Principles Chemistry II
Principles Chemistry II CHM 212
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Chapter 19 Chemical Thermodynamics In Chapter 5 we examined the relationship between heat and reactions In Chapter 15 we saw that the rate and extent of reactions depend on energy considerations In this chapter we will examine energy relationships again and begin to tie things together At this point it would be a very idea for you to review the enthalpy sections of Chapter 5 in particular sections 1 7 4 and 6 7 7 This chapter attempts to be as correct as possible about some dif cult topics We will be skipping over some of that material because the purpose here is to give you a basic introduction to chemical thermodynamics In some places the notes expressly say that you won39t be responsible for certain material If you have any questions about topics I don39t go over in class make sure you ask me about them 191 Spontaneous Processes Many things in nature appear to always proceed in the same direction The book offers a falling egg or melting ice at room temperature as examples Other processes include iron rusting and a re burning It would be shocking for you to observe any of these processes occurring in reverse Processes that occur without outside intervention are called spontaneous Processes that are spontaneous in one direction are nonspontaneous in the reverse direction NB Spontaneity says nothing about rate Some spontaneous processes occur quite readily and rapidly for example a brick falls immediately and rapidly when you let go Others occur quite slowly A container lled with hydrogen and oxygen will remain essentially unchanged for centuries unless a spark is added to start the reaction Both are spontaneous however The latter example has a high activation energy Two factors affect reaction spontaneity within a system The first you have already encountered enthalpy As you know from both experience e g fire and Chapter 5 exothermic processes tend to be spontaneous There is a second major factor entropy which we will discuss shortly Reversible and Irreversible Processes 7 Skip this section 192 Entropv and the Second Law ofTher J If you think about it energy tends to disperse For example a hot object will warm the things around it and when a glass drops on the oor the energy will spread throughout it frequently causing the glass to shatter Energy never spontaneously concentrates Entropy S is a measure of how much energy is dispersed as a function of temperature Another example that is easy to see and that we ll return to in a bit is that of a liquid and gas of the same substance at the same temperature eg watersteam at 100 The gas can both hold more energy internally and because it is more mobile carry it away from a source more readily For this reason the entropy of the gas would be higher than that of the liquid Another way to look at this is by examining a gas at two different temperatures Figure 1018 p 387 shows the range of velocities of nitrogen gas molecules at 0 and 100 0C You can see that the molecules at the higher temperature are spread out over a much wider range of speeds If each velocity is viewed as a state then there is a wider range of states available to the higher temperature molecules We ll build on this idea of molecular motion in Section 193 Two examples of spontaneous processes that are not exothermic are expanding a gas into a vacuum and the melting of ice Consider two identical bulbs that are connected with a valve inbetween one containing a gas the other under a vacuum What happens if the valve is opened Of course some of the gas moves to the side with no gas I say quotof coursequot because no one ever guesses that the gas just stays in its original bulb for that matter no one would guess that gas in two bulbs would spontaneously move into just one of them This happens even if the bulbs are kept at constant temperature so heat evolution can39t be the answer Nor can work because the gases expand into a vacuum Yet the gas always moves into the vacuum Why When ice melts heat must be added an endothermic process Yet it always melts at 0 0C Why These processes have something in common the final state is more energetically dispersed or disordered than the initial condition This is easier to see in the second example so we39ll begin there In ice the water molecules are held in a rigid lattice unable to move On melting to liquid water the molecules are free to move about In the liquid the molecules are also able to spin or tumble with much greater freedom than they could in the solid state Thus liquid water can spread energy out within itself more effectively eg by spinning faster or by moving around to other locations ie carrying the energy away from its source For the gas bulbs the argument is similar The larger volume allows the energy stored by each molecule to be spread out further Entropy increases as dispersal of energy increases Although absolute entropies can be determined the difference in entropy between initial and final conditions is usually of greater interest to us AS Sf Si Hence AS gt 0 indicates that the final state has spread out energy relative to the initial state Skip Relating Entropy to Heat Transfer and Temperature The Second Law of Thermodvnamics What you have just seen for the two given examples is generally true Processes that result in an increase in entropy tend to be favored over those in which entropy decreases There are two ironclad rules regarding entropy however The first is the Second Law of Thermodynamics For any spontaneous process the entropy of the universe increases Recall that the quotuniversequot refers to the combination of the systems and surroundings ASuniV ASSys ASsm Also for systems at equilibrium ASumv 0 What the second law means is that if the energy within your system becomes less dispersed e g freezing water to ice the surroundings must become even more dispersed The example the book chooses to illustrate this point is a good one Iron rusts according to the reaction 4 Feg 3 02 g gt 2 Fe203 S The entropy change here is negative First 7 species combine to form 2 but more importantly the reaction coverts 3 gas molecules to the solid state How is the entropy of the surroundings increased By the heat driven off in the reaction AHrxn rs 1645 1d That heat makes molecules in the surroundings move faster Those in the solid state vibrate more rapidly and liquid and gas molecules move about at a higher velocity The spreading out of energy in the surroundings more than offsets the loss in the system when the gas molecules are moved into the solid state You don t feel the heat given off when iron rusts because the process is so slow We shall see shortly that this exothermicity is what makes the reaction spontaneous 193 The M 39 39 Interpretation of EntropV There is a third and nal law of thermodynamics The entropy of a perfect crystal of any substance at absolute zero is zero A perfect crystal is one that is 100 pure and has no defects In the last section we described entropy in terms of molecular motion For example gas molecules move about more than molecules in the solid state and so have more entropy Likewise for equal numbers of molecules in different volumes the molecules in the larger volume possess higher entropy because they can move into more locations It is relatively easy to see that a solid is more ordered than a gas And not too difficult to see that liquids are more ordered than gases The molecules move more slowly in liquids ie there is a smaller range of velocities so entropy is lower in liquids But why should entropy change in the solid state as temperature changes In order to understand how energy is dispersed by molecules we need