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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. Nature of Solutes in Solutions. Solutes Spread evenly throughout the solution. Cannot be separated by filtration.
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Solutions: Solute and Solvent Solutions • Are homogeneous mixtures of two or more substances. • Consist of a solvent and one or more solutes.
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.
Examples of Solutions • The solute and solvent in a solution can be a solid, liquid, and/or a gas.
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 2) H2O solvent = 1) KCl 2) H2O
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
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.
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.
Examples of Like Dissolves Like SolventsSolutes Water (polar) Ni(NO3)2 (ionic) CH2Cl2 (nonpolar) I2 (nonpolar)
Learning Check Which of the following solutes will dissolve in water? Why? 1) Na2SO4 2) gasoline (nonpolar) 3) I2 4) HCl
Solution Which of the following solutes will dissolve in water? Why? 1) Na2SO4Yes, ionic 2) gasolineNo, nonnpolar 3) I2 No, nonpolar 4) HClYes, polar Most polar and ionic solutes dissolve in water because water is a polar solvent.
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.
Equations for Solution Formation • When NaCl(s) dissolves in water, the reaction can be written as H2O NaCl(s) Na+(aq) + Cl- (aq) solidseparation of ions
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.
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.
Electrolytes Electrolytes • Produce positive (+) and negative (-) ions when they dissolve in water. • In water conduct an electric current.
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)
Learning Check Complete each of the following dissociation equations for strong electrolytes dissolved in water: H2O A. CaCl2 (s) 1) CaCl2 2) Ca2+ + Cl2- 3) Ca2+ + 2Cl- H2O B. K3PO4 (s) 1) 3K+ + PO43- 2) K3PO4 3) K3+ + P3- + O4-
Solution Complete each of the following dissociation equations for strong electrolytes dissolved in water: H2O A. 3) CaCl2 (s) Ca2+ + 2Cl- H2O B. 1) K3PO4 (s) 3K+ + PO43-
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 NH4+(aq) + OH- (aq)
Nonelectrolytes Nonelectrolytes • Form only molecules in water. • Do not produce ions in water. • Do not conduct an electric current.
Equivalents • An equivalent (Eq) is the amount of an ion that provides 1 mole of electrical charge (+ or -).
Electrolytes inBody Fluids • In replacement solutions for body fluids, the electrolyte amounts are given in milliequivalents per liter (mEq/L).
Examples of IV Solutions • In intravenous solutions, the total mEq of positively charged ions is equal to the total mEq of negatively charged ions.
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
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
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
Saturated Solutions A saturated solution • Contains the maximum amount of solute that can dissolve. • Has some undissolved solute at the bottom of the container.
Unsaturated Solutions An unsaturated solution • Contains less than the maximum amount of solute. • Can dissolve more solute.
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.
Solution A. 2Amount is less than the solubility. B. 1In 100 g of water, 100 g KBr exceeds the solubility at 40C. C. 2This would be 50 g KBr in 100 g of water, which is less than the solubility at 40C.
Solubility of Solids Changes with Temperature • The solubility of most solids increases with an increase in temperature.
Solubility of Gases and Temperature • The solubility of gases decreases with an increase in temperature.
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?
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.
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.
Next Time • We will continue with Chapter 8
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.
Solubility Rules • A soluble salt dissolves in water. • Insoluble salts do not dissolve in water.
Using the Solubility Rules • The solubility rules predict whether a salt is soluble or insoluble in water.
Learning Check Indicate if each salt is (1) soluble or (2) insoluble. A. ______ Na2SO4 B. ______ MgCO3 C. ______ PbCl2 D. ______ MgCl2
Solution Indicate if each salt is 1) soluble or 2) insoluble. A.1 Na2SO4 B. 2 MgCO3 C. 2 PbCl2 D. 1 MgCl2
Formation of a Solid • When solutions of salts are mixed, a solid forms when ions of an insoluble salt combine.
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
Solution A. BaCl2 + Na2SO4 1) BaSO4 B. AgNO3 + KCl 3) AgCl C. KNO3 + NaCl 4) none; all combinations are soluble.
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 concentrationdescribes the amount of solute that is dissolved in 100 parts of solution. amount of solute 100 parts solution
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
Mass of Solution grams of solute + grams of solvent 50.0 g KCl solution
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