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Notes After Midterm 1 and before Midterm 2

by: Paige Anderson

Notes After Midterm 1 and before Midterm 2 Chemistry 1220

Marketplace > Ohio State University > Chemistry > Chemistry 1220 > Notes After Midterm 1 and before Midterm 2
Paige Anderson
GPA 3.4
General Chemistry 1220
Dr. Stoltzfus

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These notes were taken in Dr. Fus' class. They include definitions, explanations of different concepts and worked-out practice problems. I hope they are helpful!
General Chemistry 1220
Dr. Stoltzfus
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Date Created: 06/04/15
Sign lia ritmin iEil39iHzigjrrlm Hemri e Midi MIH l HIE3 I336 E JITI til5H lawn75 HTSi Ella M fl 155315 Elal ii TiEEz H39E39 I ji E JHJI 5H5 Ehil Ei EJ HuiE all MEI EliFIE ag CI Int Tcr rn 39ia ua 5F Huif i 39 gwl HM ij39naf 55533 Iasleight Empirer Ail 3513ng m4 Fi 724 mama mam5 H sjtHaiz g HEM We al 39i mnmm H H Mis t Mitzi mm handimJnul I39IE39 Midterm 1 SP15 High Score 100 Low Score 1484 Mean Score 6189 Std Dev 1418 Here is how we assign the grades at the end of the semester We arrange the points from highest to lowest and get rid of the outliers at the top and on the bottom Then we take the high score in this case 9275 and subtract the mean 92756398 2877 There are six grades from C to A so we divide this number by 6 28776 4795 This is the point value increment for each letter grade above the mean C 6398 B 6878 B 7357 B 7837 A 8316 A 8796 There are 4 grades below a C so we take the mean and subtract the low grade 639840 2398 and divide that by 4 23984 6 C 5798 C 5198 D 4598 D 40 E below 40 This is how we assign letter grades to you in this class It is done by a statistical distribution There are no quotpoints addedquot to anyone39s score and the scale of 90100 8090 etc does not even enter the conversation Letter grades are meaningless until all points out of 1000 at the end of the semester are accounted for This is my prediction based only on the rst exam When labs recitation and online homework are added in these values will go up Prelecture 9 Temperature and Rate aAbB products BU rate ALxZ kZ Experimentally it is found that virtually all rate constants show an exponential increase with temperature 0 Rate Constants often double with every 10 increase in temperature The collision Model 0 For a reaction to occur collision energy z E activation energy 0 Higher temperaturehigher kinetic energy or molecules reactingarger percentage of molecules react to form products The Orientation Factor 0 The probability of a particular high energy collision resulting in a reaction depends on the molecular orientation during the reaction Transition State Theory and Activation Energy Focuses on a hypothesized intermediate species called an activated complex which forms during an energetic collision This exists very brie y and dissociates either back to the reactants or on to products The Arrhenius Equation o 2 Requirements for Products 0 Collision energy 2 activation energy 0 Proper orientation Ea 0 k A eF 39 A constant frequency factor 0 Ink a i lnA R in if o If we have 2 data points 1 Cl I k lnA Ea I lnkz lnklz R T k E 1n LL2Lk1 F 6 Intermediate a species that participates in elementary steps of a reaction but are not involved in the overall reaction Steady State Approximation Solve for the concentration of the intermediate by assuming that equilibrium is established in the fast step that precedes the slowest step like Elli f jg 11313 f jai if m gt fl Catalysis Catalyst speeds up a reaction by reducing the activation energy If reduces the activation energy