to look at the different types of motions molecules undergo The motion discussed exclusively to this point is translational motion This is motion that results in movement between different locations e g moving from here to there Translational motion in gases is high moderatetolow in liquids and very neartoexactly zero in solids When solids crystallize they inevitably have defects One such defect is a missing molecule at a site As a solid warms to near its melting point it is possible for a molecule to hop from its existing site to an adjacent hole Nevertheless this does not happen often and translational motion does not generally contribute significantly to the entropy of solids Translational motion disperses energy by carrying it to another site and through the range of velocities available to the molecule When a molecule spins about an internal axis it undergoes rotational motion Like all motions rotational motion decreases with temperature Small highly symmetrical molecules frequently have signi cant rotational motion Consider a sphere e g a noble gas atom As it spins about any internal aXis there is nothing to bump into a nearby atom to slow its rotation Likewise linear HCECH could spin in place much easier than H3CCECCH3 because the latter has CH bonds that project out of the molecular aXis that could easily bump into a neighboring molecule Replacing the hydrogens in the latter molecule with CH3 groups would slow rotation even more at a given temperature H H a H cc H HIc cc calH A and H s1de v1ew Vlew s1de v1ew end Finally all atoms in molecules engage in vibrational motion The location of atoms in molecules is not static Bound atoms vibrate much like balls connected by a spring At higher temperatures there is more vibration Both bond vibration and rotation allow the dispersal of energy m a molecule We ll spend some time in class discussing this concept The entropy of a crystal is zero when it has no defects and all of the atoms and molecules cease to engage in translational and rotational motions and are limited to a single set of vibrations A few generalizations are a For a collection of substances at the same temperature the gases will usually possess greater entropy than the solids or liquids regardless of chemical makeup b For a single substance Sg gt S gt 85 0 V In a given physical state entropy increases with temperature d In a chemical reaction entropy will increase if the number of gas molecules increases 194 Calculation of Entropy Changes The entropy change associated with reactions or physical changes is calculated much like enthalpy changes are calculated The generic equation one uses is Asorxn 2nAS products 2nAS reactants where n are the stoichiometric coefficients from the reaction Remember they have units moles and that the superscript 0 means the entropy value was determined under standard conditions 25 0C 1 atm for gases pure liquids and solids l M concentration if in solution An important difference between enthalpies and entropies is that while elemental enthalpies are defined as zero all entropies have nonzero values Also note from their units JmolK that entropies are temperature dependent Example What is the entropy change associated with burning one mole of ethane CH4g 2 02 g gt C02g 2 H20 ASquot 1 molCO22l36 Jmol K 2 molH206991 Jmol K 1 molCH41863 Jmol K 2 m01022050 JmolK 2428 JK 195 Gibbs Free Energy If enthalpy and entropy independently tend to cause reactions to proceed spontaneously then as you might expect there is a relationship between them that always predicts spontaneity We know that enthalpy is favorable when it is negative and entropy is favorable when it is positive Furthermore we know entropy increases with increasing temperature We can combine these observations to yield a new quantity Gibbs free energy AG where AG AH iTAS For any process If AG lt 0 the process is spontaneous as written AG gt 0 the process is spontaneous in the reverse direction AG 0 the process is at equilibrium Standard Free Energy Changes To make free energy easier to use we will employ the same standard conditions used in enthalpy That is 25 0C 1 atm for gases pure liquids and solids and l M for solutions Elements are de ned as having AGo 0 Remember the superscript 0 means standard state Just as for AH and AS AGODm ZnAGf products ZnAGf reactants Using Appendix C we can determine whether or not a reaction is spontaneous or not at a particular temperature It is important to state again that AGo tells us nothing about reaction rate 196 Free Energy and Temperature The temperature dependence of AG can be signi cant Your experience tells you that reactions accelerate when the temperature increases What may surprise you is that whether the reactions occurs or not Is it spontaneous or not may depend on temperature There are four possible enthalpy entropy pairs AH lt 0 amp AS gt 0 AG is always negative so the reaction is always spontaneous e g 203 g gt 302 9 AHquot 285 kJmol ASquot 140 JmolK AH gt 0 amp AS lt 0 AG is always positive so the reaction is never spontaneous eg The reverse of the previous reaction AH lt 0 amp AS lt 0 AG lt 0 only at low temperatures Low temperatures make the entropy term smaller and less important eg CaOS C02 g gt CaCO3 5 high temperature drives the C02 off the product AH gt 0 amp AS gt 0 AG lt 0 only at high temperatures High temperatures make the entropy term larger and more important Think of it as the extra energy required to drive endothermic reactions The words low and high in the preceding sentences appeared in quotation marks Why Because they are relative terms For example assume the AH for two dilTerent reactions is 10 kJmol Reaction 1 has a AS of10 JmolK while AS for reaction 2 is 100 JmolK In the rst case the reaction isn t spontaneous until 1000 K 727 0C a high temperature indeed However the second reaction is spontaneous over 100 K l73 0C a temperature that no one would consider toasty Thus high temperature means that if one begins warming a reaction mixture from 1 K the reaction is initially nonspontaneous and at some temperature will become spontaneous Low temperature means that warming will eventually cause a spontaneous reaction to become nonspontaneous The following example will show how such a calculation proceeds Ex Is the reaction 2 H2 g 02 g gt 2 H20 spontaneous at 25 0C At what temperature does it become spontaneous AH 2 molH2028585 7 2 molH2x0 m K 1 m01020 m K 1 57l70 kJ ASrm 2molH206996 i 2molH2l3058 I 1m01022050 I K mo mol K mol K 32624 JK lkJ AG 0 57l70 k 7 298 K 32624 JK 7 m IOOOJ 474 kJ AG lt 0 so the reaction is spontaneous at 25 0C Since AHIXHo lt 0 and ASDmo lt 0 the reaction is favored at low temperatures The crossover point for spontaneity is when AGIXHo 0 Below this temperature the reaction is spontaneous above it nonspontaneous In this calculation we assume AH and AS don t change with temperature They actually do but not very much 0 k 57170 k 7 TX32624 JK 17k 1000 T 17524 K or 14792 0C 197 Free Energv and the Equilibrium Constant The book derives a very useful equation but all you need to do is to remember it AG RT1nKeq where R 8314 JmolK and Keq KC for solutions KP for gases use atm and X for liquid mixtures Thus the free energy of an equilibrium can simply be determined by measuring the amount of each substance present Not surprisingly if Keq is greater than one the equilibrium lies to the right AG is negative January 10 2005 Chapter 20 Electrochemistry We will now examine in depth reactions in which electrons are passed from one species to another These reactions were first introduced in Section 44 and involve changes in oxidation numbers In particular you should go back now and review how to assign oxidation numbers Much of the work in this