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by: Brady Spinka


Marketplace > University of Texas at Austin > Chemistry > CH 302 > PRINCIPLES OF CHEMISTRY II
Brady Spinka
GPA 3.98

James Holcombe

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James Holcombe
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This 13 page Class Notes was uploaded by Brady Spinka on Monday September 7, 2015. The Class Notes belongs to CH 302 at University of Texas at Austin taught by James Holcombe in Fall. Since its upload, it has received 19 views. For similar materials see /class/181861/ch-302-university-of-texas-at-austin in Chemistry at University of Texas at Austin.




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Date Created: 09/07/15
66 Applications of the Equilibrium Constant Equilibrium Constant allows us to predict the tendency of the reaction to occur not the speedrate 2 whether a given set of concentrations represent an equilibrium condition 3 the equilibrium position that will be achieved from a given set of initial concentrations the Extent of a Reaction the tendency for a reaction to occur is indicated by the magnitude of the equilibrium constant eX Kgtl further towards completion the size of K and the time required to reach equilibrium are not related Reaction Quotient A system not at equilibrium will shift in the direction that produces the missing component To determine we use Reaction Quotient Q Use law of mass action but use initial concentrations 7 not equilibrium concentrations Aa Bb Cc Q 7 C Ac Ara BJAb 1 Q K equilibrium no shift 2 QgtK products more than reactants shift will go to the left reactants 3 QltK reactants more than products shift to right products Initial concentrations is equal to the molarity When finding Q compare against K To figure out the change in concentration required to get to equilibrium Initial concentration 7 change equilibrium concentration to find equilibrium concentrations plug in mass action law 67 Solving Equilibrium Problems 1 write the balanced equation for the reaction 2 write the equilibrium expression using the law of mass action using initial concentrations 3 calculate Q and determine shift 4 Define the change needed to reach equilibrium initial concentration 7 change equilibrium concentration 5 Define the equilibrium concentrations by inputting the above into mass action equation 68 Le Chatelier s Principle if a change in conditions a stress is imposed on a system at equilibrium the equilibrium position will shift in a direction that tends to reduce that change in conditions The ElTect of a Change in Concentrations Changing concentration of reactants to where total reactants is greater than products shifts toward products and vice versa If a component is added to a system The system will shift away from the added component if a component is removed from the system the system will shit towards the removed component the ElTect of a Change in Pressure 3 Ways to Change Pressure 1 add or remove gaseous reactant or product constant volume 2 add an inert gas constant volume no effect on equilibrium 1 and 2 it increases the total pressure but has no e ect on the concentrations or partial pressures of the reactants or products 3 change the volume of the container concentrations and thus partial pressures change when the volume of the container is reduced the system responds by reducing its own volume thus decreasing total number of gaseous molecule 1 Ideal Gas Law V RTPn l at constant T and P V proportional to n So in other words Volume decreased shift towards the side with less moles Volume increased shift towards side with more moles Only with Gas the ElTect of a Change in Temperature Heat is a product in an exothermic reaction Heat is a reactant in an endothermic reaction K decreases w increased temperature Chapter 7 Acids and Bases 