cl– (aq) Ca(OH)2 (s) Ca2+(aq) + 2 OH– (aq) AlCl3 (s) Al3+ (aq) + 3 Cl– (aq) (NH4)2SO4 (s) 2 NH4 + (aq) + SO42– (aq) 32 Writing An Equation for a Solution When NaCl(s) dissolves in water, the reaction can be written as H 2O NaCl(s) solid Na+ (aq) + Cl- (aq) separation of ions in water 33 Learning Check SF5 Solid LiCl is added to some water. It dissolves because A. The Li+ ions are attracted to the 1) oxygen atom(-) of water 2) hydrogen atom(+) of water B. The Cl- ions are attracted to the 1) oxygen atom(-) of water 2) hydrogen atom(+) of water 34 Solution SF5 Solid LiCl is added to some water. It dissolves because A. The Li+ ions are attracted to the 1) oxygen atom(-) of water B. The Cl- ions are attracted to the 2) hydrogen atom(+) of water 35 Rate of Solution You are making a chicken broth using a bouillon cube. What are some things you can do to make it dissolve faster? Crush it Use hot water (increase temperature) Stir it 36 How do I get sugar to dissolve faster in my iced tea? Stir, and stir, and stir Fresh solvent contact and interaction with solute Add sugar to warm tea then add ice Faster rate of dissolution at higher temperature Grind the sugar to a powder Greater surface area, more solute-solvent interaction 37 Learning Check SF6 You need to dissolve some gelatin in water. Indicate the effect of each of the following on the rate at which the gelatin dissolves as (1) increase, (2) decrease, (3) no change A. ___Heating the water B. ___Using large pieces of gelatin C. ___Stirring the solution 38 Learning Check SF6 You need to dissolve some gelatin in water. Indicate the effect of each of the following on the rate at which the gelatin dissolves as (1) increase, (2) decrease, (3) no change A. 1 Heating the water B. 2 Using large pieces of gelatin C. 2 Stirring the solution 39 Solubility Percent Concentration Colloids and Suspensions 40 Solubility The maximum amount of solute that can dissolve in a specific amount of solvent usually 100 g. g of solute 100 g water 41 Saturated and Unsaturated A saturated solution contains the maximum amount of solute that can dissolve. Undissolved solute remains. An unsaturated solution does not contain all the solute that could dissolve 42 Solubility UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form increasing concentration Factors Affecting Solid Solubility Polarity Temperature Surface Area Stirring Factors Affecting Solubility Polarity Temperature Pressure Intramolecular Bonding • Intramolecular bonding refers to the chemical bonding that holds atoms together within a molecule of a compound Covalent bonding and ionic bonding are the two main types of intramolecular bonding Covalent bonding involves the sharing of valence electrons involves the sharing of valence electrons between two atoms. POLAR- unequal sharing of electrons NON POLAR – equal sharing of electrons Ionic bonding involves the transference of valence electrons SOLUTE POLAR SOLVENT NONPOLAR SOLVENT Ionic Soluble Insoluble Polar Soluble Insoluble Nonpolar Insoluble soluble 47 Learning Check S1 At 40C, the solubility of KBr is 80 g/100 g H2O. Indicate if the following solutions are (1) saturated or (2) unsaturated A. ___60 g KBr in 100 g of water at 40C B. ___200 g KBr in 200 g of water at 40C C. ___25 KBr in 50 g of water at 40C 48 Solution S1 At 40C, the solubility of KBr is 80 g/100 g H2O. Indicate if the following solutions are (1) saturated or (2) unsaturated A. 2 Less than 80 g/100 g H2O B. 1 Same as 100 g KBr in 100 g of water at 40C, which is greater than its solubility C. 2 Same as 60 g KBr in 100 g of water, which is less than its solubility 49 Temperature and Solubility of Solids Temperature 0° 20°C 50°C 100°C Solubility (g/100 g H2O) KCl(s) NaNO3(s) 27.6 74 34.0 88 42.6 114 57.6 182 The solubility of most solids (decreases or increases ) with an increase in the temperature. 