Week of 2/22 Notes
Week of 2/22 Notes CHEM - 10060 - 001
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This 10 page Class Notes was uploaded by Nick Manning on Sunday February 28, 2016. The Class Notes belongs to CHEM - 10060 - 001 at Kent State University taught by TBA in Fall 2015. Since its upload, it has received 18 views. For similar materials see GENERAL CHEMISTRY I in Chemistry at Kent State University.
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Date Created: 02/28/16
nd rd th th Notes for February 22 , 23 , 25 & 26 REVIEW OF ACIDS/BASES Acids- are molecular – form ions in aqueous solutions – produce H3O+ Bases- are ionic or molecular – form ions in aqueous solution – produce OH- Aqueous solution means what? Solution where the solvent (thing that does the dissolving) is water H30+ + OH- 2H20 Neutralization Reactions HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) salt ionic compound BREAKDOWN OF REACTION Molecular EquatioHCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) Total Ionic EquatiH+ (aq) + Cl-(aq) + Na+ (aq) + OH-(aq) Na+ (aq) + Cl- (aq) + H2O (l) Soluble salt dissociated neutral molecule intact Net Ionic Equation) H+ (aq) + OH- (aq) H2O (l) Spectator Ions) Cl-, Na+ *There are some equations where there will be no spectator ions or even only one spectator ion; I also highly recommend looking over solubility rules.* *Weak Acids do not dissolve all the way; we keep it together in the total ionic equation; next example shows this ACIDS & CARBONATES - Acids react with a carbonate (CO3) to produce and release CO2 gas Example: Molecular Equation) HCN (aq) + KHCO3 (aq) H2O (l) + CO2 (g) + KCN (aq) Weak Acid!! Example of a carbonate CO2 Gas!! Total Ionic EquationHCN (aq) + K+ (aq) + HCO3- (aq) H2O (l) + CO2 (g) + K+ (aq) + CN- (aq) Weak acid stays together Net ionic Equation) HCN (aq) + HCO3- (aq) H2O (l) + CO2 (g) + CN- (aq) Spectator Ions) K+ An acid base indicator used to track a neutralization reaction. Acid-Base reaction complete when all H+ & OH- react. *titration is the proper word* Example: 50.00 mL of an H2SO4 solution mixed with 35.42 mL of NaOH with a molarity of .152 M. What is the molarity of the H2SO4? 1 : Set up balanced equation H2SO4 (aq) + 2NaOH (aq) 2H20 (l) + Na2SO4 (aq) nd 2 : Find mols of H2SO4 35.42 ml NaOH x (152 mol NaOH/1000ml) x (1 mol H2SO4/2 mol NaOH) = 2.69 x 10^-3 mol H2SO4 3 : Calculate Molarity using newfound mols H2SO4 M= (2.69x10^-3)/.05000 L = .0538 M REDOX REACTIONS Redox = reduction + oxidation We already know ions are formed by losing or gaining an electron Oxidation) when an atom loses one or more electron DOES NOT MEAN REACTION WITH OXYGEN SO FORGET ABOUT THAT Reduction) gain of one or more electron Remember: OIL RIG Oxidation Is Losing Reduction Is Gaining Redox Reaction- reaction that involves electron transfer Na gives an electron to Cl Na+Cl- (NaCl) Na is oxidizing. Cl is reducing. It It is also the the Oxidizingdered Reducing Agent because it Agent because allows for Na to reduction in Clhe oxidize. to happen. Reducing agent- the one being OXIDIZED. It allows for reduction to happen Oxidizing agent- the one being REDUCED. It allows for oxidation to happen *THEY ARE OPPOSITES. The one oxidizing = Reducing agent and the one reducing = Oxidizing agent.