WSU Chem 106 Exam #2 Study Guide
WSU Chem 106 Exam #2 Study Guide Chem 106
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This 12 page Study Guide was uploaded by Maranda Butterfield on Tuesday October 20, 2015. The Study Guide belongs to Chem 106 at Washington State University taught by Nathalie Wall in Fall 2015. Since its upload, it has received 127 views. For similar materials see Principles of Chemistry II in Chemistry at Washington State University.
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Date Created: 10/20/15
WSU Chemistry 106 Fall 2015 Exam 2 Study Guide Chapters 146 168 Experiments 15 and 16 To improve your score I would highly suggest you do these few things 1 2 3 4 Read the key concepts of each chapter at least once a day until the exam Carry this study guide with you and read through it before your professors begin lecture Do one practice exam one night the other the next night Only give yourself an hour to complete them Grade them and re read and make sure you understand the ones you got wrong Read the two labs The introduction your notes on reactions and your full post lab Trust me If you just do these four things your score will be improved immensely DO THEM 146 147 concentrations within Kc should always be written in moles per liter M equilibrium concentrations of the reactants and products depend on the initial concentrations equilibrium constant is always the same at a given temperature we need only know the initial concentrations of the reactants and the equilibrium concentration of any one reactant or product other equilibrium concentrations can be deduced through stoichiometry the initial conditions changes and equilibrium conditions are summarized in an ICE table i initial c change e equilibrium 0 Ms gt 2Bg o A B Initial 1 00 000 Change x 2x Equilibrium 075 2x 0 Using the above ice table you can solve for x and then find the final concentrations for K o KB2A 0 Reaction quotient O aAbB gtcCdD O Q ClCDldAlaBlb 148 149 Basically Q is calculated the same as K We are able to compare Q and K to see how far Q is from K O QltK reaction goes to the right 0 QgtK reaction goes to the left 0 QK reaction is at equilibrium Just a whole bunch of examples of finding equilibrium concentrations I would highly suggest looking at them Le Chatelier s principle when a chemical system at equilibrium is disturbed the system shifts in a direction that minimizes the disturbance Effect of a concentration change on equilibrium 0 Increasing concentration of reactions causes the reaction to shift to the right 0 Increasing concentration of product causes the reaction to shift to the left 0 Decreasing concentration of reactants causes the reaction to shift to the left 0 Decreasing concentration of products causes the reaction to shift to the right N204glt gt2N02g add N02 0 Before addition QK 0 Immediately after addition QgtK 0 reaction shifts to left to reestablish equilibrium N204glt gt2N02g add N204 0 Before addition QK 0 Immediate after addition QltK 0 reaction shifts to right to reestablish equilibrium Effect of volume change on equilibrium 0 Decreasing the volume causes the reaction to shift in the direction that has the fewer moles of gas particles 0 Increasing the volume causes the reaction to shift in the direction that has the greater number of moles of gas particles N2g3H2glt gt2NH3g O lowering the volume and raising the pressure the system shifts to the right decreasing the number of gas molecules and bringing the pressure back down minimizing the disturbance 0 increasing the volume results in a lower pressure and the reactions shifts to the left which increases the pressure and minimizes the disturbance Effect of temperature change on equilibrium 0 In an exothermic chemical reaction heat is a product I Increasing the temperature causes an exothermic reaction to shift left the value of the equilibrium constant decreases I Decreasing the temperature causes an exothermic reaction to shift right the value of the equilibrium constant increases 0 In an endothermic chemical reaction heat is a reactant I Increasing the temperature causes an endothermic reaction to shift right the equilibrium constant increases I Decreasing the temperature