71 The Nature of Acids and Bases Basesalkalis 7 Arrhenius produce hydroxide ions BrownstedLowry protonH acceptor Acids 7 Arrhenius produce H ions BrownstedLowry proton H donor depending on which de nition you use de nes whether it is an acidbase reaction HCL hydrochloric acid H20 water H30 hydronium ion Cl polar water molecule pulls the proton from the acid acid base conjugate acid conjugate base conjugate base 7 only thing that remains of the acid when the proton is lost conjugate acid 7 formed when the proton is transferred to the base conjugate acidbase pair 7two substances related to each other by the donating and accepting of a single proton If two bases are present there is a competition where the stronger base will gain the proton Ka acid dissociation constant water is not included productsreactants with the acid ex H30 being equal to a proton remember solids liquids not included l 72 Acid Strength strength of an acid is de ned by the equilibrium position of its dissociation reaction strong equilibrium lies far to the right almost all the original acid is dissociated yields a weak conjugate base low affinity to protons ex H2SO4 sulfuric acid HCL HNO3 nitric acid HclO4 perchloric acid diprotic acid 7 acid with two acidic protons Sulfuric acid oxyacids 7 acidic proton is attached to an oxygen atom most common organic acids 7 carbon atom backbone typically weak commonly contain carboxyl group CO O 7 H hydrohalic acid 7 H attached to a halogen atom monoprotic acids 7 one acidic proton weak equilibrium lies far to the left very little of the original acid is dissolved yields a strong conjugate base high affinity to protons Strong acids lie so far to the right in equilibrium reactions that Ka cannot be measured accurately How to judge Strength Bases conjugate bases of weak acidsCl gt H20 gt conjugate base ofa strong acid Acids strength of an acid is inversely related to the strength of its conjugate base So ifwe know the Ka of an acid ex Ka for HF gt ka for HN02 gt Ka for HCN then base strength F lt NO2 lt CN Question How do we judge whether something is a stronger or weaker acid than H20 Water as an Acid and a Base amphoteric 7 can behave either as an acid or as a base ex water is the most common autoionization of water 7transfer of a proton from one water molecule to another to produce a hydroxide ion and a hydronium ion 2H2O OH H30 the water acts as a base and acid proton donator and acceptor equilibrium expression Kw HOH this is also called the ionproduct constantdissociation constant referring to the autoionization of H20 25 C Kw l X 103914 why are units customarily omitted Sec 63 not matter what product of HH30 and OH is always Kw l X 103914 Three Possible solutions though 1 neutral HH30 OH 2 acidic solution HH30 gt OH 3 basic solution OH gt HH30 73 the pH Scale pH logH ex H 10 x1007 pH 700 700 significant figures 7 number of decimal places in the log number of significant in the original number in the above example 10 2 significant gures 700 2 decimal places other quantities that can be calculated pOH logOH pK logK pH decreases as H increases and vice versa General Strategies for Solving AcidBase Problems 1 Think Chemistry focus on the solution components and their reactions Choose the most important 2 Be systematic acidbase problems require a stepbystep approach 3 Be exible treat each problem as a separate entity 4 Be patient pick the problem apart into workable steps 5 Be con dent look within the problem for the solution let the problem guid you Do not concern yourself so much with equations as much as with understanding the problem 74 Calculating the pH of Strong Acid Solutions EX What species are present in a HCL solution HCLi strong acid so it nearly completely dissociates So the solution is actually primarily composed of H and Cl ions Major species H Cl and H20 very acidic solution