4 - Rxn and solution lecture

4 - Rxn and solution lecture - Chemical Reactions&...

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Unformatted text preview: Chemical Reactions & Solution Stoichiometry Zumdahl Chapter 4 Water Water Dissolve many different substances Salt Sugar Water Vshaped "bent" HOH angle 105 degrees Solution Process Solution Solute Substance dissolved in a liquid Dissolving medium in a solution Solute and solvent Solvent Solution Ions in Solution Electrical conductivity ability to conduct electrical current Solution is the switch Ions present, No ions present, switch is "off" switch is "on" Conductivity Electrolytes: Substances that conduct an electric current when dissolved in water Movement of ions toward electrodes Chemical reactions at electrodes Electrolyte Strength Strong Weak Ionizes completely when dissolved in water 90% or more ionized Only partly ionized in water Less than 90% ionized Does not produce ions when dissolved in water Nonelectrolyte Electrolyte Strength Strong NaCl(s) Weak CH3CO2H(l) Na+(aq) + Cl(aq) CH3CO2 (aq) + H+(aq) Glucose(aq) Nonelectrolye Glucose(s) Molarity Number of moles of solute per liter of solution. Units: mol/L moles of solute molarity = liters of solution mol M = V Molarity as a Conversion Factor For the balanced equation: aA + bB cC + dD Volume of solution of A Moles of A Moles of B Volume of solution of B GIVEN Use molarity as a conversion factor. Use coefficients to find mole ratios. Use molarity as a conversion factor. FIND Molarity Example What is the molarity of a solution made by dissolving 2.355 g of sulfuric acid (H2SO4) in water and diluting to a final volume of 50.0 mL? Molar mass of H2SO4 is 98.1 g/mol. Molarity Example What is the molarity of a solution made by dissolving 2.355 g of sulfuric acid (H2SO4) in water and diluting to a final volume of 50.0 mL? Molar mass of H2SO4 is 98.1 g/mol. Know mass of H2SO4 = 2.355g Volume of solution = 50.0 mL Molarity Example Know mass of H2SO4 = 2.355g Volume of solution = 50.0 mL Molar mass of H2SO4 is 98.1 g/mol. Moles H 2SO 4 Molarity = Liters of solution Molarity Example Know mass of H2SO4 = 2.355g Volume of solution = 50.0 mL Molar mass of H2SO4 is 98.1 g/mol. = 0.0240 mol H 2SO 4 mol H 2SO 4 1 mol of H 2SO 4 = (2.355 g H 2SO 4 ) 98.1 g H 2SO 4 Molarity Example Know mass of H2SO4 = 2.355g Volume of solution = 50.0 mL Molar mass of H2SO4 is 98.1 g/mol. = 0.0240 mol H 2SO 4 mol H 2SO 4 1 mol of H 2SO 4 = (2.355 g H 2SO 4 ) 98.1 g H 2SO 4 1L Liter of solution = (50.0 mL) = 0.0500 L 1000 mL Molarity Example mol H 2SO 4 1 mol of H 2SO 4 = (2.355 g H 2SO 4 ) 98.1 g H 2SO 4 = 0.0240 mol H 2SO 4 1L Liter of solution = (50.0 mL) = 0.0500 L 1000 mL 0.0240 mol H 2SO 4 Molarity = = 0.480 M 0.0500 L Dilution Standard Solution Dilution + = Dilution Dilution Moles of solute after dilution = moles of solute before dilution number of moles = Molarity (mol/L) X Volume (L) Dilution number of moles = Molarity (mol/L) X Volume (L) moles of solute = M1V1 = M 2 V2 Dilution number of moles = Molarity (mol/L) X Volume (L) moles of solute = M1V1 = M 2 V2 1 V M 2 = M1 V 2 Dilution Example What is the final concentration if 75 mL of a 3.5 M glucose solution is diluted to a volume of 450 mL? 1V1 M M2 = V2 Dilution Example What is the final concentration if 75 mL of a 3.5 M glucose solution is diluted to a volume of 450 mL? M1 = 3.5 M Know: V1 = 75 mL V2 = 450 