Week of 2/22 Notes
Week of 2/22 Notes CHEM - 10060 - 001
Popular in GENERAL CHEMISTRY I
Popular in Chemistry
This 0 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 29 views. For similar materials see GENERAL CHEMISTRY I in Chemistry at Kent State University.
Reviews for Week of 2/22 Notes
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 02/28/16
Notes for February 22 23rd 25th amp 26th REVIEW OF ACIDSBASES Acids are molecular form ions in aqueous solutions produce H30 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 OHltfgt 2H20 Neutralization Reactions HCI aq NaOH aqgt NaCI aq H20 I v salt ionic compound BREAKDOWN OF REACTION Molecular Equation HCI aq NaOH aq l NaCI aq H20 I Total Ionic Equation H aq CIaq Na aq OHaq Na aq Cl aq Soluble salt dissociated neutral molecule intact Net Ionic Equation H aq OH aq l H20 I Spectator Ions C 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 gtltWeak Acids do not dissolve all the way we keep it together in the total ionic equation next example shows this ACIDS amp CARBONATES Acids react with a carbonate C03 to produce and release C02 gas Example Molecular Equation HCN aq KHC03 aq I H20 I C02 9 Kc N a q Weak Acid Example of a carbonate C02 Gas Total Ionic Equation HCN aq K aq HC03 aq g H20 I coz g K aq CN aq Weak acid stays together Net ionic Equation HCN aq HC03 aq l H20 I C02 g CN aq Spectator Ions K An acid base indicator used to track a neutralization reaction AcidBase reaction complete when all H amp 0H react titration is the proper word Example 5000 mL of an H2S04 solution mixed with 3542 mL of NaOH with a molarity of 152 MWhat is the molarity of the H2S04 1 Set up balanced equation H2S04 aq 2Na0H aq 2H20 l Na2S04 aq 239 Find mols of H2S04 3542 ml NaOH x 152 mol NaOHlOOOml x 1 mol HZSO42 mol NaOH 269 x 10quot3 mol H2504 3 Calculate Molarity using newfound mols H2504 M 269x10quot305000 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 Oxidation Reduction atom loses an electron a tomo gains an electron a39quot39 Remember 0 L RI G Oxidatioiniils Losing Reduction s Gaining Redox Reaction reaction that involves electron transfer Na gives an electron to Cl l NaCl NaCl Cl is reducing lt Na is oxidizing It is also the Reducing Agent because it allows for the reduction in CI to happen is also considered the Oxidizing Agent because it allows for Na to oxidize 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 ln covalent bonds oxidations numbers re ect 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 710 important wise RULES FOR OXIDATION NUMBERS Pure Element O 0 Ca 5 O Monatomic lon the charge on the atom o Ca2 2 0 02 2 Oxygen in compounds except peroxides amp uorine 2 0 Oxygen in peroxides X202 1 0 H202 Hydrogen bonded to nonmetals except Boron 1 0 HCI 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 0 Ca2O 2 O Polyatomic lons sum of oxidation numbers Basic Rules Oxidizing Reducing loses electrons gains electrons is oxidized is reduced becomes the reducing becomes the oxidizing agent agent lncreases its oxidation decreases its oxidation number number BALANCING REDOX REACTIONS ln 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 amp total number of electrons transferred through the reaction This is usually shown through half reactions EXAMPLE Redox Reaction Zn 5 Cu2 aq l Zn2aq Cu 5 One side Zn 5 l Zn2 in solution Half Reaction Zn 5 l Zn2aq 2 electrons Oxidation Numbers 0 l O Another portion Cu2aq l Cu 5 Half Reaction Cu2aq 2 electrons l Cu 5 Oxidation Numbers 0 l O 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 elementsspecies 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 reacUons Important 7 Steps Calculate Oxidation Numbers Separate the reaction out into two half reactions Balance all atoms except Oxygen and Hydrogen Balance the Oxygen by adding HZO to the other side Balance the H atoms in the H20 you just added by adding H to the other side Balance the charge by adding the electrons to the other side a You might have to multiply the halfreactions by integers to balance electrons 7 Add the two half reactions together so the electrons canceL WBWN Ch 8 Check the charges In Basic Solutions a few more steps are needed 1 Complete steps 17 from above 2 Add OH to both sides of the nal equation to neutralize H and cancel out H20 as necessary 3 Check the Charges Example I2s HNO3 aq l HO3 aq N02 g H20 l Ox 5 0 1 6 1 6 4 2 2 BY oxidation rules we know all these charges For 0 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 nd N and l l2s HNo3 aq HO3 aq N02 g H20 I 5 5 4 I2 D HO3 aning HNO3 D N02 that it Balance the l2 by Add the H205 mUIt39p39y39 g HNo3 g N02 H20 I2 D 2HO3 Add the corresponding H s H HNo3 g N02 H20 6H20 392 l 2quot 03 Add the electrons Add the H for the H20 H HNO3 1 e I N02 6H20 l2 2HIO3 10H H20 6 Oxygens so 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 l e I N02 H20 Then combine the equations and cancel out any of the same things on both sides to end up with l2 s 10HNO3 aq l 2HO3 aq 10 N02 g 4 H20 l