lots of H very little OH WRITE THE MAJOR SPECIES H2O H OH Le Chatelier s Principle 7 react to a change in equilibrium therefore since there is an overabundance of H the autoionization of water will be driven far to the left So contribution of H ions by H2O is negligible Therefore H in solution is nearly the whole solution N 10 M of H pH logH logl0 000 eXtremely acidic 75 Calculating the pH of Weak Acid 1 write the major species in the solution eX HF Ka 72 X 1004 100 M solution H20 small Ka weak acid only slight dissociation HF H F Ka 72 X 1004 H20 H OH Kw 10 X 10014 HF will dissociate more than water because it is still the stronger acid compared to water This is indicated by the K Therefore we will be looking at the HF as the dominant determinant of pH with water adding little Ka 72 x 10A4 HF HF H HF I 100M 0 0 C X X X E l 7 X X X Ka 72 X 10A4 XAZl 72 X 10A4 XAZ square root both sides solve for X X H 27 X 10A2 pH log27X10A2 157 pH logH HOW DO WE CHECK OUR ANSWER Solving Weak Acid Equilibrium Problems list major species choose the species that produces the H then write the reaction that produces it compare K and see which species will dominate write equilibrium eXpression for dominant reaction ICE chart plug into equilibrium equation solve for X verify calculate H and pH pW aMrbE N The pH ofa MiXture of Weak Acids Same General Steps EX 1M HCN Ka 62 X lOAIO 5 M HN02 Ka 40 X 10A4 Major Species HCN HN02 H2O Compare K s HN02 is much stronger than the others so HN02 H N02 Ka 40 X 1024 productsreactants HNO2HNO2 HN02 H N02 1 5M 0 0 C X X X E 5X X X weak acid 5X X is negligible 40 x 10A 4 xx5 XAZS 20 X 1005 XAZ square root both sides solve for X X 45 X 1002 H x pH logH When there s two acids you must still look at the weaker acid and determine the concentration of the other nonH ions 7 in this case HCN and thus CN HCN H CN Ka 62 X 10010 productsreactants HCNHCN 45 X 1002CN1 solve for X X 14 X 1008 CN very small amount dissociates PERCENT DISSOCIATION often used to specify the amount of weak acid that has dissociated in achieving equilibrium in an aqueous solution percent dissociation amount dissociatedinitial concentration X 100 EX Above 100 M solution of HF H 27 X 1002 In other words for the system to reach equilibrium 27 X 1002 moles of the original 100 M HF dissociates So Percent Dissociation 27 X 10021 X 100 27 For a weak acid the percent dissociation increases as the acid becomes more dilute Percent association gt in a 01 M solution than a 10 M solution because even though the amount dissociate is slightly smaller the initial concentration is leven ess Using Percent Dissociation To Calculate Ka EX 0100 M solution lactic acid is 37 dissociated calculate K Major species HC3H503 H2O lactic acid is a weak acid but much stronger than H2O HC3H503 H C3H503 K productsreactants HC3H503HC3H503 H C3H503 HC3H503 I 0 0 0100 M C X X X E X X 01007X percent dissociation amount dissociatedinitial concentration 37 XHC3H503 X 100 X010 X 100 37 X 010 37100 X 37100 X 010 37 X 1003 molL Now plug in X to calculate K to equilibrium eXpressions 76 Bases Arrhenius base iproduces OH in aqueous solution BronstedLowry 7 base is a proton acceptor Strong Bases Group 1A NaOH 2A BaOH2 they dissociate completely when dissolved in water complete dissociation means for eXample a 100 M solution of NaOH actually contains 100 M of Na and 100 M of OH the same concept applies for strong acids Calculating the pH of a Strong Base 50 X 1002 M NaOh solution NaOH H20 NaOH is a strong base completely dissociates NaOH Na OH OH 50 X 1002 M H KwOH 10 X 1001450 X 1002 20 X 10013 pH logH log20 X 10013 1270 Other compounds can be proton acceptors and not necessarily release OH EX NH3 H20 NH4 OH REMEMBER Base Acid Conjugate Acid Conjugate Base Conjugate Acid gained the proton Conjugate Base lost a proton Kb reactantsproducts Kb irefers to the