Chapter 8 Solutions Solutions: Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one or more solutes.
Download ReportTranscript Chapter 8 Solutions Solutions: Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one or more solutes.
Chapter 8 Solutions 1 Solutions: Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one or more solutes. 2 Nature of Solutes in Solutions Solutes Spread evenly throughout the solution. Cannot be separated by filtration. Can be separated by evaporation. Are not visible, but can give a color to the solution. 3 Examples of Solutions The solute and solvent in a solution can be a solid, liquid, and/or a gas. 4 Learning Check Identify the solute and the solvent in each. 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 solvent = 1) KCl 2) H2O 2) H2O 5 Solution Identify the solute and the solvent in each. A. brass: 20 g zinc + 50 g copper solute = 1) zinc solvent = 2) copper B. 100 g H2O + 5 g KCl solute = 1) KCl solvent = 2) H2O 6 Water Water is the most common solvent. The water molecule is polar. Hydrogen bonds form between the hydrogen atom in one molecule and the oxygen atom in a different water molecule. 7 Water 8 Like Dissolves Like A solution forms when there is an attraction between the particles of the solute and solvent. A polar solvent such as water dissolves polar solutes such as sugar and ionic solutes such as NaCl. A nonpolar solvent such as hexane (C6H14) dissolves nonpolar solutes such as oil or grease. 9 Examples of Like Dissolves Like Solvents Solutes Water (polar) Ni(NO3)2 (ionic) CH2Cl2 (nonpolar) I2 (nonpolar) 10 Learning Check Which of the following solutes will dissolve in water? Why? 1) Na2SO4 2) gasoline (nonpolar) 3) I2 4) HCl 11 Solution Which of the following solutes will dissolve in water? Why? 1) Na2SO4 Yes, ionic 2) gasoline No, nonnpolar 3) I2 No, nonpolar 4) HCl Yes, polar Most polar and ionic solutes dissolve in water because water is a polar solvent. 12 Formation of a Solution Na+ and Cl- ions on the surface of a NaCl crystal are attracted to polar water molecules. In solution, the ions are hydrated as several H2O molecules surround each. 13 Equations for Solution Formation When NaCl(s) dissolves in water, the reaction can be written as H2O NaCl(s) Na+(aq) + Cl- (aq) solid separation of ions 14 Learning Check Solid LiCl is added to 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. 15 Solution Solid LiCl is added to 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. 16 Electrolytes Electrolytes Produce positive (+) and negative (-) ions when they dissolve in water. In water conduct an electric current. 17 Strong Electrolytes Strong electrolytes ionize 100% in solution. Equations for the dissociation of strong electrolytes show the formation of ions in aqueous (aq) solutions. H2O 100% ions NaCl(s) Na+(aq) + Cl-(aq) H2O CaBr2(s) Ca2+(aq) + 2Br- (aq) 18 Learning Check Complete each of the following dissociation equations for strong electrolytes dissolved in water: H2O A. CaCl2 (s) 1) CaCl2 2) Ca2+ + Cl23) Ca2+ + 2ClH2O B. K3PO4 (s) 1) 3K+ + PO432) K3PO4 3) K3+ + P3- + O419 Solution Complete each of the following dissociation equations for strong electrolytes dissolved in water: H2O Ca2+ + 2Cl- A. 3) CaCl2 (s) H2O B. 1) K3PO4 (s) 3K+ + PO43- 20 Weak Electrolytes A weak electrolyte Dissolves mostly as molecules in solution. Produces only a few ions in aqueous solutions. Has an equilibrium that favors the reactants. HF + H2O H3O+(aq) + F- (aq) NH3 + H2O OH- (aq) NH4+(aq) + 21 Nonelectrolytes Nonelectrolytes Form only molecules in water. Do not produce ions in water. Do not conduct an electric current. 22 Equivalents An equivalent (Eq) is the amount of an ion that provides 1 mole of electrical charge (+ or -). 