for both the forward and reverse reactions Lecture 9 2615 Which of the following statements is or are true A The activation energies for the forward and reverse directions of a reaction can be different B Assuming that A is constant if both Ea and T increase then k will increase C For two different reactions the one with the smaller value of Ea will necessarily have the larger value for k AnswerA Maxwellbolzman distribution Consider the following data for the conversion of methyl isonitrile to acetonitrile CH3NC gt CH3CN Ea 160 kJmol and k 10 x 10396 s1 at 4300 K What is the rate constant at 320 C Please discuss your response with Abigail Rubins behind you and to your left Rachel Teater to your left lllll Sophie Carus behind you A 8110 15 5 1 B 22gtlt1013 5 1 C 27gtlt10 9 s 1 D 2310 1 5 1 E 92gtlt103 5 1 AnswerD A B gt X ratekAB2 Select an acceptable mechanism a A e 33 b A C fast Y slaw fast fast alw g fast A E L zr A it slow E 4quot Slaw C A E fast E t C fst C A Y BMW X Answer E Prelecture 11 Chemical Equilibrium Most chemical reactions are reversible aAbB El cCdD when the rate of the forward reaction is equal to the rate of the reverse reaction there is no net change in the concentration of the reactants and products with time The system is said to be in dynamic equilibrium The Equilibrium Constant aAbB El cCdD the ratio of concentrations is called an equilibrium constant expression Kproductsreactants raised to power of their coef cients DAquot 3 BL A ai 3 Gift 3 Kczi 0 A Ab are reactants 6 Ad are products 0 Kc depends only on the particular reaction and temperature ex 2NOg 02gC2N02g N0222 Z 0 NOZ202 Z Z Kczi Equilibrium Expressions Involving Gases Gases are typically expressed in terms of pressures so when gases are involved the K can be expressed in terms of partial pressures This is known as the K1 2N08Clz8EIZNOCl8 Kp19x10 3at25 c Noczz2 z NOZ2C12 4 6 KZZ c PNOICZZ 2 6 PNOZ 2PCZZ 4 6 szz 0 Relationship between K c K P o PVnRT o PnV x RT Nmol VL o PconcentrationRT RTLA szKcZ An the difference in stoichiometry of gaseous reactants and products productsreactants for example equation RTLA szKcZ RTZ23 szKcZ 00821 298z 1 19x103KcZ 0 K54 6 gtlt102M1 The Magnitude of Equilibrium Constants Kproductsreactants o if K is large products gtgtlreactamsl if K is small reactants gtgtpr0ducts a very large K means a reaction is mostly completeso is mostly products 0 a very small K means a reaction does not proceed forward readilyso is mostly reactants r de lesale We akaquot lquot a a 39 3 l 3925 5 Manipulating Equilibrium Constants Consider the Following Reaction N283H28E2NH38 0 At 127 C the equation constants are 0 NH3l31x10 2M o H2l31x10 3M N 0 LL285gtlt101M o WhatistheKc 6 NH322 r 3H2Z3 Nle r Kczi 31x10 222 6 31 gtlt103Z3 85x101 Z K626 0 6238 gtlt104M2 o What is the K cEthe reverse reaction 3 H28 3 NH322 0 6 Nzla chzi K 1 0 c Kc 1 1 6 38x104 0 Ilt 26gtlt10 5M2 o WhatistheK c for 2 N2g6H2gEI4NH38 0 NH3Z4 z 3H226 0 szfz z z Kfzo 38 104M2Z214 109M 4 0 K X 2 X Kco 2o o What is the Kc for 1 3 NH38D N 8 H28 H 2 i6 Kc 51 x10 3M Kizz Combining Equilibrium Constant Expressions When you combine individual Equilibria you must multiply the equilibrium constants in order to obtain the equilibrium constant for the next reaction N28028E2N08 Nziglgoziglgzvzoigl At 25 C 0241x10 31 6224 gtlt1018 What is the Keat25 C for Nzoiglgoziglgzzvoigl 0 Add equations together This is EquationlEquation 2 Equation 2 s new Kc is 1 olch To Find net Kc multiply the two new KC 5 together 417 x1017z41x10 31 Kc l7 gtlt1013 Lecture 111219115 llll I I I The following mechanism is proposed SEEah Hlfaqli re lI acgl fast Eri aiaq j Haggai S f iaqi H j a l