reaction of a base with water to form the conjugate acid and the hydr0Xide ion small Kb weak bases more acid large Kb strong bases less acid Calculating pH of a weak base and weak acid are very similar EX pH ofa 10 M solution methylamine Kb 438 X lOA4 CH3NH2 H20 H20 loses the proton because it is not a great source of OH CH3NH2 H2O CH3NH3 OH CH3NH2 CH3NH3 OH I 10 M 0 0 C X X X E 10M 7 X X X Kb 438 X 1044 CH3NH3OHCH3NH2 X X lX XAZlX the X is considered negligible because a weak base will not dissociate much Kb 438 X 1044 XAZl XAZ square root both sides X 21 X 10A2 NOT FINISHED YET pOH logOH log2l x 10A 2 168 HOH Kw or 10 x 10A 14 Therefore pH pOH l4 l4 7 168 pH 77 Polyprotic Acids Polyprotic 39 Some acids H2SO4 and H3PO4 can furnish more than one proton always dissociates in a stepwise manner one proton at a time Carbonic Acid H2CO3 l H2CO3 H HCO3 Kal HHCO3H2CO3 43 X 10 A7 2 HCO3 H CO3A2 Ka2 HCO3A2HCO3 48 X lOAll the conjugate base ofthe step 1 becomes the acid of step 2 Carbonic acid is formed when carbon di0Xide is dissolved in water so you can technically add an additional step for formation though not part of proton release CO2 H20 H2CO3 or CO2 H2O H HCO3 since H2CO3 is a strong acid and basically completely dissolves So a solution of carbonic acid can contain various amounts of all three species 7 H2CO3 CO2 H20 HCO3 and CO3A2 EX calculate the fraction of H2CO3 HCO3 and CO3A2 at pH 900 FRACTION amount of one speciestotal species fHCO3 HCO3H2CO3 HCO3 CO3A2 fHCO3 lH2CO3HCO3 1 CO3A2HCO3 Use K s to calculate ratios Kal HHCO3H2CO3 H2CO3HCO3 HKal and we know Kal 43 X 1007 and pH 900 so H 100 X 1009 plug in Answer 7 23 X 1003 HOW TO DO THE SIMPLE MATH DIVISION MOVING THINGS AROUND Ka2 HCO3A2HCO3 Ka2H CO3A2 HCO3 plug in answer 48 X 1002 Then plug into original equation fHCO3 l 23 X 10A3 1 48 X 10A2 solve l105 095 So at pH 900 95 ofthe solution is made of HCO3 continue to do this for the other two but the of those is negligible lt5 Phosphoric acid 7 triprotic acid three protons EX H3PO4 phosphoric acid H3PO4 77 H H2PO4 Kal 7 HH2PO4H3PO4 7 75 x mm H2PO4 7 H HPO4A2 Ka2 7 n1 HPO4A2H2PO4 7 62 x 10mg HP04A2 77 H PO4A3 Ka3 7 n1 PO4A3HPO4A2 7 48 x 10A 13 For a typical weak polyprotic acid Kal gt Ka2 gt etc The acid gets weaker and weaker greater the negative charge becomes the harder it is to pull that positive proton off Most weak acids only the rst dissociation step makes an important contribution to H EX Calculate pH of a 50 M H3PO4 and determine equilibrium concentrations of the species H3PO4 H2PO4 HPO4A2 P0403 Major species H3PO4 H20 after the rst dissociation step none of the others make signi cant contributions H3PO4 H H2PO4 Kal 75 X 1003 HH2PO4 H3PO4 H3PO4 H H2PO4 I 50 M 0 0 C X X X E 50 7 X X X Kal 75 X 1003 XX5X XAZ 5 square root both sides X 19 X 1001 check and see ifX is less than 510 of original concentration in this case 50M pH logH logl9X 1001 072 So H H2PO4 019M X and H3PO4 5X 48M Continue with H2PO4A2 Ka2 62 X 1008 HHPO4A2H2PO4 we know H H2PO4 019M note values for concentrations of species remain constant through equilibrium calculations plug in Ka2 62X 1008 HPO4A2 Do it again with dissociation of HPO4A2 to nd P0403 Numbers for HPO4A2 and PO4A3 show that the second and third dissociations had very little input PHOSPHORIC ACID EX2 0200 mol ofNaP H2O HCL H3PO4 H2PO4 H2PO4A2 PO4A3 1L solution pH of 4630 Calculate the concentration of all phosphate containing species 1 work backwards from pH to determine H 234X10A5 2 the different species must add up to 0200 mols 3 turn everything to one variable rearrange equilibrium eXpressions H2PO4 H3PO4Ka1H substitute in amounts gured from previous example except H H 234 x 1005 Kal 75 x 1003 do the same for the following expressions Need to get a further explanation SULFURIC ACID Sulfuric acid is unique strong acid with the rst dissociation 1 H2SO4 H HSO4 