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by: Carmela Kilback


Carmela Kilback
GPA 3.92

William Zoller

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William Zoller
Class Notes
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This 24 page Class Notes was uploaded by Carmela Kilback on Wednesday September 9, 2015. The Class Notes belongs to CHEM 142 at University of Washington taught by William Zoller in Fall. Since its upload, it has received 18 views. For similar materials see /class/192554/chem-142-university-of-washington in Chemistry at University of Washington.




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Date Created: 09/09/15
Chapter 4 The Major Classes of Chemical Reactions 41 The Role ofWater as a Solvent 42 Precipitation Reactions and AcidBase Reactions 43 Oxidation Reduction Redox Reactions 44 Counting Reactants and Products in Precipitation AcidBase and Redox Processes 45 Reversible Reactions An Introduction to Chemical Equilibrium The Role of Water as a Solvent The solubility of Ionic Compounds Electrical conductivity 7 The ow of electrical current in a solution is a measure of the solubility of ionic r a measurement of the presence of ions in solution Electrolyte 7 A substance that conducts a current When dissolved in Water Soluble ionic compound dissociate completely and may conduct a large current and are called Strong Electrolytes NaClltsgt H200 Q Nay Cl m When Sodium Chloride dissolves into Water the ions become solvated and are surrounded by Water molecules These ions are called aqueous are free to move through out the solution and are conducting electricity or helping electrons to move through out the solution Fig 41 P135 Electrical Conductivity of lonic Solutions A mmmacmcsg m a Wmquot WM caudwl muenl W m a mu an m Electron Distribution in Molecules of H2 and H20 67 A A B VIE 7 5 3 2 x D 6 w Determining Moles of Ions in Aqueous Solutions of Ionic Com ounds I Problem How many moles of each ion are in each of the following a 4 0 moles of sodium carbonate dissolved in Water b 46 5 g of rubidium uoride dissolved in Water 0 5 14 x 1021 formula units of iron III chloride dissolved in Water d 75 0 ml of 0 56M scandium bromide dissolved in Water e 7 8 moles of ammonium sulfate dissolved in Water H20 a Nazcos S ta 2 Naa1 CO W moles ofNa 4 0 moles NaZCO3 x 2 quot 01 Nai 1 mo NaZCO3 Determining Moles orons in Aqueous Solutions oronic Compounds II o b Rme L qum F1 M molesobeF 465gRbe10447 gRbF H20 0 FeCls ts gt WSW 3 C11 moles of FeCl3 9 32 x 1021 formula units 1 mol FeCl3 6 022 x 1023 formula units FeCl3 0 0155 mol FeCl3 3 mol Cl39 moles ofCl 0 0155 mol FeCl3 x lmol Fecl3 Determining Moles of Ions in Aqueous Solutions of Ionic Compounds III Hp d ScBrs ts scal 3 Br W Converting from volume to moles Moles of ScBr3 75 0 ml x 1L x0 56 mol SCBr 0 042 mol ScBr3 103 ml 1 L Moles ofBr39 0 042 mol ScBr3 x 3m 1 r39 1 mol ScBr3 H o e N39H hsom 1quot 2NHW SO39Zaq Moles of NH4 7 8 moles NH42804 i o NH 2 4 The Dissolution of an Ionic Compound The Solubility 0ronic Compounds in Water Ihe solubility of Ionic Compounds in Water depends upon the relative strengths of L 39 ion in th 39 39 r A 4 L 39 39 I umui in the solvent There is a tremendous range in the solubility of ionic compounds in Water The solubility of so called insoluble compounds may be several orders of magnitude less than ones that are called soluble in Water for example Solubility ofNaCl in Water at 20 C 365 gL Solubility ofMgCl2 in Water at 20 C 542 5 gL Solubility of AlCl3 in Water at 20 C 699 gL Solubility obeCl2 in Water at 20 C 9 9 gL Solubility of AgCl in Water at 20 C 0 009 gL Solubility of CuCl in Water at 20 C 0 0062 gL The Solubility ofCovalent Compounds in Water 39Ihe covalent compounds