l39ffl Hr gfaqj 5433quot fang r Hl iaq selfam fast Hiaql sfquot fag amtiaqi Hljaq1 rliaqi fast Which of the following statements isare true A H is the catalyst for the mechanism B HSO339 HBrO2 and HBrO are intermediates C The rate law is rate lerO339llHllSO3239l D The overall reaction is 3 SO3239aq Br0339 2 SO4239aq Br m1 E The ratedetermining step has the highest activation energy AnswerABE Explanation In 1912 Fritz Haber developed the Haber process for making ammonia from nitrogen and hydrogen His development was crucial for the German war effort of World War I providing the Germans with ample xed nitrogen for the manufacturing of explosives Which of the following graphs could accurately re ect the establishment of equilibrium In 1912 Fritz Haber developed the Haber process for making ammonia from nitrogen and hydrogen His development was crucial for the German war effort of World War I providing the Germans with ample xed nitrogen for the manufacturing of explosives Which of the following graphs could accurately re ect the establishment of equilibrium N2g 3H2g 2NH3g AnswerA l Explanation C is incorrect because H2 should decrease in Aquot if Q concentration 3 C D times faster i i N 13 than N2 I H d a 8 arm n E mm 3 A SH I if 739s l 1quot I 1 Time b I X2 Y2 x Time I Tweak I For the reaction shown above from the prelecture K 272 l The magnitude of the equilibrium constant gives us iportent information about the equilibrium mixture l if K gtgtgt 1 large K Equilibrium lies to the right products predominate l if K ltltlt 1 small K Equilibrium lies to the left reactants predominate D iii ail ii iiiquot i i fail PI 2quot quot g 395 quot j l i39l iiiquot In 539 I Ii I 1392 hair i 1 i9 quotquot Iquot E in quotI The equilibrium constant for the reaction N204g2N02 at 2 C is K 20 If each yellow sphere represents 1 mol of N204g and each gray sphere 1 mol of N02 which of the following 10 L containers represents the equilibrium mixture at 2 C lll lllll M E Q M AnswerB N0222 822 Explanation 20 A not right Kczi Kczz Prelecture 12 Heterogeneous Reactions 0 Because their concentrations do not vary do not include concentration terms for pure solids or pure liquids in equilibrium expressions Ex CacoglleICaOlslCozlgl hm RT LA szKcZ C02lKPPCO2 Lecture 122111115 Consider the equilibrium at 298 K Mg 029 Br2g 2 NOBrg Calculate the equilibrium constant K given 2 NOg Br2g 2 NOBrg K 20 2 NOg x N2g 029 Kc 2 X 1030 A 11 x 1030 B 24 x 103931 C 95 x 103931 D 42 x 1030 E 39 x 103932 AnswerE Explanation Find Kc then plug into equation llll KpKcRTquotchange of moles Given the following reaction CO 9 2H2g CH3OH g In an experiment 042 mol of CO and 042 mol of H2 were placed in a 100L reaction vessel At equilibrium there were 029 mol of CO remaining Kc at the temperature of the experiment is A 280 B 0357 C 145 D 175 Answer D Explanation For the Equilibrium 2Brg 2g Br2g K 85 x 10393 at 150 C If 27gtlt10 2 atm of IBr is placed in a 20L container what is the partial pressure of IBr after equilibrium is reached A 23 x 10392 atm B 26 x 10392 atm C 29 x 102 atm D 17 x 10392 atm E 27 x 10392 atm Answer A Explanation Prelecture 13 The Reaction Quotient O N283H28E2NH38 NH3L2 z 3 k H26 all concentrations are equilibrium concentrations Q szP Nzla a 0 K is a speci c value of Q at equilibrium concentrations can be used at any time o The values of Q and K can be compared to determine which direction a system will shift to reach equilibrium E o If in this shown equilibrium QgtK then a net shift to the left must occur to reach equilibrium o If in this shown equilibrium QltK then a net shift to the right must occur to reach equilibrium Calculating Equilibrium