Chapter 11 Solutions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.

Download Chapter 11 Solutions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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<ul><li><p>Chapter 11 SolutionsCopyright 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings</p></li><li><p>Solute and SolventSolutionsAre homogeneous mixtures of two or more substances.Consist of a solvent and one or more solutes.</p></li><li><p>Nature of Solutes in SolutionsSolutesSpread 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.</p><p>Copyright 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings</p></li><li><p>Examples of Solutions</p></li><li><p>WaterWaterIs the most common solvent.Is a polar molecule.Forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule.Copyright 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings(Intermolecular Forces)</p></li><li><p>Formation of a SolutionNa+ and Cl- ions,On the surface of a NaCl crystal are attracted to polar water molecules.In solution are hydrated as several H2O molecules surround each.</p></li><li><p>Equations for Solution Formation</p><p>When NaCl(s) dissolves in water, the reaction can be written as:</p><p> H2ONaCl(s) Na+(aq) + Cl- (aq)</p><p> solid separation of ions </p></li><li><p>Learning CheckSolid 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.</p><p>B. The Cl- ions are attracted to the 1) oxygen atom ( -) of water. 2) hydrogen atom (+) of water.</p></li><li><p>Like Dissolves LikeTwo substances form a solution: When there is an attraction between the particles of the solute and solvent.When a polar solvent, such as water, dissolves polar solutes such as sugar and ionic solutes such as NaCl.When a nonpolar solvent, such as hexane (C6H14), dissolves nonpolar solutes such as oil or grease.</p></li><li><p>Learning CheckWhich of the following solutes will dissolve in water? Why?1) Na2SO42) gasoline (nonpolar)3) I24) HCl</p></li><li><p>Solutes and Ionic ChargeIn water, Strong electrolytes produce ions and conduct an electric current. Weak electrolytes produce a few ions. Nonelectrolytes do not produce ions.</p><p>Copyright 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings</p></li><li><p>Strong ElectrolytesStrong electrolytes, Dissociate in water producing positive and negative ions.Produce an electric current in water.In equations show the formation of ions in aqueous (aq) solutions. H2O 100% ionsNaCl(s) Na+(aq) + Cl (aq) H2OCaBr2(s) Ca2+(aq) + 2Br (aq)</p></li><li><p>Weak Electrolytes A weak electrolyte,Dissociates only slightly in water.In water forms a solution of only a few ions and mostly undissociated molecules.</p><p>HF(g) + H2O(l) H3O+(aq) + F- (aq)</p><p> NH3(g) + H2O(l) NH4+(aq) + OH- (aq)</p></li><li><p>SolubilitySolubility Is the maximum amount of solute that dissolves in a specific amount of solvent. Can be expressed as grams of solute in 100 grams of solvent, usually water. g of solute100 g water</p></li><li><p>Unsaturated solutions Contain less than the maximum amount of solute. Can dissolve more solute. </p><p>Saturated solutions Contain the maximum amount of solute that can dissolve. Have undissolved solute at the bottom of the container. </p></li><li><p>Soluble and Insoluble SaltsIonic compounds thatDissolve in water are soluble salts.Do NOT dissolve in water are insoluble salts. </p><p>***</p></li><li><p>Equations for Forming SolidsA molecular equation shows the formulas of thecompounds.Pb(NO3)(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq)</p><p>An ionic equation shows the ions of the compounds.Pb2+(aq) + 2NO3(aq) + 2Na+(aq) + 2Cl(aq) PbCl2(s) + 2Na+(aq) + 2NO3(aq) </p><p>NOTE: the spectator ions</p><p>A net ionic equation shows only the ions that form a solid. Pb2+(aq) + 2Cl(aq) PbCl2(s) </p></li><li><p>Learning Check A precipitate forms in the following reaction. Write the molecular, ionic and net ionic equations for the reaction.