50 Temperature and Solubility of Solids Temperature 0° 20°C 50°C 100°C Solubility (g/100 g H2O) KCl(s) NaNO3(s) 27.6 74 34.0 88 42.6 114 57.6 182 The solubility of most solids increases with an increase in the temperature. 51 Temperature and Solubility of Gases Temperature 0°C 20°C 50°C Solubility (g/100 g H2O) CO2(g) O2(g) 0.34 0.17 0.076 0.0070 0.0043 0.0026 The solubility of gases (decreases or increases) with an increase in temperature. 52 Temperature and Solubility of Gases Temperature 0°C 20°C 50°C Solubility (g/100 g H2O) CO2(g) O2(g) 0.34 0.17 0.076 0.0070 0.0043 0.0026 The solubility of gases decreases with an increase in temperature. 53 Learning Check S2 A. Why would a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun ? B. Why would fish die in water that gets too warm? 54 Solution S2 A. Gas in the bottle builds up as the gas becomes less soluble in water at high temperatures, which may cause the bottle to explode. B. Because O2 gas is less soluble in warm water, the fish may not obtain the needed amount of O2 for their survival. 55 Gas Solubility CH4 2.0 Solubility (mM) O2 CO 1.0 He 0 10 20 30 40 50 56 Solubility Curves Show the conditions that affect states of the solution: unsaturated, saturated, supersaturated. 57 Solubility vs. Temperature for Solids 140 KI 130 Solubility Table shows the dependence of solubility on temperature Solubility (grams of solute/100 g H2O) 120 NaNO3 110 gases solids 100 KNO3 90 80 HCl NH4Cl 70 60 NH3 KCl 50 40 30 NaCl KClO3 20 10 SO2 0 LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517 10 20 30 40 50 60 70 80 90 100 How to determine the solubility of a given substance? • Find out the mass of solute needed to make a saturated solution in 100 cm3 of water for a specific temperature(referred to as the solubility). • This is repeated for each of the temperatures from 0ºC to 100ºC. The data is then plotted on a temperature/solubility graph, and the points are connected. These connected points are called a solubility curve. How to use a solubility graph? A. IDENTIFYING A SUBSTANCE ( given the solubility in g/100 cm3 of water and the temperature) • Look for the intersection of the solubility and temperature. Learning Check SG1: What substance has a solubility of 90 g/100 cm3 in water at a temperature of 25ºC ? Learning Check SG2: What substance has a solubility of 200 g/100 cm3 of water at a temperature of 90ºC ? Look for the temperature or solubility •Locate the solubility curve needed and see for a given temperature, which solubility it lines up with and visa versa. Learning Check SG3: What is the solubility of potassium nitrate at 80ºC ? • What is the solubility of potassium nitrate at 80ºC ? Learning Check SG4: At what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ? Learning Check SG4: At what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ? Learning Check SG5: At what temperature will potassium iodide have a solubility of 230 g/100 cm3 ? Learning Check SG5: At what temperature will potassium iodide have a solubility of 130 g/100 cm3 ? Using Solubility Curves: What is the solubility of sodium chloride at 25ºC in 150 cm3 of water ? From the solubility graph we see that sodium chlorides solubility is 36 g. Place this in the proportion below and solve for the unknown solubility. Solve for the unknown quantity by cross multiplying. Solubility in grams = unknown solubility in grams 100 cm3 of water other volume of water ___36 grams____ = unknown solubility in grams 100 cm3 of water 150 cm3 water The unknown solubility is 54 grams. You can use this proportion to solve for the other volume of water if you're given the other solubility. C. Determine if a solution is saturated, unsaturated,or supersaturated. • If the solubility for a given substance places it anywhere on it's solubility curve line it is saturated. • If it lies above the solubility curve line, then it's supersaturated, • If it lies below the solubility curve line it's an unsaturated solution. Remember though, if the volume of