mL Dilution Example Know: M1 = 3.5 M V1 = 75 mL V2 = 450 mL 1V1 (3.5 M glucose)(75 mL) M M2 = = = 0.58 M glucose 450 mL V2 Dilution number of moles = Molarity (mol/L) X Volume (L) moles of solute = M1V1 = M 2 V2 C1V1 = C 2 V2 Precipitation Reactions Solubility Table Nitrates Chlorides Sulfates (Handout) soluble soluble, except Ag+, Hg22+ Sodium Potassium Ammonium soluble, except Ca2+, Ba2+ Hg2+, Pb2+, Ag+ soluble soluble soluble Predicting Precipitation cation1 + anion1 + cation2 + anion2 Predicting Precipitation cation1 + anion1 + cation2 + anion2 2AgNO3(aq) + Na2CO3(aq) Predicting Precipitation cation1 + anion1 + cation2 + anion2 2AgNO3(aq) + Na2CO3(aq) Ag2CO3(s) + 2NaNO3(aq) Net Ionic Equations Pb(NO3)2(aq) + 2KI(aq) 2KNO3(aq) +PbI2(s) Net Ionic Equations Pb(NO3)2(aq) + 2KI(aq) 2KNO3(aq) +PbI2(s) Pb2+(aq) + 2NO3(aq) + 2K+(aq) + 2I(aq) 2K+(aq) + 2NO3(aq) + PbI2(s) Net Ionic Equations Pb(NO3)2(aq) + 2KI(aq) 2KNO3(aq) +PbI2(s) Pb2+(aq) + 2NO3(aq) + 2K+(aq) + 2I(aq) 2K+(aq) + 2NO3(aq) + PbI2(s) Net Ionic Equations Pb(NO3)2(aq) + 2KI(aq) 2KNO3(aq) +PbI2(s) Pb2+(aq) + 2NO3(aq) + 2K+(aq) + 2I(aq) 2K+(aq) + 2NO3(aq) + PbI2(s) NET IONIC EQUATION: Pb2+(aq) + 2I(aq) PbI2(s) Net Ionic Equations KOH(aq) + HNO3(aq) H2O(l) + KNO3(aq) Net Ionic Equations KOH(aq) + HNO3(aq) H2O(l) + KNO3(aq) Ionic Equation: K+(aq) + OH(aq) + H+(aq) + NO3(aq) H2O(l) + K+(aq) + NO3(aq) Net Ionic Equations KOH(aq) + HNO3(aq) H2O(l) + KNO3(aq) Ionic Equation: K+(aq) + OH(aq) + H+(aq) + NO3 (aq) H2O(l) + K+(aq) + NO3 (aq) Net Ionic Equations KOH(aq) + HNO3(aq) H2O(l) + KNO3(aq) Ionic Equation: K+(aq) + OH(aq) + H+(aq) + NO3 (aq) H2O(l) + K+(aq) + NO3 (aq) NET IONIC EQUATION: OH(aq) + H+(aq) H2O(l) Precipitation Stoichiometry Precipitation Stoichiometry Aqueous solutions Na2SO4 and Pb(NO3)2 Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed. are mixed, PbSO4 precipitates. Precipitation Stoichiometry Precipitation Stoichiometry Precipitation Stoichiometry Precipitation Stoichiometry 1.25 L x 0.0500 mol Pb2+ = 0.0625 mol Pb2+ L 2.00 L x 0.0250 mol SO42 = 0.0500 mol SO42 L Precipitation Stoichiometry Precipitation Stoichiometry Precipitation Stoichiometry 0.0500 mol PbSO4 x 303.3 g PbSO4 = 15.2 g 1 mol PbSO4 Precipitation Stoichiometry Arrhenius Acid and Base Acid forms hydrogen ions (H+) in water HCl H+ + Cl HCl + H2O H3O+ + Cl forms hydroxide ions (OH)in water NaOH Na+ + OH Base BronstedLowery Acid and Base Acid Gives a hydrogen ion (H+), proton Proton donor Base Accepts a hydrogen ion (H+), proton Proton acceptor AcidBase Reaction Neutral Base Reaction Adding H+ creates positive charge H H H H Negatively Base Reaction .. .. .. .. Neutralization Reaction Neutralization Reaction What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? Neutralization Reaction Neutralization Reaction Neutralization Reaction Neutralization Reaction Key Titration Terms Titrant solution of known concentration used in titration. Analyte substance being analyzed. Equivalence point enough titrant added to react exactly with the analyte. Endpoint the indicator changes color so you can tell the equivalence point has been reached. Titration Titration Example When a 5.00 mL sample of household vinegar is titrated, 44.5 mL of 0.100 M NaOH solution is required to reach the end point. What is the acid concentration of vinegar? Titration Example When a 5.00 mL sample of household vinegar is titrated, 44.5 mL of 