chapter requires you to do this Reactions involving a change in oxidation numbers of two or more substances are called oxidationreduction reactions They are also called redox or electron transfer reactions In general the study of reactions involving the movement of electrons is called electrochemist 201 OxidationReduction Reactions In a redox reaction m an oxidation and a reduction must occur After all the gained or lost electron must come from or go to somewhere The substance that loses electrons is called the reducing agent or reductant It is oxidized The species gaining electrons is called the oxidizing agent or oxidant It is reduced 4 1 0 4 2 1 2 4 oxidationnumbers CH4 g 2 02g gt C02 g 2 H20 reductant oxidant There are a few things worth remembering about redox reactions Again oxidation and reduction must both occur in a redox reaction The second ties into the previous observation One must be careful when using oxidation states In some cases the oxidation number of a species represents the actual charge on a species eg l on chloride in sodium chloride in others it does not eg l on C1 in CCl4 The difference is that in the former case we have an ion while in the latter chlorine is in a molecule In general do not assume the oxidation number on an atom is the same as the actual charge on the atom 202 Balancin Oxidnti quot J quot Penntinnq For all redox reactions the oxidation and reduction steps can be extracted into separate equations explicitly showing the movement of electrons These are called halfreactions For the previous example oxidation CH4 2 H20 gt C02 8e39 8 H reduction 02 4 e39 gt 2 0239 The oxidation half of this pair looks a little involved but you ll see shortly how it comes about There is a set of rules that if followed exactly M yields balanced redox reactions In the stoichiometry chapter you balanced some redox reactions by inspection or by the method shown in the text This will work for some redox reactions with relative ease however others will be very very difficult to solve using these methods For that reason if you can t balance very quickly using the other methods use this one For the purpose of practice you should use only this method in this chapter or else you won t become adept enough to use it when you have to Method of HalfReactions 1 Divide the unbalanced overall reaction into the oxidation and reduction parts you ll need the oxidation numbers to do this 2 Balance the elements other than H and O by any method 3 Balance H and O by adding H and H20 respectively ifnecessary 4 Balance the charge by adding electrons 5 Multiply the coefficients of each halfreaction by the smallest factor such that the number of electrons consumed in the reduction equals the number of electrons produced in the oxidation 6 Add the halfreactions and simplify 7 If the reaction was done in basic solution add enough OH39 to each side to exactly react with the H in the balanced equation This can also be done after step 4 or at the very end whichever you prefer EX Balance the reaction Fe2aq Cr207239aq gt Fe3aq Cr3aq acidic solution a oxid Fe2 gt Fe3 redn Cr207239 gt Cr3 b oxid Fe2 gt Fe3 redn Cr207239 gt 2 Cr3 c oxid Fe2 gt Fe3 redn Cr207239 14H gt 2Cr3 7H20 d oxid Fe2 gt Fe3 e39 redn Cr207239 14H 6e39 gt 2Cr3 7H20 e oxid 6 X Fe2 gt Fe3 e39 redn 1X Cr207239 l4H 6e39 gt 2 Cr3 7H20 f 6 Fe2aq Cr207239aq 14Haq gt 6 Fe3aq 2 Cr3aq 7H20 At this point you should check to make sure that the element totals are the same on both sides E that the total charge on each side is the same This is very important If this wasn t necessary part a oxidation would already be balanced Yet the two species are clearly not identical To balance reactions in basic solution initially balance it just like an acidic reaction At the end there will be an additiona step g that will convert the reaction from acidic to basic solution EX Balance SZ39aq MnO439 gt 85 Mn02s a oxid S239 gt S redn MnO439 gt Mn02 since S amp Mn are already balanced skip b c oxid S239 gt S redn MnO439 4H gt Mn02 2 H20 d oxid S239 gt S 2e39 redn MnO439 4H 3e39 gt Mn02 2H20 e oxid 3 X S239 gt S 2e39 redn 2 X MnO439 4H 3e39 gt MnO2 2 H20 f 3S239 2Mn0439 8H gt 3S 2Mn02 4H20 g Now add 8 OH39 to each side The OH39 will react with H to yield water 3S239 2Mn0439 8H20 gt 3S 2Mn02 4H20 80H39 3 S239aq 2 MnO439aq 4 H20 gt3 Ss 2 Mn02s 8 OH39aq 203 Voltaic Cells The electrons that ow in redox reactions can do work The most common deVice for harnessing this energy is the battery see Section 207 Batteries are a type of voltaic or galvanic cell These are deVices that require the electrons to move along an external pathway instead of directly between reacting species In normal reactions the reactants are mixed in the same container they couldn t react if they weren t In redox reactions the reactants can actually be placed in separate containers that are connected such that electrons can ow between them Consider the reaction Cu2aq Zns gt Cus Zn2aq The gure below shows a fairly typical voltaic cell Zn gt Zn2 2e39 Cu2 2e39 gt Cu Devices inserted into solution to act as sites of electron exchange are called electrodes In the picture above the solid zinc and copper blocks are electrodes It is important to remember that electrodes need not be made of the same material as the species undergoing redox processes In fact the most common substances used for electrodes are platinum and graphite The electrode at which oxidation occurs is called the m while reduction occurs at the cathode In the picture the zinc strip is the anode while the copper piece is the cathode The anode is labeled with a negative charge the cathode with a positive charge Why When oxidation occurs electrons are left on this electrode They then ow towards the cathode which is labeled positive because if negatively charged electrons move in that direction it must be positive relative to the anode For current to ow there must also be a salt bridge between the vessels This is because when Zn2 forms there must be a counterion present as well Likewise when Cu2 reduces to copper metal there will be an excess of nitrate ions The salt bridges keep each reaction container electronically neutral by removing or adding ions as appropriate A 39 39 View ofFlectrode Processes 7 Read on your own 204 Cell EMF In redox reactions electrons move because as in all reactions the products are more stable than the reactants We could use AG but that must be calculated It would be far easier to use a directly measurable quantity The M V measures the potential energy difference between the electrodes One volt is the potential difference imparting one joule of energy to a one coulomb C charge The electromotive force emf is the potential difference between two electrodes in an electrochemical cell It is also frequently called the cell potential E0811 When reactions are done under standard state conditions 1 atm l M 25 0C one obtains a standard emf or standard cell potential E00811 For the cell in the picture on the previous page the emf is measured as Zns Cu2aq gt Cus Zn2aq E00811 110 V But this reaction is just the sum of two halfreactions The following discussion is different from the textbook but I find it easier to understand The two halfreactions for this cell are shown below oxid Zn gt Zn2 2e39 Eooxid redn Cu2 2e39 gt Cu Eoredn Since the reduction looks just like the halfreaction in the reduction potentials table Appendix E we copy the number down exactly The oxidation is reversed from the table Zn2 2e39 gt Zn Eoredn 076 V and in this case we change the sign on the value from the tables Now we can add the reactions to get the net reaction and add the oxidation and reduction halfcell potentials to get the overall cell potential oxid Zn gt Zn2 2e39 Eooxid076 