water isn't 100 cm3 to use a proportion first as shown above. Solubility how much solute dissolves in a given amt. of solvent at a given temp. SOLUBILITY CURVE Solubility (g/100 g H2O) KNO3 (s) KCl (s) HCl (g) Temp. (oC) unsaturated: solution could hold more solute; below line saturated: solution has “just right” amt. of solute; on line supersaturated:solution has “too much” solute dissolved in it; above the line Solids dissolved in liquids Sol. Gases dissolved in liquids Sol. To As To , solubility To As To , solubility Sometimes you'll need to determine how much additional solute needs to be added to a unsaturated solution in order to make it saturated. For example,30 grams of potassium nitrate has been added to 100 cm3 of water at a temperature of 50ºC. How many additional grams of solute must be added in order to make it saturated? How many additional grams of solute must be added in order to make it saturated? From the graph you can see that the solubility for potassium nitrate at 50ºC is 84 grams If there are already 30 grams of solute in the solution, all you need to get to 84 grams is 54 more grams ( 84g-30g ) Solubility vs. Temperature for S 140 KI 130 Solubility Table shows the dependence of solubility on temperature Solubility (grams of solute/100 g H2O) 120 NaNO3 110 gases solids 100 KNO3 90 80 HCl NH4Cl 70 60 NH3 KCl 50 40 30 NaCl KClO3 20 10 SO2 0 LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 517 10 20 30 40 50 60 70 80 90 100 Solubility vs. Temperature f 140 KI Classify as unsaturated, saturated, or supersaturated. per 100 g H2O 45 g KCl @ 60oC =saturated 50 g NH3 @ 10oC =unsaturated 70 g NH4Cl @ 70oC =supersaturated 120 Solubility (grams of solute/100 g H2O) 80 g NaNO3 @ 30oC =unsaturated 130 110 NaNO3 gases solids 100 KNO3 90 80 HCl NH4Cl 70 60 NH3 KCl 50 40 30 NaCl KClO3 20 10 SO2 0 10 20 30 40 50 60 70 80 90 100 Describe each situation below. (A) Per 100 g H2O, 100 g NaNO3 @ 50oC. (B) Cool solution (A) very slowly to 10oC. (C) Quench solution (A) in an ice bath to 10oC. Unsaturated; all solute dissolves; clear solution. Supersaturated; extra solute remains in solution; still clear. Saturated; extra solute (20 g) can’t remain in solution, becomes visible. Soluble and Insoluble Salts A soluble salt is an ionic compound that dissolves in water. An insoluble salt is an ionic compound that does not dissolve in water 84 Aqueous Solutions How do we know ions are present in aqueous solutions? The solutions:_________ They are called ELECTROLYTES HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. 85 Solubility Rules 1. A salt is soluble in water if it contains any one of the following ions: NH4+ Li+ Na+ K+ or NO3- Examples: soluble salts LiCl Na2SO4 KBr Ca(NO3)2 86 Cl- Salts 2. Salts with Cl- are soluble, but not if the positive ion is Ag+, Pb2+, or Hg22+. Examples: soluble not soluble(will not dissolve) MgCl2 AgCl PbCl2 87 SO42- Salts 3. Salts with SO42- are soluble, but not if the positive ion is Ba2+, Pb2+, Hg2+ or Ca2+. Examples: soluble not soluble MgSO4 BaSO4 PbSO4 88 Other Salts 4. Most salts containing CO32-, PO43-, S2and OH- are not soluble. Examples: soluble not soluble Na2CO3 CaCO3 K 2S CuS 89 Learning Check S3 Indicate if each salt is (1)soluble or (2)not soluble: A. ______ Na2SO4 B. ______ MgCO3 C. ______ PbCl2 D. ______ MgCl2 90 Solution S3 Indicate if each salt is (1) soluble or (2) not soluble: A. _1_ Na2SO4 B. _2_ MgCO3 C. _2_ PbCl2 D. _1_ MgCl2 91 Solutions Molarity 92 Molarity (M) A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute 1 liter solution 93 Units of Molarity 2.0 M HCl 6.0 M HCl = 2.0 moles HCl 1 L HCl solution = 6.0 moles HCl 1 L HCl solution 94 Molarity Calculation NaOH is used to open stopped sinks, to treat cellulose in the making of nylon, and to remove potato peels commercially. If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution? 95 Calculating Molarity 1) 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH 40.0 g NaOH 2) 500. mL x 1L_ 1000 mL 3. 