Concentration Example fame ail at m H rsl if r at nk is it l Him g R If Emil ER 93 l quotElfL 1E4 at at E Ftlfl 3139 l e l on 5 r E if ix 3E mtg isst 7 1 7 3 EC Hill1 339 1 I Eli lf j m W Elijqarx 5 H3 7 p a Ea 39 l rs Lawn1 flit 7 Ar 4 Ef p in a flit Usquot if MEEH Le Chatlier s Principle 0 Used to qualitatively predict the changes occurring when a system at equilibrium is disturbed If a change is imposed on a system at equilibrium the position of the equilibrium will shift in a direction that tends to reduce that change 0 Examples of things that will change equilibria Concentration Pressure Temperature Inert gases Catalysts The effects of Changim Concentration Ml r313 in Meta if g 17 WT Emjlfll With an increase in N2 QltK 0 To reach equilibrium a shift from reactants to products must occur shift to the right With an increase in NH3 the equilibrium will shift left With removal of H2 the equilibrium will shift left With removal NH3 the equilibrium will shift right m xiii3 gel mt Effects of Pressure Changes When the pressure of an equilibrium mixture involving gases is increased the reaction will shift in the direction in which the number of moles of gases becomes smaller Mtg it Effzf l a Lle rl39g With an increase of pressure the equilibrium will shift right With an increase of volume the equilibrium will shift left FEE 53 1 er IlfemIrqjll With an increase of pressure there is no effect on equilibrium because the moles of gas on both sides of the reaction are equaL Effect of an lnert Gas 0 When an inert gas is added to a system at constant volume there is no effect on the equilibrium I I c When an inert gas is added at constant pressureincrease volume the equilibrium will shift to the direction with the greater number of moles of gas Effect of Temperature 0 The effect of temperature not only changes the equilibrium position but also will change the equilibrium constant K because K depengs on temperature MW T HEquot 513 E M a i h l lf Pl Elim fHJWL An increase of temperature in exothermic reaction will increase the amount of heat present and the equilibrium will shift to the left 0 A decrease in temperaturein exothermic reaction will shift the reaction to the right Effect of Catalyst 0 A catalyst speeds up a reaction therefore it has no effect on the direction of the equilibrium Lecture 1321315 The K for the following reaction is 393 at 1200 K COg 3H2g CH4g H20g Given the concentrations CO 00450 M H2 0132 M CH4 0020 M and H20 00550 M one can conclude that A the system is not at equilibrium and the reaction will proceed to the right B the system is not at equilibrium and the reaction will proceed to the left C the system is at equilibrium and no net change will occur Answer C Explanation For the reaction 4 NH3g 5 029 4 NOg 6 H20g AH 904kJ which of the following changes will shift the equilibrium to the llll right toward the formation of more products You can pick more than one answer A Adding 1 atm of Neg to the reaction vessel B Adding more water vapor C Removing 029 D Increasing the temperature E Increasing the volume of the reaction vessel AnswerE Explanation o A No effect Ne doesn t react with the products B shifts to left the equation shifts left to balance equilibrium C shifts to left D shifts to left E shifts to right Prelecture 14 Acid Base equilibria In chapter 4 strong acids and strong bases were introduced a Ef e J m HI Hm acid Hail 2 Isn39t ijhlgl s H l Hiquot 33943jo Wig fl H illlw imam lainll A i3 a qul r i g ii Strong acids and bases fully dissociate in a solution 0 Weak acids and bases partially dissociate in a solution Acids KC 0 Strong acid has large KC 0 Weak acid has small KC Bases Kb 