V redn Cu2 2e39 gt Cu Eo 034 V Zns Cu2aq gt Cus Zn2aq E00811 110 V Thus E00811 Eoredn Eooxid if you write the halfcells as above Earlier we said that oxidations and reductions cannot happen independently of one another so how can be measure halfcell potentials Actually we can t Using halfreactions requires us to de ne the potential of one reaction then measure all others relative to it This is a lot like how AH is treated The reaction used to de ne emfs is 2 Haq 2 e39 gt H2 g Eoredn 0 V Thus whenever the hydrogen electrode is used the cell potential equals the potential for the other halfreaction or its reverse depending on the direction of the hydrogen reaction EX Find the reduction potential for the oxidation of zinc if the emf for the following reaction is 070 V Zns 2 HQ gt Zn2aq H2 g oxid Zn gt Zn2 2e39 Eooxid redn 2H 2e39 gt H7 Equotr an0V net Zn 2 H gt Zn2 H2 E00811 076 V Eocell Eoredn Eooxid V 0 Eooxid Eooxid 076 V But we want the reduction potential of zinc the reverse of the reaction show above so we need to change the sign on our answer Zn2aq 2e39 gt Zns E redn 076 V This is generally true when a halfreaction is reversed the sign on emf changes EX What is E00811 for the reaction Zns Cu2aq gt Cus Zn2aq using the values in Table 201 on p 791 oxid Zn gt Zn2 2e39 Eooxid076V redn Cu2 2e39 gt Cu Equotr Aquot 034 V net Zn Cu2 gt Cu Zn2 Eoceu 110 V An important feature of electrochemistry is that cell potential doesn t depend on reaction stoichiometry Is this reasonable Consider the reaction 3 Cu2 2 Al gt 3 Cu 2 Al3 The function of the extra copper ions is simply to take away more electrons The energy of the copper hasn t changed OXidizing and Reducing Agents Open your book to Table 201 before going further This is a list of standard reduction potentials The first reaction the reduction of uorine to fluoride is one that you would expect to proceed spontaneously It has a positive Eoredn and this can be a convenient way to remember F2 2e39 gt 2 F39 Eoredn 287 V the sign for spontaneity when you look at a reduction potential table Remember that any reaction spontaneous in one direction is nonspontaneous in the other Thus at the other end of the table the reduction of lithium ions to lithium metal is very unfavorable very negative Hence any reaction on the table coupled with a reaction below it run backwards will yield a Li gt Lit e39 Eyed 305 v spontaneous net reaction It is also important to note that the same species can act as a reductant in one reaction and an oxidant in another For example Fe2 Zn gt Fe Zn2 Eocell032 V 2Fe2 C12 gt 2 Fe3 2 Cl39 Eoce11059V Finally the more positive a reduction potential the better oxidant the species is Conversely the more negative a reduction potential the better the species to the right of the arrow in a halfreaction is as a reductant 205 Spontaneity of Redox Reactions It is important to remember that any reaction with a positive Eo value is spontaneous We know from Chapter 19 that reactions are spontaneous when AG0 is negative Therefore the two must be mathematically related The equation that accomplishes this is AGquot n rFEo where n is the number of moles of electrons transferred and F Faraday constant 96500 Cmol 96500 JVmol Ex What is the standard free energy change for the reaction Zns Cu2aq gt Cus Zn2aq AGquot 2 molwjl10 V1 kJ 193 k V mol 1000 J Equilibrium constants may also be calculated from cell potentials using the equations A 0 RTaneq and A 0 n FEo combine to yield n FEo RTaneq EX What is the equilibrium constant for the reaction Zns Cu2aq gt Cus Zn2aq 8314 Jmol K298 K aneq 96500 JV mol2 mol 110 V aneq 857 Keq 164 X 1037 Of course this is hardly an equilibrium in the real sense of the word Indeed to have a single Cu2 ion left in the solution after adding zinc to a l M Cu2 solution the container size would have to be around 27 trillion liters a cube 186 miles on a side For a reaction with one electron transferred and aEo 010 V the equilibrium constant would be around 50 for 020 V it is 2400 As you can see equilibrium constants increase quite rapidly with cell potential Thus this equation is of limited value 206 Effect of Cell quot on EMF 7 skip 207 Batteries This section discusses four types of voltaic cells The first three are batteries that you may well be familiar with the lead storage car battery the alkaline battery and rechargeable batteries The final entry is for fuel cells which are not currently commercially very important today but will become so in the reasonably near future They have the advantage of providing large amounts of electricity with no direct pollution Automobile manufacturers are looking to fuel cells as a way of powering electric cars Read this section on your own 208 Corrosion Corrosion is the electrochemical breakdown of metals by the environment This process is thermodynamically favorable for most metals under ambient conditions Only gold will not corrode for thermodynamic reasons Other factors prevent platinum and palladium from corroding All other metals corrode Corrosion may either be very destructive as in the cases of iron and zinc or hardly noticeable as for aluminum or chromium The reason for the variation results from the nature of the products In all cases a metal oxide results from air oxidation of the metal Aluminum oxide A1203 adheres strongly to aluminum metal and thus provides a protective coating on the metal Further oxidation does not occur because neither oxygen nor water vide infra can penetrate it In contrast when destructive oxidation occurs the metal oxide akes off This constantly provides fresh metal for further oxidation until finally all of the metal oxidizes Interestingly while corrosion certainly requires oxygen acidic water is also a requirement The more acidic the water the faster corrosion occurs The halfreactions for this process follow oxid Ms gt M aq ne39 redn 02 g 4Haq 4e39 gt ZHZOQ One reason corrosion is such a problem is that not all of the metal need be exposed to the environment In normal reactions the reactants must collide with each other or intermediates for the reaction to occur Since metals conduct electricity exposure of relatively small areas of the metal can cause significant corrosion There are several ways to prevent corrosion A common way is to seal the surface with paint or plastic so that air and water don t have access to the metal This works unless the paint chips or peels off The other common method is to either coat the surface with a metal that is 12 easier to oxidize galvanization or to attach the metal by wire to a block of easier to oxidize metal located elsewhere For example iron is usually protected with either zinc or magnesium In these cases the second metal is sacri ced Note the magnesium and zinc are not chosen because they are less expensive than iron In fact they cost more per pound The philosophy for doing this goes as follows The manufacturer of a product containing sheet metal can t know where the paint will be scratched or ake off A thin coat of zinc over the whole sheet will protect against a scratch anywhere because of the ability of the metal sheet to conduct electricity The zinc coating costs very little and will protect the sheet until the scratch is painted over from rust that could eat all the way through the sheet If nothing is done the scratch will result in rust going all the way through a panel and requiring its replacement This is much more expensive than the cost of the coating The same sort of logic governs using zinc or magnesium blocks to protect pipes or bridges 209 Electrolysis All of the reactions discussed in this chapter to this point occur naturally spontaneously in the forward direction They are nonspontaneous in the reverse direction