0.10 mole NaOH 0.500 L = 0.500 L = 0.20 mole NaOH 1L = 0.20 M NaOH 96 Learning Check M1 A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution? 1) 8 M 2) 5 M 3) 2 M 97 Solution M1 A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution? 2) 5 M M = 2 mole KOH = 5 M 0.4 L 98 Learning Check M2 A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 0.20 M 2) 5.0 M 3) 36 M 99 Solution M2 A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 72 g x 1 mole x 180. g 1 = 2.0 L 0.20 M 100 Molarity Conversion Factors A solution is a 3.0 M NaOH.. Write the molarity in the form of conversion factors. 3.0 moles NaOH and 1 L NaOH soln 1 L NaOH soln 3.0 moles NaOH 101 Learning Check M3 Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution? 1) 15 moles HCl 2) 1.5 moles HCl 3) 0.15 moles HCl 102 Solution M3 3) 1500 mL x 1 L = 1000 mL 1.5 L 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1L (Molarity factor) 103 Learning Check M4 How many grams of KCl are present in 2.5 L of 0.50 M KCl? 1) 1.3 g 2) 5.0 g 3) 93 g 104 Solution M4 3) 2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl 1L 1 mole KCl 105 Learning Check M5 How many milliliters of stomach acid, which is 0.10 M HCl, contain 0.15 mole HCl? 1) 150 mL 2) 1500 mL 3) 5000 mL 106 Solution M5 2) 0.15 mole HCl x 1 L soln x 1000 mL 0.10 mole HCl 1L (Molarity inverted) = 1500 mL HCl 107 Learning Check M6 How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 1) 12 g 2) 48 g 3) 300 g 108 Solution M6 2) 400. mL x 1 L = 0.400 L 1000 mL 0.400 L x 3.0 mole NaOH x 40.0 g NaOH 1L 1 mole NaOH (molar mass) = 48 g NaOH 109 Solution Percent Concentration 110 Percent Concentration Describes the amount of solute dissolved in 100 parts of solution amount of solute 100 parts solution 111 Mass-Mass % Concentration mass/mass % = g solute x 100% 100 g solution 112 Mixing Solute and Solvent + Solute 4.0 g KCl Solvent 46.0 g H2O 50.0 g KCl solution 113 Calculating Mass-Mass % g of KCl g of solvent g of solution = = = 4.0 g 46.0 g 50.0 g %(m/m) = 4.0 g KCl (solute) x 100 = 8.0% KCl 50.0 g KCl solution 114 Learning Check PC1 A solution contains 15 g Na2CO3 and 235 g of H2O? What is the mass % of the solution? 1) 15% (m/m) Na2CO3 2) 6.4% (m/m) Na2CO3 3) 6.0% (m/m) Na2CO3 115 Solution PC1 mass solute = 15 g Na2CO3 mass solution = 15 g + 235 g = 250 g %(m/m) = 15 g Na2CO3 x 100 250 g solution = 6.0% Na2CO3 solution 116 Mass-Volume % mass/volume % = g solute x 100% 100 mL solution 117 Learning Check PC2 An IV solution is prepared by dissolving 25 g glucose (C6H12O6) in water to make 500. mL solution. What is the percent (m/v) of the glucose in the IV solution? 1) 5.0% 2) 20.% 3) 50.% 118 Solution PC2 1) 5.0% %(m/v) = 25 g glucose x 100 500. mL solution = 5.0 %(m/v) glucose solution 119 Writing Factors from % A physiological saline solution is a 0.85% (m/v) NaCl solution. Two conversion factors can be written for the % value. 0.85 g NaCl and 100 mL NaCl soln 100 mL NaCl soln 0.85 g NaCl 120 % (m/m) Factors Write the conversion factors for a 10 %(m/m) NaOH solution NaOH and NaOH soln NaOH soln NaOH 121 % (m/m) Factors Write the conversion factors for a 10 %(m/m) NaOH solution 10 g 100 g NaOH NaOH soln and 100 g 10 g NaOH soln NaOH 122 Learning Check PC 3 Write two conversion factors for each of the following solutions: A. 8 %(m/v) NaOH B. 12 %(v/v) ethyl alcohol 123 Solution PC 3 Write conversion factors for the following: A. 8 %(m/v) NaOH 8 g NaOH and 100 mL 100 mL 8 g NaOH B. 12 %(v/v) ethyl alcohol 12 mL alcohol and 100 mL 100 mL 12 mL alcohol 124 Using % Factors How many grams of NaCl are needed to prepare 250 g of a 10.0% (m/m) NaCl solution? Complete data: ____________ g solution ____________% or (______/_100 g_) solution ____________ g solute 125 Clculation Using % Factors 250 10.0% ? g solution or (10.0 g/100 g) solution g solute 250 g NaCl soln x 10.0 g NaCl = 25 g NaCl 100 g NaCl soln 126 Learning Check PC4 How many grams of NaOH do you need to measure out to prepare 2.0 L of a 12%(m/v) NaOH solution? 