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CHM1045 Exam #2 Study GuideChapter 4: Reactions in Aqueous SolutionsSolutions-Solutions: Homogeneous mixtures-Solute: The component of the solution that gets dissolved-Solvent: The component that dissolves the solute-To describe solutions accurately, we must describe how much of each component is present-Concentration: Quantitatively, the relative amount of solute in the solution-Ex) Table salt mixed with water, seems to disappear  homogeneous-If both components start in the same state, the major component is the solvent-Dilute Solutions: Have a small amount of solute compared to solvent-Concentrated Solutions: Have a large amount of solute compared to solventSolution Concentration Molarity-Molarity: Moles of solute per 1 liter of solution-Used because it describes how many molecules of solute in each liter of solutionPreparing 1 L of a 1.00 M NaCl Solution1 L)(in solution ofamount moles) (in solute ofamount M molarity, CHM1045 Exam #2 Study Guide-Practice Problem #4.1oFind the molarity of a solution that has 3.21 g of magnesium iodide dissolved in 180 mL of solution0.0641 M-Most solutions are generally between 0 and 18 MUsing Molarity in Calculations-Molarity shows the relationship between moles of solute and liters of solution-If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugaro2 liters = 4.0 moles sugaro0.5 liters = 1 mole sugaro2 mol sugar / 1 L solutiono1 L solution / 2 mole sugaro1 L solution : 2 mole sugar-Practice Problem #4.2oHow many liters of a 0.4 M unknown solution contains 0.15 moles of the unknown?0.38 LDilution-Solutions are often stored as concentrated stock solutions-To make solutions of lower concentrations from these stock solutions, more solvent is addedoThe amount of solute doesn't change, just the volume of solution2CHM1045 Exam #2 Study GuideMoles solute in solution 1 = moles solute in solution 2-The concentrations and volumes of the stock and new solutions are inversely proportionaloM1•V1 = M2•V2-Practice Problem #4.3oA student needs to dilute 100mL of a concentrated 6.0 M LiOH aqueous solution to obtain a 1.5 M solution. How much water does he have to add?0.3 LSolution Stoichiometry-Since molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction-Practice Problem #4.4oWhat volume of 0.150 M Li2S is required to completely react with 125 mL of 0.150 M Co(NO3)2?Li2S(aq) + Co(NO3)2(aq)  LiNO3(aq) + CoS(aq)-0.125 LElectrolytes and Nonelectrolytes-Electrolytes: Materials that dissolve in water to form a solution that will conduct electricity-Nonelectrolytes: Materials that dissolve in water to form a solution that will not conduct electricityMolecular View of Electrolytes and Nonelectrolytes-In order to conduct electricity, a material must have charged particles (ions or electrons) that are able to flow-Electrolyte solutions all contain ions dissolved in wateroIonic compounds are mostly electrolytes because they tend to dissociate into their ions when they dissolve-Nonelectrolyte solutions contain whole molecules dissolved in the wateroGenerally, molecular compounds do not ionize when they dissolve in wateroThe notable exception = molecular acids-General rule – ionic compounds dissolve to give electrolyte solutions and molecular compounds (except acids) give nonelectrolytes-Ex) NaOH – electrolyte-Ex) C2H5OH – nonelectrolyte3CHM1045 Exam #2 Study Guide-When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions-Result is a solution with free moving charged particles able to conduct electricityStrong and Weak Electrolytes-Strong Electrolytes: Materials that dissolve completely as ionsoMost ionic compounds and strong acidsoSolutions conduct electricity as well-Weak Electrolytes: Materials that dissolve mostly as molecules, but partially as ionsoWeak acidsoSolutions conduct electricity, but not well-When compounds containing a polyatomic ion dissolve, the polyatomic ion stays togetheroNa2SO4 (aq)  2 Na+ (aq) +SO42- (aq)oHC2H3O2 (aq)   H+ (aq) + C2H3O2- (aq)Acids-Acids ionize when they dissolve in wateroThe molecules are pulled apart by their attraction for the wateroWhen acids ionize, they form H+ cations and anions-The percentage of molecules that ionize varies from one acid to another-Strong Acids: Acids that ionize virtually 100%oHCl (aq)  H+ (aq) + Cl- (aq)oStrong acids & strong bases are strong electrolytes-Weak Acids: Acids that only ionize a small percentageoHF (aq)   H+ (aq) + F- (aq)oWeak acids & weak bases are weak electrolytes4CHM1045 Exam #2 Study GuideSolubility of Ionic Compounds-Some ionic compounds dissolve very well in water at room temperatureoEx) NaCl-Some ionic compounds dissolve hardly at all in water at room temperatureoEx) AgCl-Soluble Compounds: Compounds that dissolve in a solventoNaCl = soluble in water-Insoluble Compounds: Compounds that don't dissolve in a solventoAgCl = insoluble in water-The degree of solubility depends on the temperature-Even insoluble compounds dissolve, just not enough to be meaningful-Solubility = a spectrum (lots of gray areas)Solubility RulesCompounds that are Generally Soluble in WaterCompounds Containing the Following Ionsare Generally SolubleExceptions – When combined with the ionson the left, the compound is insolubleLi+ Na+ K+ NH4+NoneNO3- C2H3O2-NoneCl- Br- I-Ag+ Hg22+ Pb2+SO42-Ag+ Ca2+ Sr2+ Ba2+ Pb2+Compounds that are Generally Insoluble5CHM1045 Exam #2 Study GuideCompounds Containing the Following Ionsare Generally InsolubleExceptions – When combined with the ionson the left, the compound is soluble or slightly solubleOH-Li+ Na+ K+ NH4+Ca2+ Sr2+ Ba2+S2-Li+ Na+ K+ NH4+ Ca2+ Sr2+ Ba2+CO32- PO43- CrO42-Li+ Na+ K+ NH4+-Practice Problem #4.5oStrong, weak, or non-electrolyte?NaI – Strong electrolyteMg(OH)2 - NonelectrolytePbS – NonelectrolyteLi2S – Strong electrolyteHNO2 – Weak electrolytePrecipitation Reactions-Precipitation Reactions: Reactions between aqueous solutions of ionic compounds that produce an ionic compound that is insoluble in water-Precipitate: Insoluble product of a precipitation reaction-2 KI (aq) + Pb(NO3)2 (aq)  PbI2 (s) +2 KNO3 (aq)-No precipitate formation = no reactionoKI (aq) + NaCl (aq)  KCl (aq) + NaI (aq)All ions still present, no reactionPredicting the Products of a Precipitation Reaction1. Determine what ions


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FSU CHM 1045 - Exam #2 Study Guide

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