that are very soluble in Water are the ones With OH group in them and are called Polar and can have strong A po ar vviLu water quot r r such as table sugar sucrose CuHuOl l beverage alcohol ethanol CZHSOH and ethylene glycol CZH OZ in antifreeze H c OH MethanolMethyl Alcohol Other covalent compounds that do not contain a polar center or the OH group are considered NoniPolar and have little or no interactions With Water molecules Examples are the hydrocarbons in Gasoline and Oil This leads to the obvious problems in Oil spills Where the oil Will not mix With the Water and forms a layer on the surface Octane CBH18 and or Benzene C H Acids A group of Covalent molecules which loose Hydrogen ions to water molecules in solution When gaseous hydrogen Iodide dissolves in Water the attraction of the oxygen atom of the Water molecule for the hydrogen atom in HI is greater that the attraction of the of the Iodide ion for the hydrogen atom and it is lost to the Water molecule to form an Hydronium ion and an Iodide ion in solution We can Write the Hydrogen atom in solution as either H7 or as HKOaq they mean the same thing in solution The presence of a Hydro en atom that is easily lost in solution is an Acid and is called an acidic solution The Water H20 could also be Written above the arrow indicating that the solvent was Water in Which the HI Was dissolved leo H200 117m I w HI HZOO H307 I 39aq H 0 HI 11729 sq The Hydrated Proton Strong Acids and the Molarity of H Ions in A ueous Solutions of Acids Problem In aqueous solutions each molecule of sulfuric acid will loose two protons to yiel two Hydronium ions an one sulfate ion What is the molarity of the sulfate and Hydronium ions in a solution prepared by dissolving 155g of concentrate sulfuric acid into sufficient water to produce 2 30 Liters of acid solution Plan Determine the number of moles of sulfuric acid quot 39 by the volume to get the molarity ofthe acid and the sulfate ion The in dtuuium39 39 wiu 39 L quot 39 Solution Two moles of H are released for every mole of acid HZSO a 2 H10 2 H30aq SOquot 1m 1 1 H so Moles st0A 7155 g st04 x W 71 58 moles stoA 1 58 mol SO 391 r 2 Malmty Of 04 2 30 1 solution Molarity of H 2 x 0 687 mol Ht 2 30 liters fh h ai ti n 39 of PbNO32 and Nal m Mwmu m m mm n rm Precipitation Reactions A solid product is formed When ever two aqueous solutions are mixed there is the possibility of forming an insoluble compound Let us look at some examples to see how we can predict the result of adding two different solutions together PbNOK W NaIW n Pb zw 2 Now Na w 1W When we add These two solutions together the ions can combine in the way they came into the solution or they can exchange partners In this case we could have Lead Nitrate and Sodium Iodide or Lead Iodide and Sodium Nitrate formed to determine which will happen we must look at the solubility tableP 141 to determine what could form The table indicates that Lead Iodide will be insoluble so a precipitate will form PbNOKZaq 2 NaIW gt PbIZ S 2 NaNOKW Precipitation Reactions Will a Precipitate form If We add a solution containing Potassium Chloride to a solution containing Ammonium Nitrate Will We get a precipitate KCIW N39lLNO3 Sq Kym Clkaq NHAQW N031 By exchanging cations and anions We see that We could have Potassium Chloride and Ammonium Nitrate or Potassium Nitrate and Ammonium Chloride In looking at the solubility table it shows all possible products as soluble so there is no net reaction KCIW N39H4NO3 W No Reaction If We mix a solution of Sodium sulfate With a solution of Barium Nitrate Will We get a precipitate From the solubility table it shows that Barium Sulfate is insoluble therefore We Will get a precipitate Nazso4W Eamon gt Basom 2 NaNOKW i mafia1 gt