0.100 M NaOH solution is required to reach the end point. What is the acid concentration of vinegar? Calculate moles of OH 0.100 mol OH- (44.5 mL OH-) = 4.45 x 10-3 mol OH 1000 mL Titration Example When a 5.00 mL sample of household vinegar is titrated, 44.5 mL of 0.100 M NaOH solution is required to reach the end point. What is the acid concentration of vinegar? Calculate moles of H+ from vinegar mol H + 1 -3 (4.45 x 10 mol OH-) = 4.45 x 10-3 mol H + 1 mol OH- Titration Example When a 5.00 mL sample of household vinegar is titrated, 44.5 mL of 0.100 M NaOH solution is required to reach the end point. What is the acid concentration of vinegar? Calculate concentration of CH3CO2H -3 1 (4.45 x 10 mol CH3 CO 2 H) = 0.890M CH3 CO 2 H 0.00500 L Redox Reactions Oxidationreduction reactions One or more electrons are transferred Charges on atoms change Redox Reactions Mg(s) + I2(g) MgI2(s) Left side Mg oxidation state = 0 I2 oxidation state = 0 Mg oxidation state = +2 I oxidation state = 1 Right side Oxidation State Rules Element in the natural form is zero Element in a monatomic ion = charge of ion Hydrogen = +1 when bonded to nonmetals = 1 when bonded to metals Oxygen = 2 except in peroxides = 1 Sum of elements in compound = 0 Polyatomic ion = charge of ion Metals are positive, nonmetals are negative Oxidation State Examples NO2 Oxidation State Examples NO2 Oxygen = 2 Nitrogen = ? Oxidation State Examples NO2 Oxygen = 2 Nitrogen = ? Compound = 0 = 2(2) + N N = +4 Oxidation State Examples SO3 Oxidation State Examples SO3 Oxygen = 2 Total charge from oxygen = 3(2) = 6 Oxidation State Examples SO3 Oxygen = 2 Total charge from oxygen = 3(2) = 6 Sulfur = +6 Compound = (+6) + (6) = 0 Oxidation State Examples Cr2O72 Oxidation State Examples Cr2O72 Oxygen = 2 Total charge from oxygen = 7(2) = 14 Oxidation State Examples Cr2O72 Oxygen = 2 Total charge from oxygen = 7(2) = 14 Chromium = ? Oxidation State Examples Cr2O72 Oxygen = 2 Total charge from oxygen = 7(2) = 14 Chromium = 14 + (2) = +12 Two chromium = +12 Each chromium = +6 Redox Reaction 2 Fe(s) + 3O2(g) Fe2O3(s) Oxidation numbers Fe = 0 O2 = 0 Fe2O3 Oxygen = 2 Iron = +6 left side left side 3 Oxygens = 3(2) = 6 2 Irons = 2(x) = +6 x = 3 Redox Reaction 2 Fe(s) + 3O2(g) Fe2O3(s) oxidation Fe = 0 goes to Fe = +3 O2 = 0 goes to O = 2 reduction Oxidation/Reduction LEO says GER Lose electrons Gain electrons oxidation reduction OILRIG Oxidation is loss Reduction is gain Oxidation/Reduction Oxidation A2 A A A+ A A A+ A2+ + electron + electron + electron + electron Reduction Redox Example +1 charge 0 charge +2 electrons = reduced 0 charge Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) 2 electrons = oxidized +2 charge Redox Example 0 charge 1 charge +2 electrons = reduced 1 charge 2I(aq) + Br2(aq) I2(aq) + 2Br(aq) 2 electrons = oxidized 0 charge Redox Example Reducing Agent Oxidizing agent Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) 2I(aq) + Br2(aq) I2(aq) + 2Br(aq) Oxidizing agent Reducing Agent Redox Reaction Reducing Agent Loses one or more electrons Causes reduction Becomes more positive (less negative) Gains one or more electrons Causes oxidation Becomes more negative (less positive) Oxidizing Agent Redox Examples Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) Oxidation number Fe(s) Cu2+(aq) Fe2+(aq) Cu(s) = = = = Redox Examples Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) Oxidation