Electrochemical reactions can be pushed in the nonspontaneous direction by providing an external current source A processes driven by an external source of electrical energy is called electrolysis In an electrolytic cell electrodes are placed into either molten salt or a solution of the ions to be electrolyzed A current is then passed through the solution with a potential that is larger than the EquotIXH for the formation of the material This drives the reaction in reverse The book has a nice picture of how molten table salt NaCl is electrolyzed to yield elemental sodium metal and chlorine gas on p 927 Figure 237 There is also an unrealistic picture on p 813 of your book An interesting thing may happen when a salt is electrolyzed in aqueous solution It is easier to reduce water than many metal ions When this happens the metal ion remains unchanged and water is reduced to hydrogen gas and hydroxide ion NaClaq Mm NaOHWD 12 H2 g This is how commercial sodium hydroxide caustic soda is made It is one of the highest volume chemicals produced in the world today If carbon dioxide is bubbled through the solution sodium bicarbonate NaHCO3 forms NaOHW coug gt NaHCO3aq Electrolysis with Active Electrodes 7 read on your own Quantitative Aspects of Electrolysis There is great practical interest in how much electricity is required to perform an electrolysis experiment At the industrial scale the price of electricity will be a signi cant factor in the cost of the process and at both large and small scales the rate of delivery of the electrons will affect the speed at which the reaction proceeds For example all aluminum is produced by electrolysis For that reason aluminum manufacturers typically locate their foundries near inexpensive sources of electricity eg the Paci c northwest The amount of energy associated with an electrochemical process depends on the total number of electrons that ow between the electrodes After all electricity is just the ow of electrons Since the charge on the electron is fixed we can measure the ow of electrons by monitoring the charge as it passes by a fixed point The rate of charge movement is measured in terms of amperes A where lAlE 14 Recall that the coulomb a measure of charge where 1 mole of electrons carries 96500 C or 1 T of charge Thus we can calculate either the amount of electricity produced by a redox reaction or the amount required to effect an electrolysis EX How much magnesium and chlorine are produced when a current of 0452 A is passed through molten magnesium chloride for 150 hours First generate a balanced halfreaction for either element Mg2 2e39 gt Mg or 2 C139 gt C12 2e39 For each element it is important to balance using whole number coefficients This is a relatively complicated dimensional analysis problem You can begin with either the given electrochemical data or the molecular weight of the element Both methods are used below 1mol 2431 massMg0452 A i w 150 hr l T Mg A s hr 96500 C 2 F molMg 0307 gMg maSSCu 7090gc12 11110102 1 F 1C 3600s150hr0452 A molc12 2T 96500C As K hr 0897 gCl2 Electrochemical Work 7 skip January 10 2005 Chapter 16 7 Acid Base Egu ib a Most solutions that occur in nature are slightly acidic One reason for this is that when carbon dioxide dissolves in water it forms carbonic acid H2CO3 Basic solutions that exist in nature typically have limited direct exposure to air For example your blood is slightly basic and under ideal circumstances is not open to the air Likewise peach pits contain strychnine an organic base which is not protonated because of the shell that seals out the atmosphere Figure 165 p 623 lists the pHs of some common substances Note that only seawater occurs openly in nature with a pH over 7 We ll see why before this chapter is through 161 Acids and Bases A Brief Review Earlier in Chapter 2 you were given quick working definitions of acids and bases We will now expand on those de nitions Acidic solutions generally have a sour taste eg grapefruit and lemon juices and dissolve certain metals Bases have a bitter taste and feel slippery The slippery feeling comes from the base breaking down oils and fatty acids in your hands The old lye soap used a century ago cleaned by removing the top layer of skin and anything that was attached to it Modern soaps use a different type of active chemical that doesn t react with your skin The first good definition of acids and bases was proposed by Arrhenius the same person who proposed activation energy He proposed that acids were substances that produces H ions in water and bases were substances that releases OH39 ions in water However this de nition turns out to be too restrictive The book s definition is incorrect Indeed ammonia is not an Arrhenius base as stated in Section 162 but ammonium hydroxide the product of the reaction between water and ammonia is 162 BronstedLowg Acids and Bases The hydrogen ion H is nothing but a proton and can t really exist freely in aqueous solution because of its high charge density The positive charge attracts the lone pairs of electrons on nearby water molecules The result is that each proton is surrounded by water molecules with their lone pair electrons pointing at the proton Typically 3 to 4 water molecules associate with each proton HH203 or HH204 but the hydrogen ion is usually written as either Haq or H3Oaq for simplicity s sake The latter form indicates the association of water with the proton and is called the hydronium ion In 1923 two chemists Johannes Bronsted and James Lowry independently proposed new definitions of acids and bases They proposed an id is a substance that donates a proton to another substance A m is something that can accept a proton from another substance These definitions have a couple of signi cant advantages over the older Arrhenius definitions First some compounds such as alcohols clearly acted like acids some of the time but did not qualify under the Arrhenius de nition For this reason all Arrhenius acids and bases are also Bronsted Lowry acids and bases but not vice versa Second it included reactions that occur in non aqueous solutions For example the gas phase reaction HClg NH3g gt NH4ClS is a BronstedLowry acidbase reaction but not an Arrhenius acidbase reaction BronstedLowery acidbase reactions are frequently called protontransfer reactions because of the descriptive nature of the term Some compounds possessing both a polar elementtohydrogen bond e g OH or N H and a lone pair of electrons frequently are able to donate a proton or accept one depending on the conditions water provides an example of this behavior Molecules that can act both as an acid and base are called amphoteric For example water reacts with hydrogen chloride gas to yield hydronium and chloride ions H20 HClg gt H3Oaq C139aq It also reacts with ammonia to yield ammonium and hydroxide ions H20 NH3 g gt NH4aq OH39aq Conjugate AcidBase Pairs In an acidbase reaction a proton is transferred from the acid to the base HAB AA39HB H HA B gt A H acid base conjugate conjugate base acid Example HCl H20 gt C139 H3O The base that received the proton can now donate that proton to something else It has become an acid Likewise the acid that donated the proton can now accept one from somewhere else It has become a base The new base and acid from which it was formed are called a conjugate acidbase m Analogously the new acid and the base from which it was formed are also a conjugate acidbase pair Every BronstedLowry acid has a conjugate base and every BronstedLowry base has a conjugate acid Relative Strengths of Acids and Bases When writing an acidbase reaction the direction of the arrow tells us something important The stronger is an acid or base the weaker is its conjugate Consider hydrogen chloride