0 Strong base has large Kb 0 Weak base has small Kb Evolution of de nition of acids and bases 0 1880 s Arrhenius 0 acid a substance that when dissolved in H20 increases H HgtOH 0 base a substance that when dissolved in H20 increases OH OHgtH o Arrhenius de nition of acids and bases only de nes acids and bases in aqueous solutions 0 1923 BronstedLowry Acids and Bases o more general de nition 0 acid a protondonorH ion 0 base a protonacceptor ii F o H30 is the acid in reverse reaction 0 A is the base in reverse reaction Coniugate AcidBase Pairs 5 quot ag I a T T Hara quot39 5mg Lag Hill r11 Jen rel r933 h fquot a i E if 1ampny Ir Hamil l fw i 1 A Htl cl M Hg d 5 5 3quot mixHis Emil win alzwia an Sulaquot iiiquot bit If Haj grail amphiprotic a substance thatis capable of acting as either an acid or a base Relative Strengths of Acids and Bases W m it viiam ha an rial all E Hj 39 l 4 rlilgDTr 4 Him 1 g am o If H20 is a stronger base than X then the equilibrium will shift to the right 0 Most of the acid will be in the ionized form o If A is a stronger base than H20 then the equilibrium will shift to the left 0 Most of the acid HA will be in its unionized form 0 The farther the equilibrium lies to the right the stronger the acid more ions present The stronger the acid the weaker the conjugate base 0 The weaker the acid the stronger the conjugate base 0 Relative strengths of acids and bases example problem 5F H EraEh 39quot 393 gm tr HF gt gt W 39 7 Er F a 7 EUquot n A r quot39 1 mm The autoionization of water ML n Mimi 1 my j T 0 Water is amphiprotic 6 H30 6 o 6 OHquot 3 szz ll llll llll EH F39 1 Rim5 firs533E m Gilt Emil Ellaj r fqu E j ijlfwj main 2 er jfm ffj 49 I 7 Lecture 14 21615 Which of the following is not true concerning conjugate bases A Every acid has a conjugate base B The conjugate base of a BronstedLowry acid is that acid minus its acidic proton C Some conjugate bases are acidic D OH is the conjugate base of H3O Answer D Explanation D is incorrect because the conjugate base of H30 should be H20 Which of the following is not true A Strong acids have weak conjugate bases B weak acids have strong conjugate bases C The weaker the acid the stronger the conjugate base D The stronger the acid the weaker the conjugate base E All of the above are true Answer B Explanation B is not true because weaker acids have stronger conjugate bases but do not necessarily have strong conjugate bases What is the pH of a 10398 M HCI solution A 600 B 696 C 700 0 o b The dihydrogen phosphate ion H2PO439 is amphiprotic In which of the following reactions is this ion serving as a base A H3Oaq H2 PO439aq H3PO4aq H20 B H3Oaq HPO4239aq H2PO439aq H20 C H3PO4 HPO4239aq 2 H2PO439aq Answer A and C Prelecture 1521815 The pH scale 0 The pH scale provides a convenient way to represent the acidity of a solution 0 pHogH H10E7pH7 0 pOHlogOH 0 pKalogKa W1 Eli E tj V L a mi i 1 I ll Hui 4quot 1 IIlr l39 M Al sriff 391 l J1 hip l r3 imag H 3 lieital a law gramme lri39 5 3 pH of Strong Acids strong acids fully ionize in solution rmfa Examples What is the pH of a 010M nitric acid solution 39 HNO3H01OM pHlogHlog010 IOH1 What is the pH of a lElOM HCI solution H from HCllE10 H from H201E7 H totall001E7 og1001E7pH69996 pH7 llll Lecture 1521815 Consider the following equilibrium reaction HSO4 aq OH aq SO42 aq H20 Which substances are acting as acids in the reaction A H505 and OH B OH and 5042 C SO42 and H20 D H50 and H20 E None of the substances are acting as acids in this reaction Answer Explanation the transfer of a proton always involves both a donor and an acceptor Acid proton donor 0 Base proton acceptor Based on the information give in Figure 