1) 24 g NaOH 2) 240 g NaOH 3) 2400 g NaOH 127 Solution PC4 2.0 L soln x 1000 mL = 2000 mL soln 1L 12 % (m/v) NaOH = 12 g NaOH 100 mL NaOH soln 2000 mL x 12 g NaOH = 100 mL NaOH soln 240 g NaOH 128 Learning Check PC5 How many milliliters of 5 % (m/v) glucose solution are given if a patient receives 150 g of glucose? 1) 30 mL 2) 3000 mL 3) 7500 mL 129 Solution PC5 5% m/v factor 150 g glucose x 100 mL = 3000 mL 5 g glucose 130 Preparing a Solution by Dilution 131 Units of Concentrations amount of solute per amount of solvent or solution Percent (by mass) = Molarity (M) = g solute g solution x 100 = g solute x 100 g solute + g solvent moles of solute volume in liters of solution moles = M x VL 132 Solutions Colloids and Suspensions Osmosis and Dialysis 133 Solutions Have small particles (ions or molecules) Are transparent Do not separate Cannot be filtered Do not scatter light. 134 Colloids Have medium size particles Cannot be filtered Separated with semipermeable membranes Scatter light (Tyndall effect) 135 Examples of Colloids Fog Whipped cream Milk Cheese Blood plasma Pearls 136 Suspensions Have very large particles Settle out Can be filtered Must stir to stay suspended 137 Examples of Suspensions Blood platelets Muddy water Calamine lotion 138 Osmosis In osmosis, the solvent water moves through a semipermeable membrane Water flows from the side with the lower solute concentration into the side with the higher solute concentration Eventually, the concentrations of the two solutions become equal. 139 Osmosis 4% starch 10% starch H2O semipermeable membrane 140 Equilibrium is reached. 7% starch 7% starch H2OO water flow becomes equal 141 Osmotic Pressure Produced by the number of solute particles dissolved in a solution Equal to the pressure that would prevent the flow of additional water into the more concentrated solution Increases as the number of dissolved particles increase 142 Osmotic Pressure of the Blood Cell walls are semipermeable membranes The osmotic pressure of blood cells cannot change or damage occurs. The flow of water between a red blood cell and its surrounding environment must be equal 143 Isotonic solutions • Exert the same osmotic pressure as red blood cells. • Medically 5% glucose and 0.9% NaCl are used their solute concentrations provide an osmotic pressure equal to that of red blood cells H2O 144 Hypotonic Solutions Lower osmotic pressure than red blood cells Lower concentration of particles than RBCs In a hypotonic solution, water flows into the RBC The RBC undergoes hemolysis; it swells and may burst. H 2O 145 Hypertonic Solutions Has higher osmotic pressure than RBC Has a higher particle concentration In hypertonic solutions, water flows out of the RBC The RBC shrinks in size (crenation) H2O 146 Dialysis Occurs when solvent and small solute particles pass through a semipermeable membrane Large particles retained inside Hemodialysis is used medically (artificial kidney) to remove waste particles such as urea from blood 147 Colligative Properties On adding a solute to a solvent, the properties of the solvent are modified. • • • • Vapor pressure decreases Melting point decreases Boiling point increases Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not 148 on the KIND of solute particles. Change in Freezing Point Pure water Ethylene glycol/water solution The freezing point of a solution is LOWER than that of the pure solvent 149 Change in Freezing Point Common Applications of Freezing Point Depression Propylene glycol Ethylene glycol – deadly to small animals 150 Change in Freezing Point Common Applications of Freezing Point Depression Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? a) sand, SiO2 b) Rock salt, NaCl c) Ice Melt, CaCl2 151 Change in Boiling Point Common Applications of Boiling Point Elevation 152