* Electron transfer is tracked by Oxidation Numbers. OXIDIZATION NUMBERS -Charge an atom would have if all reactions involved complete electron transfer, so if all bonds were kind of like ionic bonds. - In covalent bonds, oxidations numbers reflect electron distribution. - Ox. Numbers not really a positive or negative, they’re really just the distribution of electrons on the elements. - There are rules for calculating the oxidation numbers of elements and compounds which are pretty important to know (like a 7/10 important wise). RULES FOR OXIDATION NUMBERS Pure Element = 0 o Ca (s) 0 Monatomic Ion = the charge on the atom o Ca2+ = 2+ o O2- = -2 Oxygen in compounds (except peroxides & fluorine) = -2 Oxygen in peroxides (X2O2) = -1 o H2O2 Hydrogen bonded to nonmetals (except Boron) = +1 o HCl = The Hydrogen = +1 Hydrogen bonded to metals or Boron = -1 o HNa = The Hydrogen = -1 Fluorine in all compounds = -1 Halogens except Oxygen = -1 Compounds = sum of the oxidation numbers o Ca(+2)O(-2)= 0 Polyatomic Ions = sum of oxidation numbers Basic Rules: Oxidizing Reducing: -loses electrons -gains electrons - is oxidized -is reduced -becomes the reducing -becomes the oxidizing agent agent -Increases its oxidation -decreases its oxidation number number BALANCING REDOX REACTIONS In Redox Reactions, the number of electrons lost should equal the number of electrons gained. Reactions must account for the both number and type of each atom present & total number of electrons transferred through the reaction. This is usually shown through half reactions. EXAMPLE) Redox Reaction) Zn (s) + Cu2+ (aq) Zn2+(aq) + Cu (s) One side) Zn (s) Zn2+ in solution Half Reaction) Zn (s) Zn2+(aq) + 2 electrons Oxidation Numbers) 0 0 Another portion) Cu2+(aq) Cu (s) Half Reaction) Cu2+(aq) + 2 electrons Cu (s) Oxidation Numbers) 0 0 - Add together the two half reactions to show overall reactions - Electrons must cancel - Total electrons gained = total electrons lost To write redox reactions: - Identify actual elements/species involved (aka write the total ionic equation) - Assign the appropriate oxidation numbers - Only species that change are included in the reaction To balance redox reactions with half reactions Important 7 Steps 1. Calculate Oxidation Numbers 2. Separate the reaction out into two half reactions 3. Balance all atoms except Oxygen and Hydrogen 4. Balance the Oxygen by adding H2O to the other side 5. Balance the H atoms in the H2O you just added by adding H+ to the other side 6. Balance the charge by adding the electrons to the other side. a. You might have to multiply the half-reactions by integers to balance electrons 7. Add the two half reactions together so the electrons cancel. 8. Check the charges! In Basic Solutions a few more steps are needed: 1. Complete steps 1-7 from above 2.Add OH- to both sides of the final equation to neutralize H+, and cancel out H2O as necessary 3.Check the Charges! Example) I2(s) + HNO3 (aq) HIO3 (aq) + NO2 (g) + H2O (l) Ox. #s) 0 +1 ? -6 +1 ? -6 ? -4 +2 -2 *BY oxidation rules we know all these charges. For O, we multiply -2 by the number of Oxygens that are present. Each separate compound has a neutral overall oxidation, so we can use what we know to find N and I. I2(s) + HNO3 (aq) HIO3 (aq) + NO2 (g) + H2O (l) +5 +5 +4 I2 HIO3es from 0 to +5, meaning HNO3 NO2idized and N goes from +5 to +4, meaning that it was reduced. Now we Balamake our half reactions. Add the H2Os multiplying HNO3 NO2 + H2O I2 2HIO3 Add the corresponding H’s 6 Oxygens so… H+ + HNO3 NO2 + H2O 6H2O + I2 2HIO3 Add the electrons Add the H+ for the H2O H+ + HNO3 + 1 e- NO2 6H2O + I2 2HIO3 + 10H+ + H2O Add the electrons to balance charge Now you have to notice that one side adds ten electrons and one side adds 1 electron… you can’t have this. If one is giving ten electrons one can’t be receiving only one. So you have to multiply the second equation by ten. 10 (H+ + HNO3 + 1 e- NO2 + H2O) Then combine the equations and cancel out any of the same things on both sides to end up with: I2 (s) + 10HNO3 (aq) 2HIO3 (aq) + 10 NO2 (g) + 4 H2O (l)