164 place the following equilibria in order from smallest to largest value for K Still 32513 392 i ls l fl filming acids I 13am Malia Hugligi ilr l39iasicili 3 m i E aquot in l wan 39 lam a 2112 39quot l um sniff Min E nigh l U i lln h ll 1 I1 Iil39UHT I 39iia39igal acids U Wquot HE HIM2 hastrs Ea Hiring a Hmquot M ii EH i ll I m as 1 Her Jilm if 7 H 31 1 1H quot UH 13 7 Eagli iblr acidity Hi El SET2mg 1315 In H Li MI E i Fi u E1 Il Eighth mn lhai of select conjugate millage pairs TL39I Eww Irlitjl39llfiiia n with 51 r am in rm EE Hju Eflii milfl jhiff m lira liar if39 lEII39IIT I39j CH3COOHaq H5 CH3COO39 HzSaq F39aq NH4 HFaq NH3aq H2CO3aq Cl HCO339aq HCaq Answer lll H2C03aq Cl39HCO339aqHCIaq lt F39aqNH4HFaqNH3aq lt CH3COOHaqHS39CH3COO39H25aq A solution at 25 C has pOH 1053 Which of the following statements is or are true A The solution is acidic B The pH of the solution is 347 C For this solution OH 10 1053 M Answer ABC In the titration of 3500 mL of 0737 M H2504 how much volume in mL of 0827 M KOH solution is required to neutralize the reaction A 350 mL B 112 mL C 258 mL D 624 mL E 393 mL pH example problem Prelecture 16 pH of Weak Acids 139 EH I gt u a was pH Example Problem a Hm i He raw mi 3 W WWW a 59 is p I a x r it E a 7quot l5 lee ix uh l E 1r 7 Vi IF i 39 i iilmr if r SEQX 94 Iij llfxm Eh Earth if C 3 til FW l raj l39l39F 55th 3 t1 739 Hams Ml to atl CH w 139 1 53155 I a M ii 523 Polvprotic Acid an acid with more than one proton ex HZSZ always dissociates in a stepwise manner armor h J P a newquot 1 NJ an quotM if rr quot7 Ili39 kji P quot qryn HERIE I EM smallquot 31 For a typical weak acid in 2 3 It is always easier to remove the rst proton For KC of polyprotic acids look into the 1st dissociation as it dictates the equilibrium concentrations Percent Ionization PE EE EW rwrw ww ENE I fig ag h t f y jmf a 1 311 35 if H EX ff DamA quot 5b 2 a 4quot fr39llgxf gj E E 5 K 391 3 SE b f quotm 3 371 QEMD quot Jlr39 i D 7 Weak Bases rs Ew ij MHsff i iii96 W iu f il WET g law his ma a quot11 quot as C a 195 quot 3 Hf ifquot H quot in a 435 E l i wear EH EU r12 1 PQH for 31 fit ifgl 39l 4 leigrf f 143 Lecture 1622015 What is the pH of a 01 M formic acid solution A It will be 10 which is the same as the pH of a 01 M HCI solution B It will be less than the pH of a 01 M HCI solution C It will be greater than the pH of a 01 M HCI solution Answer C because formic acid is a weaker acid than HCI The pH of a 01 M formic acid HCOOH solution is 25 What is the Ka of formic acid HA 4 1 Oil 0 C N 0003llZZ Z X 94 has awayOi 0 109314 Kai0340 How will the pH of a 01 M acetic acid HC2H302 solution compare to a 01 M formic acid HCOOH solution 0 If x is more than 5 of the original value use the quadratic formula Otherwise x can be assumed to be negligible in te denomiatr llll Ei 393939 DiFHHi l El ni I39i EEI Hai i1 H 5 hasq Mimi h Ir IE5 JannaE 451 lljlx39IEHJillILEa E39il39iglquotinil Ha IF Jaruh Hal El Lil39s quotr Eu HE Fij Iii D u39ll 39wnhl Hr hll39i 39FI i m C twill IHj H ii Mn is I l 1 Ellilurch EH3 Emiliaa HI Hint hl r in lam M ll in m 1 gnaw g IUlal 39n allerEilIquotlh lilng i a I it Farm an Elf2i i I 5 up 53 r L ITEMwinquot Fll TiI39FZiliiuilmfllilzl E1 H quota l mm Mali u guy EVEdash EAHERE39 flllj lii1lilijlu I u gifllnnl39nll3nl 39lg 1 39 5g 39 gquot m Hi 5 quot5 a jig 11539 II 39Eiii F quot Uni 39 WEGEM mailtill DJIC39ILI39EEUJ is a Hf 39 Purinea hw Equot w H a i fnjlhfuithIJJ HE 539 l3939 3 In LE iaagvat iii Ha Wu l athg EWqWI em Hindquot HI 1 IE3 I11T1Vs New 5 iE minimgm ulnahim llrm I I w leuniElnw imt39efj Ha 39 5 M3 n r in 4 It qfebuwfnauid III J39I iJ at m 39 ILgle39bl39Ecl uu dun HEELI Jug 39 air 39 