</p><p> BaCl2(aq) + Na2SO4(aq) </p></li><li><p>Mass PercentThe mass percent (%m/m) Concentration is the percent by mass of solute in a solution.mass percent (%m/m) = g of solute x 100 g of solute + g of solventIs the g of solute in 100 g of solution. mass percent = g of solute x 100 100 g of solution</p></li><li><p>Calculating Mass PercentMass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution).g of KCl = 8.00 gg of solvent (water) = 42.00 gg of KCl solution = 50.00 g</p><p> 8.00 g KCl (solute) x 100 = 16.0% (m/m) 50.00 g KCl solution</p></li><li><p>Volume PercentThe volume percent (%v/v) is:Percent volume (mL) of solute (liquid) to volume (mL) of solution.volume % (v/v) = mL of solute x 100 mL of solutionSolute (mL) in 100 mL of solution.volume % (v/v) = mL of solute 100 mL of solution</p></li><li><p>Percent Conversion FactorsTwo conversion factors can be written for each type of % value.</p></li><li><p>Learning Check Write two conversion factors for each solutions:A. 8.50%(m/m) NaOH</p><p>B. 5.75%(v/v) ethanol</p></li><li><p>Using Percent FactorsHow many grams of NaCl are needed to prepare225 g of a 10.0% (m/m) NaCl solution?STEP 1 Given: 225 g solution; 10.0% (m/m) NaCl Need: g of NaClSTEP 2 g solution g NaClSTEP 3 Write the 10.0% (m/m) as conversion factors.10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaClSTEP 4 Set up using the factor that cancels g solution.225 g solution x 10.0 g NaCl = 22.5 g NaCl100 g solution </p></li><li><p>Molarity (M)Molarity (M) is:</p><p>A concentration term for solutions.</p><p>Gives the moles of solute in 1 L solution.</p><p>moles of soluteliter of solution </p></li><li><p>Preparing a 1.0 Molar SolutionA 1.00 M NaCl solution is prepared:By weighing out 58.5 g NaCl (1.00 mol) andAdding water to make 1.00 liter of solution.</p></li><li><p>Learning CheckWhat is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3?1) 0.557 M 2) 1.44 M3) 1.71 M</p></li><li><p>Molarity Conversion FactorsThe units of molarity are used as conversion factors in calculations with solutions.</p></li><li><p>Learning Check How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution?1) 20.0 g AlCl3 2) 16.7g AlCl33) 2.50 g AlCl3 </p></li><li><p>Learning CheckHow many milliliters of 2.00 M HNO3 contain 24.0 g HNO3?1) 12.0 mL2) 83.3 mL3) 190. mL</p></li><li><p>DilutionIn a dilution,Water is addedVolume increasesConcentration decreases</p><p>M1V1 = M2V2 initial diluted</p></li><li><p>Learning CheckWhat is the final volume (mL) of 15.0 mL of a 1.80 MKOH diluted to give a 0.300 M solution?</p><p>1) 27.0 mL2) 60.0 mL 3) 90.0 mL </p></li><li><p>Using Molarity of ReactantsHow many mL of 3.00 M HCl are needed to completelyreact with 4.85 g CaCO3?</p><p>2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)</p><p>STEP 1 Given: 3.00 M HCl; 4.85 g CaCO3 Need: volume in mLSTEP 2 Plan: g CaCO3 mol CaCO3 mol HCl mL HCl</p></li><li><p>Using Molarity of Reactants (cont.)2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l)STEP 3 Equalities 1 mol CaCO3 = 100.09 g; 1 mol CaCO3 = 2 mol HCl 1000 mL HCl = 3.00 mol HCl </p><p>STEP 4 Setup 4.85 g CaCO3 x 1 mol CaCO3 x 2 mol HCl x 1000 mL HCl 100.09 g CaCO3 1 mol CaCO3 3.00 mol HCl</p><p> = 32.3 mL HCl required</p></li><li><p>Learning CheckIf 22.8 mL of 0.100 M MgCl2 is needed to completelyreact 15.0 mL of AgNO3 solution, what is the molarity ofthe AgNO3 solution?</p><p>MgCl2(aq) + 2AgNO3(aq) 2AgCl(s) + Mg(NO3)2(aq)</p><p>1) 0.0760 M2) 0.152 M3) 0.304 M</p></li></ul>

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