hello everyone mrs hansen once again hey look we're on chapter three already moving through our organic chemistry lessons the focus of this chapter is on acids and bases and a lot of this in the certainly the beginning part maybe a um hopefully a little bit of a review from your gen chem two days talking about acid-base theories we'll begin the journey talking about bronsted-lowry acid bases the ability to transfer a proton from an acid to a base so we'll all about conjugate pairs as we think about proton transfer in the bronsted-lowry conjugate acid-base theory we'll talk about excuse me the flow of electron density using those curved arrows we had used something you know the same thing curved arrow notation for resonance structures we'll now understand how to apply that same designation of an electron rich area moving towards electron deficient area for acid base reactions we'll look at a quantitative perspective of bronsted-lowry's this is a ka chart more specifically the pka chart please notice that there is a link in your e-learning that i provide you a copy of the pka i strongly urge you to print that just so you have it conveniently located through the lessons we're going to refer to that quite often and really the more familiar you become with how it's organized and the information you can glean out of that chart the quicker more proficient uh you become so definitely have a pka chart it's a link in your e-learning it comes straight out of your text which i also believe not only in chapter three but i if i'm not mistaken it might be in the front cover or the back cover of your chemistry book if you have a hard copy that's our first lesson we then pick up kind of in a second session talking about more of a qualitative perspective of acidity we'll look at the position of an equilibrium remember how we talk about equilibrium as a dynamic equilibrium pushing forward in reverse at the same time and the proper choice of reagents to influence the position of equilibrium and then we start looking at leveling effects solvation effects counter ions and lewis acid-base theory which we talked about in the lewis theory as electron transfer so with bronsted-lowry it's all about proton tracking and with lewis acid-base theory we track electrons so let's dive in we're going to begin with a definition just a reminder that a bronsted-lowry an acid donates a proton and remember a proton is simply another name for saying h plus the hydrogen ion is indeed the same thing as a proton an atom of hydrogen right has one proton and one electron it has zero neutrons so that loss of electron to form a positive ion leaves you with simply a proton left so a proton a proton is the same as saying hydrogen ion so if an acid donates a proton where does it go well it goes to the base that accepts the proton so a very classic example of bronze delaware acid-base chemistry gives me hcl hydrochloric acid highlighted in red is the proton hcl and over here it's reacting with water now notice water has a very rich electron region on top of that oxygen the polarity of the bonds are even bringing the electron density towards the oxygen the polarity of this bond is bringing the electron density to the chloride leaving the hydrogen very electron deficient now the bronsted-lowry theory says we're going to have a proton transfer a proton is transferred from an acid to a base so thinking about the mechanism this very rich area on the oxygen water acting as a base in this example will reach out to pluck off remove steel away the proton from the acid hcl and when it does so this bond which is made of two electrons right a shared pair the bond collapses on the chloride and giving it another set of dots so on the product side we now have the chloride ion notice now it has three sets of dot and the formal charge is negative one chlorine lives in group seven but it has eight assigned electrons that's why it's a negative one formal charge and since water accepted a proton notice that the pair of dots that used to be on the oxygen is now a shared pair to that with the hydrogen so here we have one shared pair the oxygen itself remember it lives in group six it has one two three four five assigned electrons and that's why we see the plus one on the hydronium ion that's the acid ion in acid-base chemistry and that plus one charge represents the formal charge on the oxygen so with an acid base proton transfer we require two arrows to represent the reaction mechanism the base reaching out for the proton and the bond collapsing to form the anion which would be the conjugate base of the acid so that's the journey in this first section is just reminding us about bronsted-lowry acid-base pairing the species that forms are known as conjugates so a conjugate acid forms when a base accepts a proton the conjugate base forms when the acid gives up its proton remember a general rule between two pairs acids always have one more hydrogen so conjugates are those two species one from the left side one to the right side that only differ by the absence or presence of a hydrogen ion for example this elk oxide anion oxide here that o alk oxide means it's got a carbon group with an oxide negative ion alk oxide anion reacting with water so now all of a sudden we have this very electron rich oxygen area and based on polarity you know that hydrogens in a water molecule are very electron deficient and so that deficient hydrogen is partially positive it's a great target for this electron rich alkoxide anion so the mechanism in this conjugate pair the alkoxide anion acting as the base reaches out for the partial positive hydrogen that's the proton and it plucks it off of that water molecule the shared pair of electrons in that covalent bond then collapse back onto the oxygen and we form the conjugate acid of the alkoxide ion notice this now is a functional group called an alcohol we have the o h so instead of three sets of dots we have two sets and now a bond leading out to that proton and the hydroxide conjugate base the conjugate base of water is hydroxide