Agzcrodal Ewe 41 Solubility Rules for Ionic Compounds in Water Soluble Ionic Compounds Insoluble lonic Compounds 1 All common compounds of Group 1 All common metal hydroxides 1A1ionsL N 39 K et 39 a c are Insoluble except those 1 ammonium ion NHH are Group 1 1 and the larger so ub e members of Group 2A2 All on man nitrates N031 beginn39n in Ca acetates C CO and most 2 All common carbonates perchlorate olon are soluble 321a p 3 mon chlorides CF bromides P0437 are insoluble except BF and lo ides 139 aresoluble oseofGroup A a 2 Cu NH 39 exceptt ose of Ag b 4 and H92 3 All common sulfides are 4 All common sulfates 504239 are insoluble except those ol soluble except those of Group 1A1 Group 2A2 a Sr Ba and PM nd NH Predicting Whether a Precipitation Reaction Occurs I Writing Equations a Calcium Nitrate and Sodium Sulfate solutions are added together 0 ecular Equation CaN032W Na7S04 W gt CaSO Total Ionic Equation Ca1a 2 NOK39W 2 Naa sore C21304 5 2 Nag 2 Nqu 4 s NaNqu Net Ionic Equation 7 Ca m SO 4 CalSO4 s Spectator Ions are Na and N0339 b Ammonium Sulfate and Magnesium CLI 394 tnnpthpr In exchanging ions no precipitates Will be formed so there Will be no Chemical reactions occurring All ions are spectator ions Acid Base Reactions Neutralization Rxns An Acid is a substance that produces H HKO ions When dissolved in Water A Base is a substance that produces OH39 ions When dissolved in Water Acids and Bases are electrolytes and their strength is categorized in terms of their degree of dissociation in Water to make hydronium or hydroxide ions Strong acids and bases dissociate completely and are strong electrolytes Weak acids and bases dissociate Weakly and are weak electrolytes The generalized reaction between an Acid and a Base is Hxaq Moer gt we HZOU Acid Base Salt Water The Behavior of Stron and Weak Electrolytes A Strong electrolyte B Weak electrolyte a m Table 42 143 Acids Bases Wong Swong Hydrochloric HCl Sodium hydroxide NaOH Hydrobromic HBr Potassium hydroxide KOH Hydroiodoic HI Calcium hydroxide CaOH2 Nitric cid HNO3 Strontium hydroxide SrOH2 Sulfuric acid HZSOA Barium hydroxide BaOH2 Perchloric acid HC104 Weak Weak Hydro uoric HF Ammonia NH3 Phosphoric acid HKPOA Acidic acid CHKCOOH or HC2HKOZ Writing Balanced Equations for net ionic for the following Chemical reactio a Calcium Hydroxideaq and Hydrolodic acidaq b Lithium Hydroxideaq and Nitric acidaq c Barium Hydr xi eaq and Sulfuric acidaq Plan These are all strong acids and bases therefore they will make water and the corresponding salm Solution 21 CaOH1aq 2mm Cal2 W 2Hzoa Problem Write balanced chemical reactions molecular total ionic and s Ca1aq 2 OH39W 2 Hy 2 1 7W a Can 2 I W 2 H10 2 on 2 Htm 2 H10 Writing Balanced Equations for Neutrali7atinn Pamfirms ll b LiOHW HNOs an LiNo W H100 L Waq 0H an HWaq N0 an In N031 H100 c BaOH1aq sto W Base S 2 H100 Ba1aq 2 OH39aq 2 Hymn sofw a Base M 2 H10 Ba1aq 2 OH39aq 2 Hymn sofw Baso M 2 H10 I An Acidbase Titration I H aq X39aq Mquot39aq OH aq gt H200 Mquotaq X39aq Finding the Concentration of Acid from an Acid Base Titration M molL of base molar ratio volume L of acid Finding the Concentration of Base from an Acid Base Titration I Problem A titration is performed between Sodium Hydroxide and Pota ium by placing 50 00 mg of solid Potassium Hydrogenphthalate in a ask with a few drops of an indicator A buret is lledwith the base and the initial buret reading is 0 55 ml at the end ofthe titration the buret reading is 33 87 ml What is the concentration of the base Plan Use the molar mass ofKHP 2 4 2 ofmoles of the acid from the balanced chemical equation the reaction is equal molar so we know the moles of base and from the difference in the buret readings we can calculate the molarity of the base