causes an endothermic reaction to shift left the equilibrium constant decreases O N2g3H2glt gt2NH3gheat 0 adding heat the reaction shifts left and K is a smaller value 0 remove heat the reaction shifts right and K is a larger value 0 N204gheatlt gt2N02g 0 adding heat the reaction shifts right and K is a larger value 0 removing heat the reaction shifts left and K is a smaller value 151 0 Introduction to acids and bases a section on heartburn interesting but nothing noteworthy 152 0 Acid properties sour taste ability to dissolve many metals ability to turn blue litmus paper red and the ability to neutralize bases 0 Common acids Hydrochloric Acid HCl Sulfuric Acid H2804 Nitric Acid HN03 Acetic Acid HC2H302 Citric Acid H3C6H507 Carbonic Acid H2C03 Hydro uoric Acid HF Phosphoric Acid H3PO4 O Carboxylic acid an acid that contains a carboxylic acid group figure in textbook They are often found in substances from living organisms 0 Base properties bitter taste slippery feel ability to turn red litmus paper blue and the ability to neutralize acids 0 Common Bases Sodium Hydroxide NaOH Potassium Hydroxide KOH Sodium Bicarbonate NaHC03 Sodium Carbonate Na2C03 Ammonia NH3 153 The Arrhenius Definition 0 Acid a substance that produces H ions in aqueous solution 0 Base a substance that produces OH39 ions in aqueous solution The Bronsted Lowry Definition 0 Acid proton H ion donor and becomes a conjugate base 0 Base proton H ion acceptor and becomes a conjugate acid 0 Amphoteric when substances can act like an acid or a base 154 0 Strong acid completely ionizes in solution Single arrow indicates this 0 Equilibrium is to the right 0 Weak acid partially ionizes in solution equilibriumdouble arrow indicates this 0 Equilibrium is to the left Monoprotic acid contains only one ionizable proton Diprotic acid contains two ionizable proton K3 is the equilibrium constant for acids Its calculated the same way as other Ks O The smaller the Ka the weaker the acid is 155 0 Summarizing KW A neutral solution contains H30 OH39 K1E 7M at 25 degrees C An acidic solution contains H30 gt OH39 K1E 7M at 25 degrees C A basic solution contains H30 lt OH39 K1E 7M at 25 degrees C K must always equal 1E 14 OOOO 156 157 158 Calculate OH for each solution and determine if its acidic basic or neutral 0 H30 75E 5 M O 75E 5OH 1E 14 0 OH 13E 10 0 Since H30 gt OH the solution is acidic pH scale From 0 to 14 pH logH30 0 pH lt 7 acidic 0 pH gt 7 basic 0 pH 7 neutral pOH scale is backwards from pH pOH logOH Since each needs OH or H30 you can always use Kw to find the other values pH pOH will always equal 14 at 25 degrees C pKa logKa O The smaller the pKa the stronger the acid Percent ionization concentration of ionized acidinitial concentration of acid X100 0 The percent ionization of a weak acid decreases with increasing concentration of the acid 0 The equilibrium H30 concentration of a weak acid increases with increasing initial concentration of the acid This is also a section with amazing examples that I would suggest you read Strong Base a base that completely dissociates in solution Kb is the equilibrium constant for bases Its calculated the same as other Ks O The lower the Kb the weaker the base The p scale note above can be applied the same way to this Conjugate Base A39 is the conjugate base of HA 0 The weaker the acid the stronger the conjugate base Multiplying the equilibrium constant for an acid and its conjugate base gives KW which is 1E 14 at 25 degrees C Conjugate Acid same 0 The conjugate acid of a weak base is a weak acid Ka X Kb Kw O pKa pr 14 0 Small highly charged metal cations form weakly acidic solutions 0 Classifying salt solutions 0 Salts in which neither the cation nor the anion acts as a acid or a base form pH neutral solutions I Cations are pH neutral I Anions are conjugate bases of strong acids 0 Salts in which the cation does not act as an acid and the anion acts as a base form basic solutions I Cations are pH neutral I Anions are conjugate bases of weak acids 0 Salts in which the cation acts as an acid and the anion does not act as a base form acidic solutions I Cations are conjugate acids of weak bases or small highly charged metal ions I Anions are conjugate bases of strong