23 Electrolytes in Body Fluids In replacement solutions for body fluids, the electrolyte amounts are given in milliequivalents per liter (mEq/L). 24 Examples of IV Solutions In intravenous solutions, the total mEq of positively charged ions is equal to the total mEq of negatively charged ions. 25 Learning Check A. In 1 mole of Fe3+, there are 1) 1 Eq 2) 2 Eq 3) 3 Eq B. 0.5 equivalents of calcium is 1) 5 mEq 2) 50 mEq 3) 500 mEq C. If the Na+ in a NaCl IV solution is 34 mEq/L, the Cl- is 1) 34 mEq/L 2) 0 mEq/L 3) 68 mEq/L 26 Solution A. In 1 mole of Fe3+, there are 3) 3 Eq B. 0.5 equivalents of calcium is 3) 500 mEq C. If the Na+ in a NaCl IV solution is 34 mEq/L, the Cl- is 1) 34 mEq/L 27 Solubility Solubility states the maximum amount of solute that dissolves in a specific amount of solvent at a particular temperature. Typically, solubility is expressed as the grams of solute that dissolves in 100 g of solvent, usually water. g of solute 100 g water 28 Saturated Solutions A saturated solution Contains the maximum amount of solute that can dissolve. Has some undissolved solute at the bottom of the container. 29 Unsaturated Solutions An unsaturated solution Contains less than the maximum amount of solute. Can dissolve more solute. 30 Learning Check At 40C, the solubility of KBr is 80 g/100 g H2O. Identify the following solutions as either (1) saturated or (2) unsaturated. Explain. A. 60 g KBr added to 100 g of water at 40C. B. 200 g KBr added to 200 g of water at 40C. C. 25 g KBr added to 50 g of water at 40C. 31 Solution A. 2 B. 1 Amount is less than the solubility. In 100 g of water, 100 g KBr exceeds the solubility at 40C. C. 2 This would be 50 g KBr in 100 g of water, which is less than the solubility at 40C. 32 Solubility of Solids Changes with Temperature The solubility of most solids increases with an increase in temperature. 33 Solubility of Gases and Temperature The solubility of gases decreases with an increase in temperature. 34 Learning Check A. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun? B. Why do fish die in water that is too warm? 35 Solution A. The pressure in a bottle increases as the gas leaves solution as it becomes less soluble at high temperatures. As pressure increases, the bottle could burst. B. Because O2 gas is less soluble in warm water, fish cannot obtain the amount of O2 required for their survival. 36 Henry’s Law According to Henry’s Law, the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid. 37 Next Time We will continue with Chapter 8 38 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. 39 Solubility Rules A soluble salt dissolves in water. Insoluble salts do not dissolve in water. 40 Using the Solubility Rules The solubility rules predict whether a salt is soluble or insoluble in water. 41 Learning Check Indicate if each salt is (1) soluble or (2) insoluble. A. ______ Na2SO4 B. ______ MgCO3 C. ______ PbCl2 D. ______ MgCl2 42 Solution Indicate if each salt is 1) soluble or 2) insoluble. A. 1 Na2SO4 B. 2 MgCO3 C. 2 PbCl2 D. 1 MgCl2 43 Formation of a Solid When solutions of salts are mixed, a solid forms when ions of an insoluble salt combine. 44 Learning Check The formula of an insoluble salt in each mixture is A. BaCl2 + Na2SO4 1) BaSO4 2) NaCl 3) Na2Cl2 4) none B. AgNO3 + KCl 1) KNO3 2) AgK 3) AgCl 4) none C. KNO3 + NaCl 1) KCl 2) NaNO3 3) ClNO3 4) none 45 Solution A. BaCl2 + Na2SO4 1) BaSO4 B. AgNO3 + KCl 3) AgCl C. KNO3 + NaCl 4) none; all combinations are soluble. 