Solution HKCBHAOWq OH W KC8H40439EHHZOE Potassium Hydrogenphthalate KHC8 Finding the Concentration of Base from an Acid Base Titration II 5000mgKHP 100g moles KHP 204 2 KHP x 1000 mg 0 00024486 mol KHP 1 mol KHP Volume of base Final buret reading Initial buret reading 33 87ml 055ml33 32ml ofbase one mole of acid one mole ofbase therefore 0 00024486 moles of acid Will yield 0 00024486 moles ofbase in avolume of33 ml 7 0 00024486 moles molarity ofbase 7 0 3332 L molarity of base An Aqueous Strong AcidStrong Base Reaction on the Atomic Scale M and x Ions rsmatn in solution as spectator tans a Aqueaur saluttans at strung am and strong base are mtxea HXW Evspnrattan at water turn and otr Chemical change is tangle n H MOHaq tram HJD ta ctr Iron base turman H30 H3039aqx39aq mix 2 o M X A 2 09 MXU 0 a r a 4 t s M39aq390H aq q 7 2 Reaction that Forms Product 2e Transfer of gt electrons Mg a V A Formation of an Ionic solid ionic compound 6 quot 39quot Shiftof H 139 Electron Electron pairs electrons airs shared shared equally unequally 6 639 Cl cI H 201 B Formation of a 39 covalent compound The Redox Process I Com ound Formatro ble 43 Rules for Assigning an Oxidation Number ON General rules For an atom in its elemental form Na 02 012 etc DN 0 For a monatomic ion DN ion char 9 The sum of DN values for the atoms in a com ound equals zero The sum or DN values for the atoms in a polyatomic ion equals the ion charge mud Rules tor specific atoms or periodic table groups 0 1 in all c mpounds 1 For Group1A1 2 For Group2A2 DN 2 in all com o 3 For hydrogen DN 1 in combination with nonmetals 71 n combination with metals and boron 4 For fluorine 71 n all compounds 5 For oxygen 1 n peroxides O 2 n all other compounds except with F 6 For Group 7A17 71 in combina ion with meta s nonmetals except 0 and other halogens lower In the group Highest and Lowest oxidation numbers of Chemically reactive main group Elements nonemetals n metalloids l metals mama A A Summary of Terminology for Oxidation Reduction Redox Reactions e m Transfer X e or shift of electrons Y gains electrons Y is reduced X loses electrons X is oxidized X is the reducing agent oxidizing agent X increases in Y decreases in oxidation oxidation number number Transition Metals Possible Oxidation States VII Determining the Oxidation Number of an Element in a Com ound Problem Determine the oxidation number Ox No of each element 39 ollowing compounds 5 Iron III Chloride b Nitrogen Dioxide c Sulfuric acid Plan We apply the rules in Table 4 3 always making sure that the x No Values in a compound add up to zero and in a polyatomic ion to the ion s charge Solution 5 FeCl3 This compound is composed of monoatomic ions The Ox No of Cl39 is 71 for a total of 3 Therefore the Fe is 3 b N02 The Ox No of oxygen is 72 for a total of 4 Since the Ox No in a compound must add up to zero the Ox No ofN is 4 c st04 The Ox No ofH is 1 so the so group must sum to 2 The Ox No of each 0 is 72 for a total of8 So the Sulfur atom is 6 Recognizing Oxidizing and Reducing Agen Problem Identify the oxidizing and reducing agent in each ofthe Rx c NiOs co gt Ni cow Plan First we assign an oxidation number 0 N to each atom or ion 39 Table 4 3 quotquot 39 39 39 i contains an atom that is oxidized O N increased in the reaction The 39 L quot39 39 39 it contains an 39 O N decreased Solution39 5 Assigning oxidation numbers Zrks 2HClW chlw Hm L Recognizing Oxidizing and Reducing Agents II b Assigning oxidation numbers p I S8 12 029 8 song is the reducing agent and is the oxidizing agent c Assigning oxidation numbers I l N10 cog p N1 cow is the reducing agent and is the oxidizing agent Balancing Redox Equations in Aqueous Acid and Base Solutions ACID You may add either H H30 or water H20 to either side