acids 0 Salts in which the cation acts as an acid and the anions acts as a base form solutions in which the pH depends on the relative strengths of the acid and the base I Cations are conjugate acids of weak bases or small highly charged metal ions I Anions are conjugate bases of weak acids 159 0 Polyprotic acids ionize in successive steps by losing an H each time each with its own K3 0 K21 gets smaller with each step 0 To find the pH we consider the first step to be the only relevant step except for with a dilute solution of sulfuric acid 0 This chapter is also amazing for examples 1510 0 Binary Acids hydrogen bonded with one other atom 0 To find the bond polarity consider the bond and use notation arrows as learned in chapter 9 If its polar with the positive on Hydrogen its an acid 0 The strength of the bond affects the strength of the corresponding acid The stronger the bond the weaker the acid 0 Oxyacids hydrogen atoms bonded to an oxygen atom which is bonded to something else 1511 1512 161 162 163 0 The electronegativity of the something else affects it The more electronegative the thing the more acidic it is O The number of oxygen atoms bonded to the something else affects it The more atoms the stronger the acid Lewis Acid and Bases 0 Acid electron pair acceptor 0 Base electron pair donor A lewis acid has an empty orbital that can accept an electron pair This is on acid rain Nothing important but it was interesting Yeah nothing here matters Buffer resists pH change by neutralizing added acid of added base It contains either 0 Significant amounts of a weak acid and its conjugate base or 0 Significant amounts of a weak base and its conjugate acid Henderson Hasselbalch equation allows us to calculate the pH of a buffer solution from the initial concentrations of the buffer components as long as the X is small approximation is valid 0 pH pKa logbaseacid There are two parts to calculating pH changes in a buffer solution 0 The stoichiometry calculation where you calculate how the addition changes the relative amounts of acid and conjugate base 0 The equilibrium calculation where you calculate the pH based on the new amounts of acid and conjugate base In order for a buffer to be reasonably effective the relative concentrations of acid and conjugate base should not differ by more than a factor of 10 O A buffer is most effective when the concentrations of acid and conjugate base are high 0 A buffer is most effective when the concentrations of acid and conjugate base are equal The effective range for a buffering system is one pH unit on either side of pKa 164 165 166 Buffer capacity the amount of acid or base that you can add to a buffer without causing a large change in pH 0 This increases as the relative concentrations of the buffer components become more similar to each other 0 It also increases as they become more concentrated Acid base titration a basicacidic solution of unknown concentration reacts with an acidicbasic solution of known concentration IndicatorpH meter a substance whose colour depends on the pH Equivalence point the point in the titration when the number of moles of base is stoichiometrically equal to the number of moles of acid Page 770 779 should really be read Indicators 0 If In HIn 1 the indicator solution will be intermediate in color pHpKa 0 If In HIn gt 10 the indicator solution will be the color of In pHpKa1 0 If In HIn lt 01 the indicator solution will be the color of HIn pHpKa 1 Ksp is the same as before Concentration of products over concentration of reactants Solving for Smolar solubility is the same as solving for the concentration Generally the solubility of an ionic compound is lower in a solution containing a common ion than in pure water Generally the solubility of an ionic compound with a strongly basic or weakly basic anion increases with increasing acidity Q is calculated same as before Ksp at equilibrium 0 If Q lt K the solution is unsaturated and more of the solid ionic compound can dissolve in the solution 0 If Q K the solution is saturated The solution is holding the equilibrium amount of the dissolved ions and additional solid does not dissolve in the solution 0 If Q gt K the solution is supersaturated Under most