lt d aia lll39lI 5 9 m D L39 he lm i uj la39hlhihiinl Ilu jl ix l iIHTquotE39 i39lEJEEiEi I I quot gr 39 TEMP311 LIN5 NIH13H i II lll i I u i Eu H I39Eii uji l39j Ml H a I i Hui inn3 ailin u visl My 39EI 9 iii MI C ELIGI F39RF ibh j EP 5 is Fs M iii 2 a quot iiing I LEi IEIEiUij Ialii ran a E at Hamil I i ii a j r32 El Era l E I39 i 3 hangs a L I ll milligraLll39 lu glij ll n at hiu er hwhnl ll i39ui u i a H n M I I ma Mining 9 El n g I Ulgl I a 5 WHEELi lieii lmgaid H r 33 3 39L s33 23 a gj H 5 5 TI ihh d l li fil ilg a h lli ggi39llrlml g E l A It will be 25 which is the same as the pH of the 01 M formic acid solution l B It will be less than the pH of a 01 M formic acid solution Answer What is the percent ionization of a 0100 M HF solution Ka 72 x 104 5 1 8 8 78 7 6 00gt 2 9 mon Prelecture 1722315 Relationship between Kc and Kb mpg draw it away I 1 W it Mrsw r 11 k i E caihiml m 39 E a a a AcidBase Properties of Salt Solutions Saltan ionic compound producing ions in solution 0 under certain conditions these ions behave as an acid or a base Ff Htlriji Hallr t Hamill Him cue A 7 I ail etng 1amp3 If the conjugate base is from a strong acid the formed reaction above does not occur and no OH ions are produced The conjugate base of a strong acid is neutral List of strong acid conjugates 3 5 7a 7 w 1 9quot E39 quotquot LT clj Ell jg 96 J El F j are The conjugate acid of a strong base is also neutral ma a sq a 1H139j T 7 Haitian m a Ci l iw E When products are both conjugates of weak acids and bases it is necessary to compare the Ka and Kb values 0 If KagtKb pHlt7acidic If KbgtKa pHgt7basic Example calculate pH given molarity of a salt Ii MntF Ar HEFH ElliEH Elem rim g jm reJHHmt mi rm i 1quotle if t h I It a 5quot kw r W has Wit 444 E rx W is E THEE I ujf 39 r 39 j H Ff39 i39 39 m quot 39 r 4 a ier 34 a l Foil 3 LE NH EEEKFHE i w Lecture 1722315 Solution A is composed of 01 M aqueous hydro uoric acid pKa 374 Solution B is composed of 01 M aqueous acetic acid pKa 474 Which solution has a higher pH I A Solution A I B Solution B I C Since both are weak acids with the same concentration they have the same pH I Answer B quot Agar 1 I Given the pKa values which of the following choices is the strongest base lll llll Acid pKa HSCN 18 HBF4 05 HIO 105 A SCN B BF4 C IO D Cl Answer C In a 0050 M solution of a weak monoprotic acid W 18 x 103 What is its Kb A 67 x 10 B 28 x 10 Which of the following salts is are basic A NaN03 BNaCN C Na2CO3 Answer B and C Explanation Salt A is comprised of a conjugate acid and base of a strong base and acid The salt therefore will not react with water and will be a neutral salt Salts B and C are comprised of a conjugate acid and base of a weak acid and strong base Therefore the conjugate bases will react with water and form a basic solution The Ka of NH4 is 56 x 1010 The Kb of CM is 2 x 105 The pH of a solution of NH4CN is A below 7 because NH4 is more acidic than CM is basic B below 7 because NH4 is more acidic than CN C above 7 because CN39 is more basic than NH4 is acidic D above 7 because CN39 is more basic than NHL Answer C Prelecture 1822515 AcidBase Behavior and Chemical Structure 0 any molecule containing a hydrogen is potentially an acid 0 many such molecules show no acidic properties ex CHCI3 CH3N02 here the CH bond is very strong and nonpolar 0 there is no tendency to produce H 0 two factors that determine the acidity of an XH bond 0 polarity the more polar the stronger the acid 0 bond strength stronger the bond the weaker the acid 0 HF v HC v HBr V Hl Polarity HFgtHCgtHBrgtH I Bond strength HlgtHBrgtHCgtHF I Actuality HlgtHBrgtHCgtHF Oxyacids HOX group 0 Acidity is directly