and water If hydrogen chloride is the stronger acid then it is better at giving up H ions than H20 If this is true then chloride ion in HCl must be poorer at holding onto its H than H20 in H3O Hence chloride ion is the weaker base and is the conjugate base of hydrogen chloride When strong acids dissolve in water they completely transfer their protons to water molecules to form hydronium ions as shown above Thus the strongest acid that can exist in water is H3O This is called the leveling effect Similarly the strongest base that can exist in water is hydroxide ion Bases stronger than hydroxide will remove a proton from water to form hydroxide ion To develop a solution that is more strongly acidic than a strong acid in water the strong acid must be dissolved in a more acidic solvent eg HCl in acetic acid Likewise dissolving a strong base in a less acidic solvent than water generates a more strongly basic solution e g NaOH in ethanol When acids and bases react the position of the equilibrium favors transfer of a proton from the stronger acid to the stronger base Thus for the reaction HAB gtA39HB there are two acids HA and HB and two bases B and A39 The stronger acid will yield the weaker conjugate base Thus all you need to do is identify the stronger of one of the pairs to determine the direction of reaction 163 The Autoionization of Water The poor electrical conductivity of very pure water surprises most people So why is water a poor conductor Because it contains very few ions In contrast tap water is a good electrical conductor because it contains dissolved saltsYou ve probably seen this when you let water evap orate from a spoon or glass when it is drying There are a few dissolved ions in pure water though and this provides water with what little conductivity it displays To produce those ions water undergoes an interesting reaction called autoionization in which it reacts with itself to form ions 2 H20 H3Oaq OH39aq or more commonly Hzom F Hltaq t 0H39ltaq The former reaction is more representative of what actually happens in solution The lone pair of electrons on one water molecule is attracted to then removes a hydrogen atom on a neighboring water molecule Since this is an equilibrium there must be an equilibrium constant associated with it H300H39 H0H39 q H20 H20 K e Usually H3O is expressed as H in these equilibrium expressions to make writing them easier Now water is a pure substance and has a xed concentration at any given temperature just like the solids presented in the previous chapter thus KeqH2O H1 OH39 Kw KW is called the ionproduct constant for water and has a value of 10 x 103914 at 25 0C You will need to memorize this number In pure water at 25 0C H OH39 10 x 10397M see Exercise 164 p 621 This very low concentration of ions accounts for the low conductivity described above for water Water in which H OH39 is called neutral It is important to remember that when an acid or base is neutralized the nal solution need not be neutral We ll return to this later When H gt OH39 the water is acidic When H lt OH39 the water is basic For example you ve worked with 01 M HCIWD in lab before As you know hydrochloric acid is a strong acid and ionizes completely in water Therefore H HCl 01M 10 X 103914 HOH39 10 X 103914 01OH39 OH39 10 X 103913M As with all equilibrium constants KW has no units so the units on H are discarded and then units for OH39 are inserted when necessary 164 The pH Scale Since the hydrogen ion concentration in water is typically quite small the pH scale was developed to simplify the eXpression of how acidic or basic a solution is It is a logarithmic scale and so eliminates the eXponential term pH 10gH The pH ofpure water at 25 0C is calculated as pH logl0 X 10397 700 There are a few points worth noting here First pH 700 is neutral only at 25 0C but unless you are told otherwise all of the eXamples you will see in this book will be at 25 OC NeXt only the zeros are signi cant The 7 provides information on the eXponent By convention pHs are written with only two digits right past the decimal Be very care ll with significant figures here A pH of 631 has only two significant digits Remember this is a logarithmic scale This means that a change in pH of one unit equals a concentration change of a factor of 10 For that reason in biological systems pH ranges are usually relatively small on the order of a few tenths of a pH unit Larger pH changes can cause cell and even organism death pH decreases as H increases Therefore in acidic solution H gt 10 X 10397 and pH lt 700 in basic solution H lt 10 X 10397 and pH gt 700 Other 2 Scales A similar scale exists for hydroxide ion not surprisingly called the pOH scale It is calculated according to the equation pOH logOH39 pH and pOH are related according to an equation derived below KW 10 X 103914 HOH39 logHOH39 logl0 X 1039 logH logOH39 1400 pH pOH 1400 Measuring pH Read this on your own You ll do a lab involving this soon 165 Strong Acids and Bases Strong Acids As stated in the previous section a strong acid completely ionizes Thus the H equals the concentration of the added acid before ionization You should remember the most common strong acids HCl HBr HI HNO3 HClO3 HClO4 and H2SO4 Only the rst proton in H2SO4 completely ionizes Assume all other acids you encounter in this class to be weak acids EXample What is the pH of020 M HClO4 H HClO4 020M pH log020 070 Strong Bases Strong bases can be thought about like strong acids They dissociate completely however since some bases produce more than one OH39 their concentrations do not necessarily equal the hydroxide concentration eg MgOH2 All Group IA and IIA metal hydroxides are strong bases except beryllium hydroxide BeOH2 Many weak bases that contain protons become strong bases when deprotonated For example ammonia is a weak base while NH239 is a strong base NH239aq H20 gt NH3aq OH39aq Sodium hydroxide and potassium hydroxide are the most commonly used strong bases This is because the other alkali metal bases are more expensive and the alkaline earth bases have low solubilities Example What is the pH of 020 M barium hydroxide OH39 020 molBa0H22mO1 H39 040 M lmolMOH2 pOH log040 040 pH 1400 7 040 1360 or Ht 13900 X1044 25 x103914M OH39 040 pH log25 x 1014 1360 166 Weak Acids Most acids are weak acids and exist as an equilibrium in water between their protonated acid form and deprotonated conjugate base form When the equilibrium is written out the transfer of the proton to water is usually written as a simple loss of a proton for clarity m HA HAM Hltaq Aaq Ka where Ka is the aciddissociation constant The larger is Ka the stronger is the acid Example 10 A 010M formic acid HCOZH solution has a pH of238 What is its Ka What is its percent dissociation a H antilog238 42 X 10393 M Note The number of signi cant gures is correct Since we assume the only source of H ions is the acid then HCOZ39 H Finally HCOZHL 010 M So HCOZHe 010M7 000420M 010M 3 3 Ka 42X10 42X10 18X104 010 H 3 b dissociation X 100 X 100 42 HCOZHi 010 Calculate Ka from pH This section is basically a demonstration on how to calculate the pH of a weak acid solution given the initial concentration of the acid and its Ka As an example the book considers a 030 M acetic acid CH3C02H or HOAc solution and calculates its pH and percent dissociation We ll work through two different solutions here 00030 M and 30 M and compare them to the book s result There is a useful mathematical tool that comes into play in these problems If the initial acid concentration is more than 100 times Ka the amount of acid that dissociates is negligible and can 10 be ignored when calculating the undissociated acid concentration If you make this approximation you must check your nal answer