circumstances the excess solid precipitates out of a supersaturated solution Selective precipitation a process involving the addition of a reagent that forms a precipitate with one of the dissolved cations but not the others 167 0 QualitativeQuantitative analysis The diagrams show this well Visual reactions versus the numbers of the reaction 168 Complex ion contains a central metal ion bound to one or more ligands O Ligand neutral molecule or ion that acts as a lewis base 0 Kf is like any other K This is the formation constant 0 L1 ot Muss Action ml 9 b8 39 39 39C 6 ID tcjftry IAIquotBI 39 It K39 lrcvcrscrcactmmst avorcd lt39K 39 I quotCille direction 5 t39nwrcd 39 lt K 39 l fomard rcactmn IS favored 0 Rclatmnshtpshctwccn K and chcmtcal equation It yuu rcwrsc the equation 1m cm the cthlvnum constant 39 A Zb 339 tl o h lmwm 39 3C A t 2 Mil I Amour It 7 Klmvutrd 39 A t 28 quot3C quot1 WW 0 HA Znh 3m wl jl39quot il Tt39 39 A 14quot1925quot 902871 4 39 AHZB K A 813 39 M 2133c K1 t 2j l8 ICI hornet Alh IA 0 qutilvnum Constant m terms ot39prcxsurc A 7 KART 39 Aquot 13th cqmlthnmn constant m terms ot conccnmon 39 Ixquot IS the equililmmn constant m terms ofpzmml pressures 0 Rcacttonquottcnt Q tri ing IAIquotZBP K 751 PX Flt pH log J Ir UH logUH 39IKT l l K rum mrurwr nquot Irnxn39u39 uu39u39x39 I m 1 0n wnwuhun PK 39 PKquot pH IK 39 10K z39 1 U0quot Hull 4 mu rah1er H a In H Maw ur39uf o lnGC330fS l39rttulfl r39ulurl 9 HfUH 339 39 39 0 I39m I 0 In lur z l ruin 39 If j 1 the indrcato39 soiut or M be mtormedato m ccior 39 If gt10tholndlcato39 aolut 0n w l be the coior of squot m Aquot 39 If I lt 01 the ndlcator seamen H be the CO O39 0 din v39quotI cat an ph sb sa n sb wa b wb sa a wb wa ka cat cation sa strong acid a acidic an anion wa weak acid ka depends on the ka sb strong base 11 neutral wb weak base b basic Experiment 15 Student 1 Results AnionTest Test I Test 11 Test 111 Test IV Cl39 Ppt yellow No Reaction No Reaction No Reaction Br39 Ppt No Reaction Lost color Acidic yellowwhite 139 White ppt No Reaction Lost Color Acidic dissolved 804239 No Reaction Ppt with Ca Lost color Acidic both C204239 No Reaction Milky ppt with Lost purple Acidic both color C03239 No Reaction Milky ppt with No Reaction Acidic both N0339 No Reaction Milky ppt with No Reaction No Reaction Note Any information in this document about labs is taken from mine and another student39s lab reports and what we personally observed in our reactions Our reactions were not always identical Our reactions COULD be wrong I would suggest you compare these to your lab first We re not completely sure what will be on the exam about the labs But the reactions seem pretty important soo here they are Lab 16 was where you found the identity of an unknown thing Alos thanks to Derek and Cassie for helping me in making this study guide Let me know if you guys have any questionsproblems Also sometime within the next few days before the exam we will be going through the practice exam questions If you re interested in joining us email me at marandabutterfield gmailcom Good luck on the exam Student 2 Results elem Teet 1 Teet 2 Teet 3 Teet It t Beeeene I JJe reeetieh Chenged There wee white elightl Wm e huhhling end e relentde met erenge netieeehle gee 111 litintte peper turned red Er Mede e Eeeeene With etirrihg Euhhled yellew end greenieh elightljf39 heeenie e releeeed gee The dieh thet elendjE with light hrewh termed yelllew end the wenldntt Be end the heet hhte h hnue peper dieeehee ehenged it te heeerne red e yeltew t39 Mede e Eeeetne With etirrihg There wee e let net yellew elightl it heeeine eieehng The gleee wee elendjr with white epeehe whdeh Be ereund The eehttien wee nieethrE hleek It ernelled end the di h h d hhte h hnue peper turned ihte e white red elendjE eeldtien with eenie yeltew ehunhe et the tee Elite Beeeene e Beeeene white I JJe reeetieh Me reeetieh red gellth end Ei tt f Er el hale reeetieh Erreeted e After 2 mine He reeetieh with heetR it leet Be heeenie the m white end elendjF with Ce E03 Me reeetieh Eeeetne white I JJe reeetieh Sizzlng white hnhhlee end elettdy Etue litintte peper herhed end hed eenie red ehieihe with He Millie Beeeene Me reeetien Me reeetieh Me reeetieh white end eledd rgE