of the chemical equation BASE You may add either OH39 or water to either side of the chemical equation Balancing REDOX Equations The oxidation number method Step 1 Assign oxidation numbers to all elements in the equation Step 2 From the changes in oxidation numbers identify the oxidized and reduced species Step 3 Compute the number of electrons lost in the oxidation and gained in the reduction from the oxidation number changes Draw tielines between these atoms to show electron changes Step 4 Multiply one or both of these numbers by appropriate factors t make the electrons lost equal the electrons gained and use the factus as balancing coef cients Step 5 Complete the balancing by inspection adding states of matter REDOX Balancing using OX No Method I l I LHHQ i 029 inQQ I I electrons lost must electrons gained therefore multiply Hydrogen reaction by 2 and we are balanced REDOX Balancing Using OX No Method 11 2 1e39 3 l I Fe2aq Mn0439aq H307 gtFe3aq Mn2aq HZOW 5 e39 2 Multiply Fe Z amp Fe 3 by ve to correct for the electrons gained by the anganese 5 Fe1aq Mn0439aq H307 5 Fe3aq Mn2aq HZOW Make four water molecules from protons from the acid and the oxy gens from the MnOA39 this will require 8 protons or Hydronium ions This will give a total of 12 water molecules formed 5 Fe w Mn0439aq 8 Hzoyaq 5 Fe w Mn2aq 12 HZOW REDOX Balancing by HalfReaction MethodI Fe zw Mn0439a F62 Mn zw acid solution Identify Oxidation and Reduction HalfReactions Fe zm F6239 e39 oxidation halfreaction Mn0439aq a Mn2aq add H to the reactanm and that will give water as a product Mn0439aq 8H30aq 5e39 a Mn1aq 1211200 I I Sum me two hal reactims reduction halfreaction Feuw gtFezw 67 x5 Mn0439aq SHKOQW 5e39 gtMnlw 1211200 Mn0439a 8H30W5e 5Fe2aq 5Fe3a 5e39Mn2a 12H200 REDOX Balancing by HalfReaction Method II Mn0439aq sogw MnOm SOAZ39W basic solution Oxidation so 30411 2e39 Add OH39 39 39 39 and mm 39 39 39 39 to balance since we have one more oxygen on sulfate than on sul te sozlw 2 OH39W 55gt 0411 H200 2e39 Reduction Mn0439a 3e39 EE Mn02 5 Add waiei quot 39 quot oxygen lost when Mn0439 goes to MnO2 and looses two oxygen atoms Mn0439aq 2HZO 3e39 5 MnOm 4 OHW Multiply the oxidation equation by 3 to make the electrons 6 Multiply th reduction equation by 2 to make the electrons 6 and add the two 3 sozrlw 2 Mn0439aq HZOU 3 SOA39ZW 2 MnOm 2 OH39W REDOX Balancing by HalfReaction MethodIII MnOA39W SOKZ39W Mn025 042139 acidic solution Oxidation 031139 Q 042139 2 e39 ndd waiei rr an nxv en and to the product side that will remain plus the two electrons sozlw H200 Q 0411 2 HQ 2 e39 Reduction MnOA39W 3 e39 Add water to the product side to t and add Hydrogens to the reactant Mn0439aq 3 e39 4mg MnOm 2HZO a Mn025 e up the extra oxygen from Mn cpds side Multiply the oxidation equation by 3 and the reduction equation by 2 l l l L I39 L I p IIIUICLUIC 33031139 2Mn04 2H7 3 sogw 2Mno2 s H200 REDOX Balancing using Ox No Method III 7 3 639 4 Acidic Solution MnOA39m 031139 MnOm sogw l l 4 2 7 5 To balance the electrons we must multiply the sul te by 3 and the We men L 4 L by 39 quot quot midwaiei quot 2 Mn0439aq 3 sogw Hzotw 2 MnOm 3 sogw Hzom1 r U bind up the oxygen atoms must be balanced and since we have called H ion hydronium ionstherefore water will be forme 2 Mn0439aq 3 sogw2 How 2 MnOm 3 sogw 3 Hzom1 REDOX Balancing by HalfReaction MethodIV Mnom magma MnOzs 304213 basic solution balance the equation as if it were in acid and then convert it to base 2Mnom ssozlw my a gt 2Mnoz 330421 H100 To convert to base add two OH39 to each side ofthe equation 2MnOA39a ssogwn H200 2Mn02 3305M HZO120H393D1 On the reactant side the H and the OH39 cancel to give water 2MnOA39a ssogwn H200 2Mn02 3305M HZO120H393D1 Cancel out the water on each side ofthe