to verify its validity 00030 M HOAC H4r OAC39 a HOAC 18 X 10395 HOAci 30M HOAce 30 7 XM Hm OAc39h 0M qu OAK xM X2 30X 18 X 10395 Since HOAci gt 100Ka assume 30 X 30 2 18 x10395 X7 30 X 73 X 10393 check 30M 7 00073M 29927 E 30 The assumption is valid pH log73 X 10393 2 l3 73X10393M dissociation X 100 024 HOAci 00030M HOAce 00030 7 XM Hm OAC39L 0M Hue OAc39 M 18 X 10395 X Z 00030 X Now 100Ka Z HOAci so assuming X is negligible might not work There are two ways to go about solving the problem here The straightforward way is to use the quadratic equation which is le to you The following is an alternative appr0Ximation method that is fairly quick and works on borderline cases It is called the method of successive approximations Begin by assuming X is negligible as we have in the past We then find X 23 X 10 This value is indeed signi cant because subtracting it from 00030 changes that value Now substitute X back into the nal equation above in the denominator only and solve for the X in the numerator 2 X 18 X 10395 00030 0000232 X 22 X 10394 Since this value is virtually unchanged from the past value we can accept it as accurate This method will work conveniently to about HOAci gt lOKa If the value of X is significantly different simply repeat until it doesn t change pH log22 X 10394 366 22 10quot39M dissociation X X 100 73 00030M Two things are noteworthy from this set of calculations On going from 0003 M to 03 M to 30 M we see two trends Not surprisingly the first is that increases acid concentration leads to decreasing pH There is something worth commenting on here however A thousandfold increase in acetic acid concentration only decreases pH from 366 to 213 the same change in strong acid concentration would cause a change of 300 units In general changes in weak acid concentration result in small changes in pH Second percent dissociation decreases dramatically in the same concentration range from 73 to 024 Thus in dilute solution a larger fraction of acid molecules ionize than in more concentrated solutions Polyprotic Acids These are acids capable of donating more than one proton Each successive dissociation is 12 designated Kal Kaz and with each constant usually 104106 times smaller than the previous one This is reasonable since it should be more dif cult to separate a proton from an anion than from a neutral molecule As long as aKa is at least a factor of 1000 from the next Ka the other Kas can be ignored when doing calculations An expanded version of Example 13 p 635 illustrates this nicely Example 13 What are the concentrations of H2CO3 HCO339 CO3239 and H in a solution that is initially 00037M in H2CO3 HHCOJ K1 a H2C03 2 43 x10397 X assume x ltlt 00037 00037 X 2 43 x10397 X 00037 x 40 x 10395 40 x 10395 ltlt 00037 so the assumption is valid Thus H2CO3 37 x 10393 M and Ht HCO339 40 x 10395 M 27 Kaz H llCOE l HCO3 75 56 x103911 assume xltlt 45 x10395 45 x10 x 56 X1011 45x10 5x 45x 1075 x 56 x103911 56 x103911 ltlt 45 x10395 so the assumption is valid Finally CO3239 56 x 1039 This brings us to an important shortcut When a polyprotic acid is dissolved in water the 13 concentration of the second and subsequent deprotonation products equals the corresponding Ka For example For H3A HAZ39 Kaz and A339 Ka3 167 Weak Bases Weak bases work much like weak acids One thing to remember is that you must treat the reaction of the weak base with water a hydrolysis reaction just like you would a comparable reaction of a weak acid with water A generic weak base reacts with water as follows Kb HBOH B BH20 HB OH39 Kb is the base dissociation constant Stronger bases have larger Kbs Many neutral weak bases are similar to ammonia These are bases with the generic formula NR3 Where R can be any combination of hydrogen and groups binding to N through a carbon atom Examples of such groups include CH3 methyl CH2CH3 ethyl C6H5 phenyl for the methyl group the possible weak bases are CH3NH2 CH32NH and CH33N The other major class of weak bases are salts of conjugate bases of weak acids Sodium acetate and sodium bicarbonate sodium hydrogen carbonate are two examples of such salts Ex What is the pH ofa 010M NH3 solution NH3 H20x NH4 OH39 NH40H Kb18x10395 NHgl NH3i 010M NH3e 0107xM NH4e 0M NH4e OH39e xM Z X 18 X 10395 just as for acids assume X ltlt 010 010 X X2 18 X 10395 010 X 134 X 10393 the assumption is valid OH39 134 X 10393M 3 pOH log134 X 10393 287 pH 1400 287 1113 168 Relationshi Between Ka and Kb Let s brie y review some of what we ve covered in this chapter The conjugate base of a weak acid is itself a weak base When dissolved in water this weak conjugate base abstracts a proton from water to yield the original weak acid and hydr0Xide ion Most weak bases are compounds of this type At first it might look awkward that the solution is basic since the weak acid is present but acidity is measured in terms of free H not undissociated acid The strength of a weak base formed from a weak acid depends on how strong the acid was in the first place Consider two acids HA and HB where HA is the stronger of the two acids Which will have the stronger conjugate base Our discussion of conjugate acidbase pairs told us that B39 will be the stronger of the two but what is the chemical justification The strength of an acid depends on how easy it forms H ions In this case it happens more readily for HA The strength of a weak base depends on the reverse process that is formation of a bond between the base and hydrogen Since the breakup of HA occurs more readily its formation cannot also be preferred Thus B39 is the stronger conjugate base The derivation of the mathematical relationship between Ka and Kb follows w HAM Hltaq t Aaq Ka HA HAHOH A an H20 HAW OH an Kb 7 Notice that A39 and HA both appear on the right side of one equation and the le side of the other Now when we add the two equations we nd something interesting occurs HAW A39aq H20 F Haq A39aq HAWD OH39aq which simplifies to H20 f Haq OH39aq KW HOH39 So how are the equilibrium constants related H OH KaKb 1m will 1 INCH 1 Kw K Em J K ix J The relationship KaKb KW is necessary to determine the pH of a solution of a weak base formed from the conjugate base of a weak acid EX What is the pH ofa 010M sodium acetate solution Kb for NaOAc is not listed in Table 164 so it must be calculated 14 Kb g 1390X10 56 x103910 Kn 18X10395 HOAcOH x2 Kb56X103910 OAc 010 x Assume X is negligible since 010 gtgt Kb 2 X 010 56 X 103910 X 74 X 10396 OH39 74 X 10396M3 pOH log74 X 10396 513 pH 1400 7 513 987 169 AcidBase Properties of Salt Solutions We have just seen that the salts of weak acids are weak bases e g NaF NaOAc etc As you ve probably guessed the reverse situation can occur as well The conjugate acid of a weak base is weakly acidic The ammonium ion NH4 is such a weak acid An interesting difference between the acidic and basic salts we ve discussed until now is the nature of their acidity and basicity All of the acids you ve seen react by releasing a proton into water In contrast while strong bases directly release hydroxide ions into water all weak bases react with water by abstracting a proton and releasing a proton In general reactions that break water up into Hf and OH39 ions are called hydrolysis reactions A question that follows from this discussion is Is it possible for such a reaction to result in an acidic solution The answer is yes When most metal ions are dissolved in water the solution becomes acidic The attending hydrolysis reaction is shown below Mnf H20 MOHH391 H We ll discuss this reaction in more detail in the last section of this chapter All cations except the alkali metal ions and Ca2 Sr2 and Ba2 yield acidic aqueous solutions In fact these solutions can be fairly acidic eg Al3 solutions can be as