Presentation on theme: "1 Solutions Properties of Water Solutions 2 Predict the % water in the following foods."— Presentation transcript:
1 Solutions Properties of Water Solutions
2 Predict the % water in the following foods
3 88% water 94% water 85% water 86% water
4 Water in the Body water gainwater loss liquids1000 mLurine 1500 mL food 1200 mLperspiring 300 mL cells 300 mL exhaling 600 mL feces 100 mL Calculate the total water gain and water loss Total ______ mL _____ mL
5 Water Most common solvent A polar molecule O - a hydrogen bond H +
6 Hydrogen Bonds Attract Polar Water Molecules
7 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
8 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?
9 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
10 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
11 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
12 Types of Solutions air O 2 gas and N 2 gas gas/gas soda CO 2 gas in water gas/liquid seawater NaCl in water solid/liquid brass copper and zinc solid/solid
13 Discussion Give examples of some solutions and explain why they are solutions.
14 Learning Check SF1 (1) element (2) compound (3) solution A. water123 B. sugar 123 C. salt water 123 D. air123 E. tea123
15 Solution SF1 (1) element (2) compound (3) solution A. water2 B. sugar 2 C. salt water 3 D. air3 E. tea3
16 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 H 2 O + 5 g KCl solute = 1) KCl 2) H 2 O solvent = 1) KCl 2) H 2 O
17 Solution SF2 A. brass: 20 g zinc + 50 g copper solute= 1) zinc solvent = 2) copper B. 100 g H 2 O + 5 g KCl solute = 1) KCl solvent = 2) H 2 O
18 Learning Check SF3 Identify the solute in each of the following solutions: A. 2 g sugar (1) mL water (2) B mL ethyl alcohol(1) and 30.0 mL of methyl alcohol (2) C mL water (1) and 1.50 g NaCl (2) D. Air: 200 mL O 2 (1) mL N 2 (2)
19 Solution SF3 Identify the solute in each of the following solutions: A. 2 g sugar (1) B mL of methyl alcohol (2) C. 1.5 g NaCl (2) D. 200 mL O 2 (1)
20 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 (C 6 H 14 ) is needed to dissolve nonpolar solutes such as oil or grease.
21 Learning Check SF4 Which of the following solutes will dissolve in water? Why? 1) Na 2 SO 4 2) gasoline 3) I 2 4) HCl
22 Solution SF4 Which of the following solutes will dissolve in water? Why? 1) Na 2 SO 4 Yes, polar (ionic) 2) gasoline No, nonnpolar 3) I 2 No, nonpolar 4) HClYes, Polar
23 Formation of a Solution Cl - Na + Cl - Na + H2OH2O H2OH2O Cl - solute Dissolved solute Hydration
24 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.
25 Electrolytes Some solutes can dissociate into ions. Electric charge can be carried.
26 Types of solutes Na + Cl - Strong Electrolyte - 100% dissociation, all ions in solution high conductivity
27 Types of solutes CH 3 COOH CH 3 COO - H+H+ Weak Electrolyte - partial dissociation, molecules and ions in solution slight conductivity
28 Types of solutes Sugar C 6 H 12 O 6 Non-electrolyte - No dissociation, all molecules in solution no conductivity
29 Types of Electrolytes Weak electrolyte partially dissociates. –Fair conductor of electricity. Non-electrolyte does not dissociate. –Poor conductor of electricity. Strong electrolyte dissociates completely. –Good electrical conduction.