46 Percent Concentration The concentration of a solution is the amount of solute dissolved in a specific amount of solution. amount of solute amount of solution The percent concentration describes the amount of solute that is dissolved in 100 parts of solution. amount of solute 100 parts solution 47 Mass Percent The mass percent (%m/m) Concentration is the percent by mass of solute in a solution. mass percent = g of solute x 100% g of solution Is the g of solute in 100 g of solution. mass percent = g of solute 100 g of solution 48 Mass of Solution grams of solute + grams of solvent 50.0 g KCl solution 49 Calculating Mass Percent Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). g of KCl = 8.0 g g of solvent (water) = 42.0 g g of KCl solution = 50.0 g 8.0 g KCl (solute) x 100 = 16% (m/m) KCl 50.0 g KCl solution 50 Learning Check A solution is prepared by mixing 15 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution. 1) 15% (m/m) Na2CO3 2) 6.4% (m/m) Na2CO3 3) 6.0% (m/m) Na2CO3 51 Solution 3) 6.0% (m/m) Na2CO3 mass solute = 15 g Na2CO3 mass solution = 15 g + 235 g = 250 g mass %(m/m) = 15 g Na2CO3 x 100 250 g solution = 6.0% Na2CO3 solution 52 Mass/Volume Percent The mass/volume percent (%m/v) Concentration is the ratio of the mass in grams (g) of solute in a volume (mL) of solution. mass/volume % = g of solute x 100% mL of solution Is the g of solute in 100 mL of solution. mass/volume % = g of solute 100 mL of solution 53 Preparing a Solution with a Mass/Volume % Concentration A percent mass/volume solution is prepared by weighing out the grams of solute (g) and adding water to give the final volume of the solution. 54 Calculation of Mass/Volume Percent Mass/volume percent (%m/v) is calculated from the grams of solute (g KCl) and the volume of solution (mL KCl solution). g of KI = 5.0 g KI mL of KI solution = 250.0 mL 5.0 g KI (solute) x 100 = 2.0%(m/v) KI 250.0 mL KI solution 55 Learning Check A 500. mL samples of an IV glucose solution contains 25 g glucose (C6H12O6) in water. What is the mass/volume % (%m/v) of glucose of the IV solution? 1) 5.0% 2) 20.% 3) 50.% 56 Solution 1) 5.0% Mass/volume %(m/v) = 25 g glucose x 100 500. mL solution = 5.0 %(m/v) glucose solution 57 Volume Percent The volume percent (%v/v) Concentration is the percent volume (mL) of solute (liquid) to volume (mL) of solution. volume % (v/v) = mL of solute x 100% mL of solution Is the mL of solute in 100 mL of solution. volume % (v/v) = mL of solute 100 mL of solution 58 Percent Conversion Factors Two conversion factors can be written for any type of % value. 59 Learning Check Write two conversion factors for each solutions: A. 8%(m/v) NaOH B. 12%(v/v) ethyl alcohol 60 Solution A. 8%(m/v) NaOH 8 g NaOH and 100 mL solution 100 mL solution 8 g NaOH B. 12%(v/v) ethyl alcohol 12 mL alcohol and 100 mL solution 100 mL solution 12 mL alcohol 61 Using Percent Factors How many grams of NaCl are needed to prepare 250 g of a 10.0% (m/m) NaCl solution? 1. Write the 10.0 %(m/m) as conversion factors. 10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl 2. Use the factor that cancels given (g solution). 250 g solution x 10.0 g NaCl = 25 g NaCl 100 g solution 62 Learning Check How many grams of NaOH are needed to prepare 2.0 L of a 12%(m/v) NaOH solution? 1) 24 g NaOH 2) 240 g NaOH 3) 2400 g NaOH 63 Solution 2) 240 g NaOH 2.0 L x 1000 mL = 2000 mL 1L 2000 mL x 12 g NaOH = 240 g NaOH 100 mL 12 % (m/v) factor 64 Learning Check 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 65 Solution 2) 3000 mL 150 g glucose x 100 mL = 3000 mL 5 g glucose 5% m/v factor (inverted) 66 Molarity (M) Molarity is a concentration unit for the moles of solute in the liters (L) of solution. Molarity (M) Examples: 2.0 M HCl 6.0 M HCl = moles of solute = moles liter of solution L = 2.0 moles HCl 1L = 6.0 moles HCl 1L 67 Preparing a 1.0 Molar Solution A 1.0 M NaCl solution is prepared by weighing out 58.5 g NaCl ( 1.0 mole) and adding water to make 1.0 liter of solution. 