equation and you are done 2MnOAka ssogw H200 gt 2Mnoz 330421 20H REDOX Balancing Using OX No MethodIV Zinc metal is dissolved in Nitric Acid to give Zn and the ammonium ion from the reduced Nitric acid write the balanced chemical equation HY N0339aq Zn aq NHXW Oxidation method 2 e l I Zn H7 N0339aq Zn w N39H4aq 5 8 e39 3 Multiply Zinc and Zn by 4 and ammonia by unity Since we have no oxygen 39 add 3 Walt 39 requiring 10 H on the reactant side 4 2 10 HQ Nozw p 4 my Nngw 3 H200 REDOX Balancing by HalfReaction MethodV Given Zn H307 NOKW gt my N39H m Oxidation Zn Zn 2 e39 Reduction H307 NOR 8 e39 gt NHJW H200 We will need three waters to pick up the oxygens from the nitrate ion and for the hydrogens we will need to have drogen ions Because the Hydrogen ions came as hy onium ions we will need 10 more water molecules 10 H307 Now 8 e39 gtNHAW 13 H200 Finally if we are to add the two equations we must multiply the Ox one by 4 to be able to cancel out the electrons so the nal balanced equation is 10 H307 Now 4 an 4 Zn w N39Hfm 13 H100 REDOX Balancing by HalfReaction Method In acid Potassium dichromate reacts with ethanolC2HSOH to yield the quot39 ofCr K 39 quot ui ei H307 CrzO7139m CZHSOH gt Cr aq co2 9 H200 Oxidation CZHSOH gt cow L add watcl ballance Hydrogen by adding protons to the product side CZHSOH 3 Hzo gt 2 com 12 HQ Since we wish to consider H as the Hydronium ion H30 we must add 1 2 water molecules to the reactant side and make the H into H30 CZHSOHQ 15 H200 2 cow 12 H207 12 e39 REDOX Balancing by HalfReaction Method Reduction CrzO7139aq gt can have two Cr and 3 electrons per atom The oxygen atoms from the quot r a wavei quot adding protons to the reactant s39de 14mm 0907113 Grim 7 H200 Each Chromium atom changes oxidation from a 6 to a 3 there by accepting 6 electrons so we add 6 electrons to the reactant side 6e 7 14 H307 0907113 2 Crew 21 H200 Adding the two equations will give the nal equation Ox CZHSOH 15 H200 gt2 co2 9 12H30W 12 e39 Rd 6e 7 14 H307 Crzogw p 2 Cram 21 HZO1x 2 CZHSOH 15 H303 2 0907113 2 com 4 Grim 27 qul REDOX Balancing by HalfReaction Method VTT A Silver is reclaimed from ores by extraction using basic Cyanide ion Aglts CN39W OH AgCN239aq Oxidation mm Ags AgCN239aq Since we need two cyanide ions to form the complex add two to the reactant side ofthe equation Silver is also oxidized so it looses an electron so we add one electron to the product side 2 New Ag AgCNz39aq t 639 Reduction 029 HZOW OH39U A A the reactant side and 2 water molecules are needed to supply the 39 s hydrogens to make hydroxide ions yielding 4 OH39 ion 4 e r ow 2 mow gt4 0H0 REDOX Balancing by HalfReaction Method Adding the Reduction equation to the Oxidation equation will require the Oxidation one to be multiplied by 4 to eliminate the electro s Ox x4 8CN39W 4 Ag 4 AgCN239W 4 e39 Rd 4 e39 oz 2Hzo 4 OH39W 8 CN39rav 4 Age 02 a 2 H200 4 AgCNfaqgt 4 OH in REDOX Balancing Using OX N0 Method V 0 1 e39 1 l l Ago CNW Ono 39 AgCN239aqgt OH w l l 0 2 To ballance electrons we must put a 4 in front ofthe Ag since ea oxygen looses two electrons and they come two at a time That requires us to put a 4 in front of the silver complex yielding 8 cyanide ions 4 Ag 8 CN39W 02 9 4 AgCN239aq OH 39aq We have no hydrogens on the reactant side therefore we must add water as a reactant and since we also add xygen we must add two wat m 0 er olecules that well give us 4 hydroxide anions giving us abalanced chemical equation 4 Ag 8 CNkaq 029 4 2 H200 4 AgCN239aq 4 OH 39aq A Redox Titration N21 ion equation 2Mn0439aq 5020239aq 16H aq gt2Mn239aq10C02g BHZOU Redox Titration Calculation outline I Problem Calcium Oxalate was precipitated from blood by the a i M molL addition of Sodium Oxalate so that solution that the precipitate