so a proton has been transferred in bronstead lowry acid base pairing if we think about just really practicing the definitions we start to see a pattern when we see electron rich areas if we think about ammonia and water well let's really consider that by drawing out the ammonia molecule which has a pair of electrons on top of the nitrogen making this region very electron rich based on the polarity of the bonds you can see that these hydrogens the electron density is being pulled away from the hydrogen and towards the nitrogen making these of course positive regions an electron rich nitrogen and so when we see that it's reacting with water and water of course we know also has its own polarity set up the polarity is being pulled towards the oxygen making this a very rich area on top of the oxygen for electrical charge and the deficient electrical charge there on hydrogens so knowing that opposites attract we see a proton transfer originating from the lone pair of the nitrogen heading out towards the positive region on the water molecule when it does so a proton is transferred so these shared pair in this covalent bond then collapse back onto the oxygen and what have we formed well the conjugate acid of ammonia is the polyatomic ion ammonium nh4 and notice that the positive formal charge on nitrogen needs to be shown there nh4 plus and the conjugate acid that forms from water when it gains its proton would be the hydronium ion whoops i did that the wrong way we took off a proton from water leaving its conjugate base silly silly this would be the o h with three sets of dots and the negative charge to represent the formal charge on hydroxide so ammonia has turned into ammonium and water has turned into the hydroxide ion a proton has been transferred ammonia was a base it accepted a proton forming its conjugate acid water acted as an acid forming its conjugate base hydroxide and any proton transfer reaction mechanism will involve two arrows this will learn is a concerted mechanism which simply means it occurs in one step even though i'm drawing two separate arrows they're happening at the same time with acetic acid and water we also have a proton being transferred acetic acid is a carboxylic acid functional group carboxylic acid and so here i can see a very rich region of electron density on this oxygen and as a matter of fact we can see it being pulled even further up towards that carbonyl oxygen making the hydrogen very positive this is the proton in this carboxylic acid that will be removed during a proton transfer it's going to react with a water molecule but in a different way than the base did of ammonia so if carboxylic acid is acting as the acid we're now looking at the other part of the molecule of water the very rich region of the oxygen the oxygen will reach out like the pair of electro is going to reach out towards the positive proton on a carboxylic acid and it's going to remove it it's going to pluck it off and when it does so the shared pair in this covalent bond collapse back onto the oxygen and what results are conjugate pairs the carboxylic acid when it loses its proton forms a conjugate base that has o negative because now the formal charge there represents a negative one this is the conjugate base of acetic acid this would be the polyatomic ion called acetate and just making sure that's going uh over on the water side because it accepted a proton we now have the conjugate acid of hydronium again oxygen with its positive formal charge makes the hydronium polyatomic ion which is the conjugate acid when water acts as a base so there's our theory kind of just getting idea of where we're heading with our acid-base chemistry so i've been showing these curved arrows as representations of electron movement and we use it to describe the flow of electron density and they're very much the similar you know same arrow describing resonance that we used in last chapter but there's more of a delocalized bond right in other words if i had an aromatic compound for instance i could use these electron arrows to show that each bond could move to a separate place and in a resonance structure we could simply move the delocalized electrons to well i didn't move them to the next position that's not done by spinning but in rather it's just done by moving the carbon double bonds and we said really that this could be represented as an equivalent structure with delocalization by just placing a circle but notice that the arrows are being used to represent electron flow in a resonant structure we're doing the same thing the arrows are representing electron flow we're using them to represent a proton transfer which is actual chemistry right so i'm actually moving electrons in a reaction mechanism they are physically movement of electrons not a resonant structure which is a hybrid but an actual movement from reactants making new products learning to draw these electron arrows is of critical importance you know every time we introduce a reaction mechanism and dive in for understanding we're tracking electron flow so these acid base electron mechanisms occur in a single step even though we have two arrows being represented they are are really thinking about this occurring simultaneously the base attacks the acid using a pair of electrons so this very rich region of electrons on the oxygen reaching out for the proton on our carboxylic acid see that functional group ku c-o-o-h a carboxylic acid this proton is then removed causing the bond to collapse on back to the oxygen the bond turns into a pair of dots putting a negative charge on that oxide the conjugate acid of hydroxide is water and so the base has a conjugate acid the acid has a conjugate base remember the acid cannot lose its proton without the base taking it all acid base reactions occur simultaneously even though there's two arrows they occur at the same time now sometimes you know in a future mechanism you might see an actual multi-step mechanism that involves multiple steps acid-base proton transfer is not multiple steps it's one step with