68 Calculation of Molarity What is the molarity of a NaOH solution prepared by adding 4.0 g of solid NaOH to water to make 0.50 L of solution ? 1. Determine the moles of solute. 4.0 g NaOH x 1 mole NaOH = 0.10 mole 40.0 g NaOH 2. Calculate molarity. 0.10 mole = 0.20 mole = 0.20 M NaOH 0.50 L 1L 69 Learning Check Calculate the molarity of an NaHCO3 solution prepared by dissolving 36 g of solid NaHCO3 in water to give a solution volume of 240 mL. 1) 0.43 M 2) 1.8 M 3) 15 M 70 Solution 2) 1.8 M 36 g x 1 mole NaHCO3 = 0.43 mole NaHCO3 84 g 0.43 mole NaHCO3 = 1.8 M NaHCO3 0.240 L 71 Learning Check 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 72 Solution 1) 0.20 M 72 g x 1 mole 180. g x 1 2.0 L = 0.20 moles 1L = 0.20 M 73 Molarity Conversion Factors The units in molarity can be used to write conversion factors. 74 Learning Check Stomach acid is 0.10 M HCl solution. How many moles of HCl are present in 1500 mL of stomach acid? 1) 15 moles HCl 2) 1.5 moles HCl 3) 0.15 mole HCl 75 Solution 3) 0.15 mole HCl 1500 mL x 1 L 1000 mL 1.5 L x = 1.5 L 0.10 mole HCl = 0.15 mole HCl 1L Molarity factor 76 Learning Check Calculate the grams of KCl that must be dissolved in water to prepare 0.25 L of a 2.0 M KCl solution. 1) 150 g KCl 2) 37 g KCl 3) 19 g KCl 77 Solution 3) 37 g KCl Determine the number of moles of KCl. 0.25 L x 2.0 mole KCl = 0.50 moles KCl 1L Convert the moles to grams of KCl. 0.50 moles KCl x 74.6 g KCl = 37 g KCl 1 mole KCl molar mass of KCl 78 Learning Check How many milliliters of 6.0 M HNO3 contain 0.15 mole of HNO3? 1) 25 mL 2) 90 mL 3) 400 mL 79 Solution 1) 25 mL 0.15 mole HNO3 x 1 L x 1000 mL 6.0 moles HNO3 1 L Molarity factor inverted = 25 mL HNO3 80 Next Time We complete Chapter 8 Review for Exam 3 81 Solutions Solutions are mixtures that Contain small solute particles (ions or molecules). Are transparent. Cannot be separated by filters. 82 Colloids Colloidal dispersions are mixtures that Contain medium-sized particles called colloids. Cannot be separated by filters. Are separated by semipermeable membranes. Scatter light (Tyndall effect). 83 Tyndall Effect A beam of light going through a colloid is visible because the light is scattered by the large solute particles. The Tyndall effect does not occur with solutions. 84 Examples of Colloids 85 Suspensions Suspensions are mixtures that Contain very large particles that are visible. Settle out rapidly. Are separated by filters. 86 Comparing the Properties of Solutions, Colloids, and Suspensions (a) Suspensions settle. (b) Filters separate suspensions, but not solutions or colloids. (c) Only solution particles go through semipermeable membranes. 87 Comparing Solutions, Colloids, and Suspensions 88 Learning Check A mixture with solute particles that do not settle, but are too large to pass through a semipermeable membrane is called a 1) solution 2) colloid 3) suspension 89 Solution A mixture with solute particles that do not settle, but are too large to pass through a semipermeable membrane is called a 2) colloid 90 Osmosis In osmosis, Water moves through a semipermeable membrane that separates two solutions with different concentrations. Water flows out of the solution with the lower solute concentration and into the solution with the higher solute concentration. The concentrations of the two solutions become equal. 91 Osmosis As water flows into the sucrose solution, the volume of the sucrose solution increases. The concentration of the sucrose solution decreases. 92 Osmosis During osmosis, water flows across the semipermeable membrane from the 4% starch solution into the 10% solution. semipermeable membrane 4% starch H2O 10% starch 93 Equilibrium Eventually, the flow of water across the semipermeable membrane becomes equal in both directions. 7% starch H2O 7% starch 94