30 Representation of Electrolytes using Chemical Equations MgCl 2 (s) Mg 2+ (aq) + 2 Cl - (aq) A strong electrolyte: A weak electrolyte: CH 3 COOH(aq) CH 3 COO - (aq) +H + (aq) CH 3 OH(aq) A non-electrolyte:
31 Electrolytes in Action
32 Strong Electrolytes Strong acids: HNO 3, H 2 SO 4, HCl, HClO 4 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) Ca 2+ (aq) + 2 OH – (aq) AlCl 3 (s) Al 3+ (aq) + 3 Cl – (aq) (NH 4 ) 2 SO 4 (s) 2 NH 4 + (aq) + SO 4 2– (aq)
33 Writing An Equation for a Solution When NaCl(s) dissolves in water, the reaction can be written as H 2 O NaCl(s) Na + (aq) + Cl - (aq) solid separation of ions in water
34 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
35 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
36 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
37 How do I get sugar to dissolve faster in my iced tea? Stir, and stir, and stir Add sugar to warm tea then add ice Grind the sugar to a powder Fresh solvent contact and interaction with solute Greater surface area, more solute-solvent interaction Faster rate of dissolution at higher temperature
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. ___Heating the water B. ___Using large pieces of gelatin C. ___Stirring the solution
39 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
40 Solubility Percent Concentration Colloids and Suspensions
41 Solubility The maximum amount of solute that can dissolve in a specific amount of solvent usually 100 g. g of solute 100 g water
42 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
Solubility SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form increasing concentration
Polarity Factors Affecting Solid Solubility Temperature Surface Area Stirring
Polarity Factors Affecting Solubility 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
48 Learning Check S1 At 40 C, the solubility of KBr is 80 g/100 g H 2 O. Indicate if the following solutions are (1) saturated or (2) unsaturated A. ___60 g KBr in 100 g of water at 40 C B. ___200 g KBr in 200 g of water at 40 C C. ___25 KBr in 50 g of water at 40 C
49 Solution S1 At 40 C, the solubility of KBr is 80 g/100 g H 2 O. Indicate if the following solutions are (1) saturated or (2) unsaturated A. 2 Less than 80 g/100 g H 2 O B. 1 Same as 100 g KBr in 100 g of water at 40 C, which is greater than its solubility C. 2 Same as 60 g KBr in 100 g of water, which is less than its solubility
50 Temperature and Solubility of Solids TemperatureSolubility (g/100 g H 2 O) KCl(s)NaNO 3 (s) 0° °C °C °C The solubility of most solids (decreases or increases ) with an increase in the temperature.
51 Temperature and Solubility of Solids TemperatureSolubility (g/100 g H 2 O) KCl(s)NaNO 3 (s) 0° °C °C °C The solubility of most solids increases with an increase in the temperature.
52 Temperature and Solubility of Gases TemperatureSolubility(g/100 g H 2 O) CO 2 (g)O 2 (g) 0°C °C °C The solubility of gases (decreases or increases) with an increase in temperature.
53 Temperature and Solubility of Gases TemperatureSolubility(g/100 g H 2 O) CO 2 (g)O 2 (g) 0°C °C °C The solubility of gases decreases with an increase in temperature.
54 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?
55 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 O 2 gas is less soluble in warm water, the fish may not obtain the needed amount of O 2 for their survival.
Gas Solubility 56 CH 4 O2O2 CO He Solubility (mM) Higher Temperature …Gas is LESS Soluble
57 Solubility Curves Show the conditions that affect states of the solution: unsaturated, saturated, supersaturated.
Solubility Table LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page Solubility vs. Temperature for Solids Solubility (grams of solute/100 g H 2 O) KI KCl NaNO 3 KNO 3 HClNH 4 Cl NH 3 NaCl KClO 3 SO 2 shows the dependence of solubility on temperature gases solids
How to determine the solubility of a given substance? Find out the mass of solute needed to make a saturated solution in 100 cm 3 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 cm 3 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 cm 3 in water at a temperature of 25ºC ?
Learning Check SG2: What substance has a solubility of 200 g/100 cm 3 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 ?
Learning Check SG4: At what temperature will sodium nitrate have a solubility of 95 g/100 cm 3 ?
Learning Check SG4: At what temperature will sodium nitrate have a solubility of 95 g/100 cm 3 ?
Learning Check SG5: At what temperature will potassium iodide have a solubility of 230 g/100 cm 3 ?
Learning Check SG5: At what temperature will potassium iodide have a solubility of 130 g/100 cm 3 ?
Using Solubility Curves: What is the solubility of sodium chloride at 25ºC in 150 cm 3 of water ? From the solubility graph we see that sodium chlorides solubility is 36 g.
Solubility in grams=unknown solubility in grams 100 cm 3 of waterother volume of water ___36 grams____=unknown solubility in grams 100 cm 3 of water150 cm 3 water Place this in the proportion below and solve for the unknown solubility. Solve for the unknown quantity by cross multiplying. 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 cm 3 to use a proportion first as shown above.