acidic as acetic acid solutions Whether a solution is acidic or basic can be determined using the following guidelines Your book presents them slightly differently a Salts made from the reaction of a strong base with a strong acid eg NaOH HCl yield neutral salts Neither ion Na or Cl39 hydrolyses water 17 b Salts made from a strong base and a weak acid e g NaOH HOAc The salt will produce basic solutions because the base cation Na doesn t hydrolyze water while the acid anion OAc39 does 0 V Salts made from a weak base and a strong acid e g NH3 HCl The salt will produce acidic solutions because the base cation NH4 releases H ions while the acid anion OAc39 doesn t hydrolyze water d Salts made from the reaction of a weak base with a weak acid In this case both counterions will hydrolyze If the weak acid is a stronger acid than the weak base is a base Ka gt Kb the solution will be acidic The converse is also true Note for these compounds one can usually expect even fairly concentrated solutions to have pHs not far from 7 1610 AcidBase Behavior and Chemical Structure One final question needs to be asked and answered regarding BronstedLowry acids and bases What factors control acidbase strengths and why do they do so On several occasions we ve touched on this now we tackle it in depth Last semester a large portion of the class was devoted to chemical structure and the electronic nature of atoms and their interaction with neighboring atoms What 3 major factors account for whether a substance is an acid or base and how readily that substance will ionize Consider the generic acid HX where X39 is any common anion The more polar the HX bond the stronger will be the acid Why The products of dissolving an acid in water are the hydrogen ion and X39 A very electronegative X atom or group will draw more electron density towards itself than will a less electronegative atom or group In doing so The H will begin to more closely approximate the nal H product and the X will begin to resemble X39 Since the starting acid 18 more closely resembles the products with more electronegative groups it takes less energy to separate the molecule into ions This results in a stronger acid Yet this can t be the only factor operating because HF is a weak acid while HCl is a strong acid even though uorine is more electronegative than chlorine HF is also much weaker than HBr and HI which are strong acids The reason for this is the HF bond is much stronger than the H Cl HBr or HI bonds The strong bond more than compensates for the bond polarity and holds the molecule largely intact when dissolved in water This also accounts for why water ionizes only very slightly 10397M at room temperature even though the 0H bond is more polar than the HCl bond Thought question If water ionized to a greater extent would it be more acidic Explain Down a group binary acids become stronger because bond strengths weaken more rapidly than bond polarity decreases Along a row bond strengths change slowly so bond polarity is more important Finally the stability of the conjugate base matters The more stable the conjugate base the more likely it is to be formed Of the three factors this is the one that is most difficult for you to predict and we won t discuss it further in this class What would happen if X instead of being very electronegative were very electropositive very small electronegativity Bond polarity would reverse and electron density would be forced onto the hydrogen instead of pulled off it The result is a series of complexes called binary metal hydrides MH These complexes are frequently bases some being quite strong 2 NaHS H20 gt 2 NaOHaq H2 g Oxyacids The other major class of acids are those in which the hydrogen atom is attached to an oxygen that is part of a complex anion e g HON02 HNO3 In these acids the bond strength factor 19 should not play a significant role Why not Because the hydrogen is always attached to an oxygen atom and while small differences in bond strengths might exist large ones should not The result is an acid whose strength depends almost solely on bond polarity Apparently small changes in an acid can change it from weak to strong For example H2803 and HN02 are weak acids while H2804 and HNO3 are strong Two factors can in uence OH bond polarity 1 For acids with the same structure acid strength increases with increasing electronegativity of the anion s central atom For example the HXO4 series HOC102 gt HOBr02 gt H0102 2 For anions with the same central atom increasing the oxidation number of that atom increases acid strength HOClO3 gt HOC102 gt HOClO gt HOCl In both cases acid strength increases because the change results in electron density being pulled away from the oxygen to which the acidic hydrogen is bound What happens when the central atom is very electropositive To answer this question it is useful to imagine an example What might one be While not a molecular compound sodium hydroxide illustrates this idea nicely Instead of pulling electron density towards itself the sodium pushes it onto the oxygen The high electronegativity of oxygen allows it to accept the electron density and ionization occurs between the Na and 0 instead of in the 0H bond Read the section on carboxylic acids p 647 on your own 1611 Lewis Acids and Bases It turns out there are many reactions that are similar to BronstedLowry acidbase reactions but do not involve proton transfer GN Lewis proposed that the common thread between these 20 reactions is that one reactant donates a pair of electrons to the other All BronstedLowry acidbase reactions are also Lewis acidbase reactions Two examples follow H HIF H H HH H H H 1F 11 IF HI H if 1 5 H Ih B E or HI H H F Ili 1l3 H 1E All nitrogenhydrogen bonds in the ammonium ion are identical covalent bonds There is something unusual in the bonding in this ion however Begin by looking at ammonia Here we begin with three identical bonds When ammonia reacts with H the new bond that results forms when both electrons come from a single atom This situation is dilTerent from a normal covalent bond where each atom contributes l electron to the bond In cases where one atom contributes both electrons to the bond the bond is called a coordinate covalent or m bond Sometimes an arrow is drawn to show which atom contributes the pair of electrons see above It is important to remember that these bonds are not necessarily weaker than standard covalent bonds The binding of oxygen to hemoglobin is an example of this type of reaction a lone pair of electrons on 02 binds to an iron ion in the hemoglobin This bond is fairly weak allowing the oxygen bound in the lungs to be released where needed in the body Carbon monoxide poisoning occurs because it binds much more strongly than oxygen Thus relatively small amounts of CO cause trouble because it so efficiently ties up hemoglobin Even exposure to a pure oxygen atmosphere only slowly frees the hemoglobin Hydrolysis of Metal Ions All metal ions are Lewis acids and Chapter 24 will deal with this topic extensively Water is a Lewis base the lone pairs on the oxygen atom 21 The oxygen atom donates a pair of electrons to the metal ion creating a coordinate covalent bond Since the positively charged metal is pulling electrons away from the electronegative oxygen the oxygen pulls strongly on the electrons in the 0H bonds This polarizes them and makes release of hydrogen ions energetically easier see p 16 of these notes Metal ions typically coordinate more than one water molecule in aqueous solution Four to six is a typical number It is rare that more than one water molecule will hydrolyze Thus MH20X gt MH20n10HltX1gt H A1H2063 gt A1H2050H2 H January 10 2005
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