was dissolved in required 2 05 ml of M01 quot1 quot 4 88 x 10394 M KMnOA to reach the endpoint a calculate the amount mol of Caq 739 Chemical Formulas b calculate the Cat2 ion conc Plan 5 Calculate the molarity of Cat2 in the H2804 solution b Convert the Cat2 concentration into units of mg Catz 100 ml blood Red ox Titration Calculation I Equation 2 KMnOA W 5 Caczom 8 st04 W gt Mnsom KZSOW 5 Casom 10 cow 8 quL a Moles of KMnOA Mol Vol x Molarity Mol 0 00205 L x 4 88 x 10394molL Mol 1 00 x 10 39 mol KMnOA b Moles of Can4 5 mol CaC204 Mol CaC204 1 00x10396 mol KMnOA x ol KMnOA Mol CaC204 mol CaCZO4 c Moles of Cat2 2 Mol Ca mol CaC204x 1 mol CaC204 Redox Titration Calculation Outline II multiply by 100 a Calc ofmol Cat2 per 100 ml M gmol b Calc ofmass of Cat2 per 100 m 1g 1000mg c convert g to mg Redox Titration Calculation II a Mo Cat2 per 100 ml Blood 2 2 Mol cw Mol cw x 100 ml Blood 100 ml Blood 1 00 ml Blood 2 Va 2 Mol Cat 2 50x 10 mol Cat x 100 ml Blood 100 ml Blood 1 00 ml Blood 2 M 2 50x104m01Ca2 100 ml Blood b mass g ofCa2 Mass Cat2 Mol Cat2 x Mol Mass Ca mol Mass Cat2 2 50 x10quotlmolCa2 x 40 08g Camol Cat2 c mass mg of Cat2 Mass Cat2 Cat2 x 1000mg CatZg Cat2 100 ml Blood Types of Chemical Reactions I 1 Combination Reactions that are Redox Reactions a Metal and aNonMetal form an Ionic compound b Two NonMetals form a Covalent compound c Combination ofan E ement and a Compound II Combination Reactions that are not Redox Reactions Metal oxide and aNonMetal form an ionic compound with lyatomic anion b Metal Oxides and water form Bases c NonMetal Oxides and water form Acids i Many ionic compounds with oxoanions form a metal oxide a gaseous nonmetal ii Many Metal oxides Chlorates and Perchlorates release Xyge b Electrolytic Decomposition Tz of Chemical Reactio 11 IV Displacement Reactions a Single Displacement Reactions Activity Series of the Metals i A metal displaces hydrogen from water or an acid ii A metal displaces another metal ion from solution iii A halogen displaces a halide ion from solution b Double Displacement Reactions i In Precipitation Reactions A precipitate forms ii In acidBase Reactions Acid Base form a salt amp water V Combustion Reactions All are Redox Processes a Combustion of an element with oxygen to form oxide b Combustion onydrocarbons to yield Water amp Carbon Dioxide Three Views of a Combination Reaction between Elements Reaction of Metals with Non metals to form Ionic ComBounds Redox Rxns Alkali Metals and Alkaline Earth Metals with the Halogens 2 Na C12 9 Ba Br2 1 BaBrm A Metal and A Nonmetal Oxygen to form ionic compounds 4 Few 3 ow 2 F6203 4 Kltsgt one Metals with the Nonmetals Sulfur and Nitrogen 16 Few 3 s8 8 F6283 5 3 Ca NHQ quot Two Non Metals combine to form a Binary Covalent ComBound Redox Rxns Halogens of nonemetals PMs 5 Fug gt 125 5 Fug ZIFSU S8 5 4 Br2 I N2 3 clw gt 2Nc1 3 Q Nitrides and Sul des PM 10 Nz gt 4 PEN Sm 2 NHQ 8P45 5 Sm SPAS Other Elements combine with Oxygen to form Oxides Metals combining with Oxygen 4 Na Om a 2 Ca Oz 2 C210 4 Al 3 OZ 2 A1203 5 Ti on a NoniMetals with oxygen N2 029 wZNOg P4s5 029 385 8 0 8 SOI 2 Fug 029 IE Combination ofA compound and an Element Nonmetal Oxides amp Halides react with additional Oxygen amp Halogens to form higher Oxides and Halides 1 nonimetal oxides with oxygen 2No 029 13406 5 2 029 ppm 2 CO 029 2 com 2 nonimetal halides with halogens 9 F2 9 PF3 Fz E pFSU IFK Fug 1150 IF50FZ Combination of Two ComEound I 1 Metal oxide with a nonimetal oxide to form an ionic compound with a polyatomic anion Nap co2 9 EB NaZCOm KZO5 SOZ CaOs sow g CaSOm 2 Metal oxides react with water to form hydroxides Nap mom B210 Hpm BaOHzs 2 Sc203 5 H200 4 ScOH3s F60 Hzo gt


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