Temp. ( o C) Solubility (g/100 g H 2 O) KNO 3 (s) KCl (s) HCl (g) SOLUBILITY CURVE Solubility how much solute dissolves in a given amt. of solvent at a given temp. below unsaturated: solution could hold more solute; below line on saturated:solution has just right amt. of solute; on line supersaturated:solution has too much solute dissolved in it; above above the line
ToTo Sol. ToTo Solids dissolved in liquids Gases dissolved in liquids As T o, 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 cm 3 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 Table LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World, 1996, page 517 shows the dependence of solubility on temperature Solubility vs. Temperature for Solids Solubility (grams of solute/100 g H 2 O) KI KCl NaNO 3 KNO 3 HClNH 4 Cl NH 3 NaCl KClO 3 SO 2 gases solids
Classify as unsaturated, saturated, or supersaturated. per 100 g H 2 O 80 g NaNO 30 o C 45 g 60 o C 50 g NH 10 o C 70 g NH 4 70 o C =unsaturated =saturated =unsaturated =supersaturated Solubility vs. Temperature for Solids Solubility (grams of solute/100 g H 2 O) KI KCl NaNO 3 KNO 3 HClNH 4 Cl NH 3 NaCl KClO 3 SO 2 gases solids
(A) Per 100 g H 2 O, 100 g Unsaturated; all solute NaNO 50 o C. dissolves; clear solution. (B) Cool solution (A) very Supersaturated; extra slowly to 10 o C. solute remains in solution; still clear. Describe each situation below. (C) Quench solution (A) in Saturated; extra solute an ice bath to 10 o C. (20 g) cant remain in solution, becomes visible.
84 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
85 How do we know ions are present in aqueous solutions? The solutions:_________ They are called ELECTROLYTES HCl, MgCl 2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. Aqueous Solutions
86 Solubility Rules 1. A salt is soluble in water if it contains any one of the following ions: NH 4 + Li + Na + K + orNO 3 - Examples: soluble salts LiCl Na 2 SO 4 KBr Ca(NO 3 ) 2
87 Cl - Salts 2. Salts with Cl - are soluble, but not if the positive ion is Ag +, Pb 2+, or Hg Examples: solublenot soluble(will not dissolve) MgCl 2 AgCl PbCl 2
88 SO 4 2- Salts 3. Salts with SO 4 2- are soluble, but not if the positive ion is Ba 2+, Pb 2+, Hg 2+ or Ca 2+. Examples: solublenot soluble MgSO 4 BaSO 4 PbSO 4
89 Other Salts 4. Most salts containing CO 3 2-, PO 4 3-, S 2- and OH - are not soluble. Examples: solublenot soluble Na 2 CO 3 CaCO 3 K 2 SCuS
90 Learning Check S3 Indicate if each salt is (1)soluble or (2)not soluble: A. ______ Na 2 SO 4 B. ______ MgCO 3 C. ______ PbCl 2 D. ______ MgCl 2
91 Solution S3 Indicate if each salt is (1) soluble or (2) not soluble: A. _1_ Na 2 SO 4 B. _2_ MgCO 3 C. _2_ PbCl 2 D. _1_ MgCl 2
92 Solutions Molarity
93 Molarity (M) A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute 1 liter solution
94 Units of Molarity 2.0 M HCl = 2.0 moles HCl 1 L HCl solution 6.0 M HCl= 6.0 moles HCl 1 L HCl solution
95 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?