proportional to the number of terminal oxygens oinltmidation number H E X lllrl Halli W F iii illllli quot WE lil x Elmll all Ml l A 1 7 1 i1 r k It II J I r v itin A i if Carboxylic Acids COOH group C BEquot 113 Hi Lecture 1822515 Which of the following can act as a Lewis Acid A HCI B CH3CH20H C BF3 Answer A B Lewis Acid electron pair acceptor Lewis Base electron pair donor Arrange the following substances in order from weakest to strongest acid HCIO3 HOI HCIO3 HBr02 HC02 HIOZ Answer HOI lt HIOZ lt HBr02 lt HC02 lt HCO3 Let39s compare the following species in aqueous solution C2H5OH CH3COOH C6H5COOH HCOOH The relative acid strength of the four compounds listed above can be related to their ability to donate H If we compare the bond dissociation energies of the 0H bond or the energy required to break the 0H bond in each compound we observe the following C2H5OH 112 kcalmol CH3COOH 96 kcalmol C6H5COOH 111 kcalmol HCOOH 105 kcalmol Based on this data one might conclude that the acid strength might be similar but in reality the acid strength is as follows C2H50H Ka 20 X 103916 CH3COOH Ka 18 X 10395 C6H5COOH Ka 63 X 10395 HCOOH Ka 18 X 10394 Lecture 1922715 Prelecture 203115 AcidBase Titrations A titration is performed to neutralize an acid by a base and the endpoint is found by the aid of an indicator 0 Strong acidbase titration Emit iii i i iitii it Ei i Milli 3 mm minair t my iiiiaii 1M1 1am my i y j aimi Limit Fi iiixi ii ii iii iiii E ii iiiii Emmi tamai r iiii iiiiii WM tit lm1 I QHPW M Weak acid Strong base titration 15mg nit Eim U5 3 ng Lam Ha 51 it1quot Fig m Ewing r rm mi it a E H 4 A E main s A H mm P 3931 mmqu mm a Mama Mm I H i r Elm 1 air Fag1f mm 1 w Him Mk MWH m may Mfg H 39 PM quotquot f H w m Lime Liatam m mm a Lquot W quot M H Mama HE 439quot kl E f 7 amw aw my a WW mimwr r k u may ii3E I HI JJF g k 1 r m w g r Ens h Jr ErmaH Lw j quot if Q Titration Curve a PH in raw JHM l PH agging Egan at PH 44th i I I I Ii L14 am i 39I All I j in La Lecture 203215 Shown below is the pH curve for 010 M HCI titrated against 010 M NaOH What differences would we observe if the 500 mL of HCI were replaced with 500 mL of 010 M HC2H302 pH 3 21 921 muwma39i Milli Wilmaiii 11a whman Em pH Im T Equmlrmc pugEn M if ctui wht i r39 quot jgr l ham maplequoti mid 7 1 7 r gaging 1 H 3533 311 4393 391 Hi Ilfl Mn i I II 0 o i o 1 w 7 em I I Q I O 5 H mmpE el i ZHi ie a i mmiml hag43 by min w ill Emit ai UirlimilipaEl 305 H39 EPH i Elli atly HEinnir a 1 I mrmtmmdym ingniiium ETI J lgrllgl rli l1fl l iil lg El ilfl illli 3 IgEHpl 1 E39JII IEI in Figure 11 tmtitm H a iilrtrngatid inth ii sshmtg hzam Elie H funF 39I tutlalnn l 503 EL ii a IELIEEI Summer at mammalian ail id la1h a EHEI w i Milan6quot n Ha Hiaqi humanly waler mEEE l ilu j Email iiiH Irrm lmd lrlj39r39l39fl II IE m gmnr Earl llll llll A The initial pH B The slope of the titration curve before the equivalence point C The pH at the equivalence point D The slope of the titration curve after the equivalence point E The volume of NaOH added Answers ABC B the acetate ion is a conjugate base of a weak acid so as the acid is titrated the solution becomes slightly more basic C The equivalence point will be basic because the conjugate base of a weak acid has the potential to react with some of the H ions so will make it a basic solution At the equivalence point the salt will dictate the pH level A 2500 mL sample of 0100 M CH3C02H is titrated with 0100 M NaOH What is the pH of the solution at the points where 245 and 255 mL of NaOH have been added Ka 18 x 105 A 644 1100 B 613 1100 C 644 985 D 613 985 E 700 800 AnswerA


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