number Fe(s) Cu2+(aq) Fe2+(aq) Cu(s) = = = = 0 Redox Examples Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) Oxidation number Fe(s) Cu2+(aq) Fe2+(aq) Cu(s) = = = = 0 2+ Redox Examples Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) Oxidation number Fe(s) Cu2+(aq) Fe2+(aq) Cu(s) = = = = 0 2+ 2+ Redox Examples Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) Oxidation number Fe(s) Cu2+(aq) Fe2+(aq) Cu(s) = = = = 0 2+ 2+ 0 Redox Examples Fe(s) + Cu2+(aq) Fe2+(aq) + Cu(s) Fe(s) Cu2+(aq) Fe2+(aq) Cu(s) = = = = 0 2+ 2+ 0 What is reduced? Oxidized? Reducing agent? Oxidizing agent? Halfreaction Method (Acidic) MnO 4 + Cl + H o Cl 2 + Mn Assign oxidation states + 2+ Halfreaction Method (Acidic) MnO 4 - + Cl- + H + o Cl2 + Mn 2+ Assign oxidation states MnO 4 +7 -2 - + Cl -1 - + H +1 + Cl 2 0 + Mn +2 2+ Halfreaction Method (Acidic) MnO 4 + Cl + H o Cl2 + Mn Reduction Halfreaction + 2+ MnO 4 Oxidation Halfreaction - o Mn - 2+ 2Cl o Cl2 Halfreaction Method (Acidic) MnO 4 - + Cl- + H + o Cl2 + Mn 2+ MnO 4 o Mn + 4H 2 O 2+ 2Cl o Cl2 - Halfreaction Method (Acidic) MnO 4 - + Cl- + H + o Cl2 + Mn 2+ 8H + MnO 4 o Mn + 4H 2 O + 2+ 2Cl o Cl2 - Halfreaction Method (Acidic) MnO 4 - + Cl- + H + o Cl2 + Mn 2+ 8H + MnO 4 o Mn + 4H 2 O + 2+ 2Cl o Cl2 2 0 - 8+ 1 2+ 0 Halfreaction Method (Acidic) MnO 4 - + Cl- + H + o Cl2 + Mn 2+ 8H + MnO 4 o Mn + 4H 2 O + 2+ 2Cl o Cl2 2 2 0 0 - 8+ 1 7+ 2+ 0 2+ Halfreaction Method (Acidic) MnO 4 - + Cl- + H + o Cl2 + Mn 2+ 8H + MnO 4 + 5e + - Mn + 4 H 2 O 2+ 2Cl o Cl2 2 2 0 0 - 8+ 1 7+ 2+ 0 2+ 8H + + MnO 4 - + 5e - Mn 2+ + 4 H 2 O 2Cl- Cl2 2e - + Halfreaction Method (Acidic) MnO 4 - + Cl- + H + o Cl2 + Mn 2+ 8H + MnO 4 o Mn + 4H 2 O + 2+ 2Cl o Cl2 2 2 0 0 - 8+ 1 7+ 2+ 0 2+ 8H + + MnO 4 - + 5e - Mn 2+ + 4 H 2 O 2Cl- Cl2 2e - + X X Halfreaction Method (Acidic) MnO 4 - + Cl- + H + o Cl2 + Mn 2+ 8H + MnO 4 + 5e + - Mn + 4 H 2 O 2+ 2Cl - Cl2 2e + - 16H + X + 2MnO 4 - +2 e - 10 2Mn 2+ + 8H 2 O X 5 5Cl 2 10e - + 10Cl- Halfreaction Method (Acidic) MnO 4 - + Cl- + H + o Cl2 + Mn 2+ 16H + + 2MnO 4 - + 10e - 2Mn 2+ + 8H 2 O 10Cl16H+ + 2MnO4 + 10Cl 5Cl 2 10e - + 2Mn2+ + 5 Cl2 + 8H2O Halfreaction Method (Basic) I (aq) + OCl (aq) o I 2 + Cl + H 2 O - Halfreaction Method (Basic) I- (aq) + OCl- (aq) o I 2 + Cl- + H 2 O (Ox) I - o I2 - (Red) OCl o Cl + H 2 O Halfreaction Method (Basic) I- (aq) + OCl- (aq) o I 2 + Cl- + H 2 O (Ox) 2I - o I2 - (Red) OCl o Cl + H 2 O Halfreaction Method (Basic) I- (aq) + OCl- (aq) o I 2 + Cl- + H 2 O (Ox) 2I - o I2 (Red) 2H + + OCl- o Cl- + H 2 O Halfreaction Method (Basic) I (aq) + OCl (aq) o I 2 + Cl + H 2 O (Ox) 2I - - - - I 2 2e + 0 - - charge 2+ 2Cl + H 2 O 10 - (Red) 2H + OCl + 2echarge 2+ 12- Halfreaction Method (Basic) (Ox) + 2I - I 2 2e + - - (Red) 2H + OCl + 2e- Cl + H 2 O 2 I- + 2H + + OCl- o Cl- + H 2 O + I 2 Halfreaction Method (Basic) I + 2H + OCl o Cl + H 2 O + I 2 + - 2OH - + 2I- + 2H + + OCl- o I 2 + Cl- + H 2 O + 2OH - Halfreaction Method (Basic) 2OH - + 2I- + 2H + + OCl- o I 2 + Cl- + H 2 O + 2OH - 2H 2 O + 2I + OCl - - o I 2 + Cl + H 2 O + 2OH - - Halfreaction Method (Basic) 2H 2 O + 2I + OCl - o - I 2 + Cl + H 2 O + 2OH - - H 2 O + 2I + OCl o I 2 + Cl + 2OH - - ...
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This note was uploaded on 11/22/2010 for the course CHM 121 taught by Professor Scheru during the Spring '09 term at Oakton.

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