Kal is very large weak acid with the second dissociation 2 HSO4 H SO4A2 Ka2 12 x 1002 Ex Calculate the pH ofa 10 M H2SO4 H2SO4 H HSO4 H2SO4 is a strong acid nearly completely dissociating major species H HSO4 and H20 So if it nearly completely dissociates we know that H and HSO4 will equal nearly 100 M but does the dissociation of HSO4 make a signi cant H contribution Ka2 12 x 1002 HSO4A2HSO4 H SO4A2 HSO4 I 10 M 0 10 M C x x X E 1 x x 1x Ka2 12 x10A2 1xx1x1x1 x x 12 x 1002 12 x 1002 is 12 of the original concentration 100M than eliminating the x39s is valid NOT DONE REMEMBER Now solve for H to see if HSO4 makes a signi cant contribution H1x 112x1002 10 M when rounded to correct signi cant digits Thus the concentration of H 10M and the pH 000 completely H For Sulfuric Acid more dilute solutions 010 M H2SO4 allow the second dissociation of HSO4 to make a bigger impact Ex2 Calculate pH of 100 x 102 001 M H2SO4 solution Major species H HSO4 and H20 H2SO4aq H HSO4 We are looking at the second dissociation in this WHY HSO4aq 7 H s04A2 SO402 H HSO4 I 0 001M 001M C X X X E X 001 X 001 7 X plug into equilibrium eXpressions Ka2 12 X 1002 HSO402HSO4 001 XX001X assume X s are negligent 001X001 12 X 1002 solve X 12 X 1002 THIS value is ABOVE 001M which is IMPOSSIBLE so we cannot make the usual assumption and instead must use QUADRATIC EQUATION 001 XX001 X get the above into an a02 bX c 0 formation X b sqrtb024ac2a remember negative roots are nonreal X 45 X 1003 Thus H 001 X 00100045 00145 pH logH 184 since X was equal to about 12 of the original concentration it is signi cant and must be added in Characteristics of Weak Polyprotic Acids 1 typically successive Ka values are so much smaller than the first that they are insigni cant so typically calculation of a weak polyprotic acid is identical to that of a weak acid 2 sulfuric acid is unique strong acid for first dissociation and weak acid for second dissociation High concentrations of sulfuric acid 100M lt the first step suppresses the second for low concentrations second step candoes make a significant contribution 78 AcidBase Properties of Salts salt 7 ionic compound Salts That Produce Neutral Solutions Why do strong acids completely dissociate For one the conjugate base has no affinity for the proton So when anions CL NO3 are placed in solution they do not combine with any H thus no effect on pH IMPORTANT Salts that consist of the cations of strong bases and anions of strong acids have no effect on H when dissolved in water EX KCl NaCl NaNO3 KNO3 7 all neutral pH 7 Salts That Produce Basic Solutions EX aqueous solution of NaC2H302 major species are Na C2H302 and H20 Na is a neutral ion C2H302 is the conjugate base of acetic acid weak acid it now acts a base accepting protons water is neutral but only source of H ions So pH controlled by C2H302 C2H302 H20 HC2H302 OH base water conjugate acid hydroxide ion Kb 7 equilibrium constant for a base reaction Kb productsreactant HC2H302OHC2H302 Relationship Between Ka Kb and Kw Ka x Kb Kw for any weak acid and conjugate base A So when Ka 01 Kb is known the other constant can be calculated Ka for an acid Kb for a base For any salt whose cation has neutral properties Na or K etc and whose anion is the conjugate base ofa weak acid the aqueous solution will be basic EX Calculate the pH of a 030 M NaF solution Ka for HF 72 X 1004 Major species Na F and H20 HF is a weak acid weak acid anions have a signi cant af nity for protons Dominant reaction F H20 HF OH So Kb HFOHF Kb KwKa 10 X 1001472 X 1004 14 X 10011 Now ICE chart to gure out OH concentration Which will then be used to calculate H concentration then pH F HF OH I 030M 0 0 C X X X E 030 7 X X X Kb xx030 x XA2030 14 x 10A 11 x 20 x 10 Check to make sure X is less than 5 of original concentration OH X 20 X 1006


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