96 Calculating Molarity 1) 4.0 g NaOH x 1 mole NaOH = 0.10 mole NaOH 40.0 g NaOH 2) 500. mL x 1 L _ = L 1000 mL mole NaOH = 0.20 mole NaOH L 1 L = 0.20 M NaOH
97 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 Drano
98 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 Drano
99 Learning Check M2 A glucose solution with a volume of 2.0 L contains 72 g glucose (C 6 H 12 O 6 ). 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
100 Solution M2 A glucose solution with a volume of 2.0 L contains 72 g glucose (C 6 H 12 O 6 ). 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 1 = 0.20 M 180. g 2.0 L
101 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
102 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
103 Solution M3 3) 1500 mL x 1 L = 1.5 L 1000 mL 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1 L (Molarity factor)
104 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
105 Solution M4 3) 2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl 1 L 1 mole KCl
106 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
107 Solution M5 2) 0.15 mole HCl x 1 L soln x 1000 mL 0.10 mole HCl 1 L (Molarity inverted) = 1500 mL HCl
108 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
109 Solution M6 2) 400. mL x 1 L = L 1000 mL L x 3.0 mole NaOH x 40.0 g NaOH 1 L1 mole NaOH (molar mass) = 48 g NaOH
110 Solution Percent Concentration
111 Percent Concentration Describes the amount of solute dissolved in 100 parts of solution amount of solute 100 parts solution
112 Mass-Mass % Concentration mass/mass % = g solute x 100% 100 g solution
113 Mixing Solute and Solvent Solute + Solvent 4.0 g KCl 46.0 g H 2 O 50.0 g KCl solution
114 Calculating Mass-Mass % g of KCl = 4.0 g g of solvent = 46.0 g g of solution = 50.0 g %(m/m) = 4.0 g KCl (solute) x 100 = 8.0% KCl 50.0 g KCl solution
115 Learning Check PC1 A solution contains 15 g Na 2 CO 3 and 235 g of H 2 O? What is the mass % of the solution? 1) 15% (m/m) Na 2 CO 3 2) 6.4% (m/m) Na 2 CO 3 3) 6.0% (m/m) Na 2 CO 3
116 Solution PC1 mass solute = 15 g Na 2 CO 3 mass solution= 15 g g = 250 g %(m/m) = 15 g Na 2 CO 3 x g solution = 6.0% Na 2 CO 3 solution
117 Mass-Volume % mass/volume % = g solute x 100% 100 mL solution
118 Learning Check PC2 An IV solution is prepared by dissolving 25 g glucose (C 6 H 12 O 6 ) 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.%
119 Solution PC2 1) 5.0% %(m/v) = 25 g glucose x mL solution = 5.0 %(m/v) glucose solution
120 Writing Factors from % A physiological saline solution is a 0.85% (m/v) NaCl solution. Two conversion factors can be written for the % value g NaCl and 100 mL NaCl soln 100 mL NaCl soln 0.85 g NaCl
121 % (m/m) Factors Write the conversion factors for a 10 %(m/m) NaOH solution NaOH a nd NaOH soln NaOH soln NaOH
122 % (m/m) Factors Write the conversion factors for a 10 %(m/m) NaOH solution NaOH a nd NaOH soln NaOH soln NaOH 10 g 100 g
123 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
124 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
125 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
126 Clculation Using % Factors 250 g solution 10.0% 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
127 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
128 Solution PC4 2.0 L soln x 1000 mL = 2000 mL soln 1 L 12 % (m/v) NaOH = 12 g NaOH 100 mL NaOH soln 2000 mL x 12 g NaOH = 240 g NaOH 100 mL NaOH soln
129 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
130 Solution PC5 5% m/v factor 150 g glucose x 100 mL = 3000 mL 5 g glucose
131 Preparing a Solution by Dilution
132 Units of Concentrations amount of solute per amount of solvent or solution Percent (by mass) = g solute g solution x 100 g solute g solute + g solvent x 100 = Molarity (M) = moles of solute volume in liters of solution moles = M x V L
133 Solutions Colloids and Suspensions Osmosis and Dialysis
134 Solutions Have small particles (ions or molecules) Are transparent Do not separate Cannot be filtered Do not scatter light.
135 Colloids Have medium size particles Cannot be filtered Separated with semipermeable membranes Scatter light (Tyndall effect)
136 Examples of Colloids Fog Whipped cream Milk Cheese Blood plasma Pearls
137 Suspensions Have very large particles Settle out Can be filtered Must stir to stay suspended
138 Examples of Suspensions Blood platelets Muddy water Calamine lotion
139 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.
141 Equilibrium is reached. water flow becomes equal 7% starch H2OOH2OO
142 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
143 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
144 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 H2OH2O
145 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 2 O
146 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) H2OH2O
147 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
148 Colligative Properties On adding a solute to a solvent, the properties of the solvent are modified. Vapor pressure decreasesVapor pressure decreases Melting point decreasesMelting point decreases Boiling point increasesBoiling point increases Osmosis is possible (osmotic pressure)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 on the KIND of solute particles.
149 Change in Freezing Point The freezing point of a solution is LOWER than that of the pure solvent Pure water Ethylene glycol/water solution
150 Change in Freezing Point Common Applications of Freezing Point Depression Propylene glycol Ethylene glycol – deadly to small animals
151 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, SiO 2 b)Rock salt, NaCl c)Ice Melt, CaCl 2 Change in Freezing Point
152 Change in Boiling Point Common Applications of Boiling Point Elevation