Chapter 4 Chemical Reactions in Solution Notes
Chapter 4 Chemical Reactions in Solution Notes Chem 111-003
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This 3 page Class Notes was uploaded by Sharneece Gary on Monday September 28, 2015. The Class Notes belongs to Chem 111-003 at University of South Carolina taught by Dmitry V Peryshkov in Fall 2015. Since its upload, it has received 74 views.
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Date Created: 09/28/15
Chapter 4 Chemical Reactions in Solution Vocabulary Solvent compound that has the same physical state as the solution Solute substance being dissolved Aqueous solution water is the solvent Strong electrolytes compound that separates completely into ions in water Weak electrolytes molecule that only partially ionizes when dissolved in water Ionization separation of a molecular compound into individual cations and anions when dissolved in water Precipitation reaction When there two reactants that are both soluble and the product that is produced is insoluble o AgNO3aq LiClaq e LiNO3aq o If you refer back to the chart you will see that that A g is an exception of Cl Complete ionic equation shows all strong electrolytes as ions in solutions Net ionic equation shows only those species in the solution that actually undergo a chemical change Spectator ions do not participate in the chemical reaction Molarity is the number of moles of solute in one liter of solution 0 Molarity Moles of soluteliters of solution Dilution making solutions of lower concentrations 0 Molarityconcentrated X Volume concentrated Molarity diluated X Volume diluted 39 MC X VCMD X VD 0 Only used for dilution problems Stoichiometric calculation molarity is used to calculate moles from volume of solution Titrations the concentration and volume of a solution of known concentration is used to determine the concentration of an unknown solution Equivalence Point the point volume in a titration where stiochiometrically equivalent amounts of the two reactants have been added Indicator a compound that changes color to signal the end of the titration 0 Changes color at the end point which the volume should be very close to the equivalence point volume Memorize the Solubility Rules Reference the two charts on Page 68 in the Lecture Notes book Simply memorizing the content will help you determine the solubility of a compound rather if it will be soluble or insoluble Precipitation Reaction 0f the two reactions which reactant will form a precipitate PbN032 2NaCl FeS04 MgN032 PbN032 2NaCl 9 2NaNO3 amp FeSO4 MgNO32 9 FeNO32 MgSO4 Net Ionic Equations Write the net ionic equation for the reaction of solutions of AgN03 and KC AgNO3aq KC1aq 9AgCls KNO3aq Agaq NO3aq Kaq Claq 9 s KaqNO3aq Answer Agaq Claq 9AgCls Molarity of Solution What is the molar concentration of sodium uoride in a solution prepared by dissolving 251g of NaF in enough water to form 2000 mL of solution Step 1 You know that molarity is molL so you want to convert the 251 grams to moles 251g NaF X 1Mol420NaF 00598 mol NaF 420 NaF came from the periodic table Step 2 convert 200 mL into Liters by moving the decimal 3 places to the left 200ml 200L Step 3 Put the moles over liters to nd the molar concentration 00598 mol NaF 200L 0299 molL Dilution Calculate the volume in mL of 600 M sulfuric acid that is needed to prepare 200L of a 0200 M solution of H2S04 Step 1 Find out what your have and what youre are missing MC 600M MD 200M VC VD 200L Step2 Since we know that the dilution formula is MC X VC MD XVD we can set up and equation to where the out two known elements are on top and the 1 element we do know is on bottom Mdil X Vdil Mconc Mdil 0200M X 200L 600M 00667 L Step3 since the problem calls for the units to be in mL convert Liters to mL by moving the decimal 3 places to the right 00667L 667 mL Stiochiometrv Calculations Calculate the mass of silver chloride formed in the reaction of 145 ml of 0123 M magnesium chloride and excess silver nitrate Step 1 Start by forming a balanced equation MgCl2 aq AgNO3 aq 9 AgCls MgN32 aq MgCl2aq 2AgNO3 aq 9 2AgCls MgNO32 aq Step 2 Convert 145mL to moles by rst converting mL to L and then L to mol taking note that they provide us with 0123 M 145mL 145L MgCl2 145L MgCl2 X 0123Mol1L 00178mol Step 3 Convert MgCl mols to AgCl mols 00178 mol MgCl2 X 2 mol AgCll Mol MgCl 00356 mol AgCl Step 4 Covert AgCl moles to AgCl grams 00356 mol AgCl X 1433 AgCl 1 mol AgCl 511g AgCl Titration Calculate the molarity of a sodium hydroxide solution if 452 mL of the solution is neutralized by 190 mL of a 0340 M solution of sulfuric acid Step 1 Start by forming a balanced equation NaOH H2SO4 9 H20 Na2SO4 2NaOH H2SO4 9 2H20 Na2SO4 Step 2 Convert 190 mL to moles by rst converting mL to L and then L to mol taking not that they provide us with 0340 M 190 mL 0019L 0019L J H2SO4 X 0340 M 1 L 646 M01 Step 3 Convert H2SO4 mols to NaOH moles 646 mol H2SO4 X 2 mols NaOH 1 mol H2SO4 1292 mol NaOH Step 4 Find the molarity by taking the moles and divide by the given volume in liters 1292 Mol NaOH 0452L 286 molL Gravimetric Analysis Calculate the molarity of Cl ions in a 2500 mL solution if addition of excess silver nitrate yielded 134 g of silver chloride Step 1 Form an equation C1 Ag 9 AgCl Step 2 Convert 134 g of AgCl into Moles AgCl 134g AgCl X lmol AgCl 14332g AgCl 000935 mol AgCl Step 3 Convert moles AgCl to moles Cl 000935 mol AgCl X 1mol Cl1 mol AgCl 000935 mol Cl Step 4 Find the molarity by taking the moles and divide by the given volume in liters 000935 mol Cl 250 L 00374 MolL
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