two arrows so here represents an example of a multi-step reaction let's kind of go through let's call this just to keep track of its structure one structure two structure three structure four you might wanna write those in your notes to kind of help track the flow of electrons now here we have a polar bond looking at the very rich region on the oxygen and this is the hydronium ion which means based on the polarity the electron density is being pulled towards the oxygen making this a very deficient region of electrons so the hydrogen is partial positive and the oxygen region is partial negative and we're looking at a transfer of a proton the oxygen reaches out for the proton and when it plucks it off it breaks this bond when the bond breaks the bond collapses and you get a pair of dots on the oxygen which used to be the bond this by mechanism is an acid-base proton transfer so now what do we have well we started with an alcohol functional group see that alcohol and all of a sudden we have a protonated alcohol following the arrows this bond which is a pair of electrons simply breaks and it takes its electrons with it we're saying minus water here the h2o gets up and leaves and i might as well introduce that it's going to be called a leaving group a leaving group leaves that's what its job is so the h2o deep just gets up and leaves by breaking the bomb that is not a proton transfer when a leaving group leaves it's a single step but only involves one arrow but what's left now is a carbocation right here we used to have you know you're looking at three carbons leading to an oh well we still have three carbons but since it's missing a bond you see the positive charge it only has three bonds instead of the four it requires for stability and thus you see the positive formal charge now look the end of this chain there is three carbons there just kind of think about what that structure is representing each one of these terminal n carbons has three hydrogens attached to make sure that each has four bonds the central carbon only has three bonds and thus a formal charge of plus one the hydrogen here which is being highlighted is now going to be plucked off by the water molecule that left a moment ago so minus h2o still out there isn't it the leaving group that just left now acts as the attacking molecule it will be the base in the next step the carbocation acts as an acid the water reaches out and abstracts that hydrogen from the terminal end of the carbon forcing this bond to collapse when we collapse this bond we now have two bonds and we formed an ene the double bond product we've undergone acid base proton transfer and in doing so we produce the alkene as our final product all right so all of that's really seeding the entire journey of organic chemistry being able to track electron flow helps us predict products by really understanding where the electrons are moving and why they're moving there don't worry about correctly used curved arrows to show acid-base reactions proton transfers for now all of the other reaction mechanisms are going to come to us so let's keep practicing i'm going to do some of your homework with you so i pulled these from your skill builders write these out with me in your notes and you'll be that much closer to having them done for for your homework it asked us to draw a mechanism for the following acid-base reaction and then label acid base and conjugate pairs so the first thing i like to do is just find one on the left and one on the right two structures that are different just by the number of hydrogens on that atom you can see water with two h's and hydroxide with only one h so they are the same structure that differ just by the presence or absence of a single hydrogen ion and we know the one with more hydrogen will always act as the acid and therefore its conjugate base must be hydroxide labeling the methoxide anion to methanol which is an alcohol the one with more hydrogen is always the acid making methoxide act as the base now that you have them labeled you realize that the arrow must always originate with the base you point towards the acid from the electron rich area of the base so if the methoxide is the base the electron rich area is going to reach out for the positively charged hydrogen ions and you can see that distribution of charge making hydrogen positive so i see that reaching out for the proton when it plucks it off the two electrons that are in this bond collapse back onto the oxygen there's the second arrow forming hydroxide and the protonated version of methoxide turns it to an alcohol known as methanol we had two arrows in a concerted mechanism means it's really one step even though there's two arrows they are occurring simultaneously we could do another from our homework oops that's just our answer just to double check we've written that correctly here we have phenol which has an aromatic benzene ring attached to the alcohol group oh we have hydroxide we have this oxide anion and then a water molecule so just kind of getting our feet wet of what we're drawing we know that even though they're not there that oxygens all have these dots around them and just based on what we see for formal charge if it's a negative one it has three sets of dots and if the formal charge is none it has two sets of dots so i can kind of fill those in hook together the conjugate pairs those that differ only by the presence or absence of a hydrogen ion the one with more hydrogen is always the acid the one with less hydrogen will be its conjugate base and between hydroxide and water water has more hydrogen so it will be the conjugate acid making the hydroxide act as the base once you have them labeled you know which direction to originate the arrow in which direction to point the arrow originates at the base so think about this polar bond making this hydrogen very positive in terms of the electron distribution the very rich oxide in the in the base here if hydroxide reaches out for this proton and when it does so when it removes the proton it makes this bond collapse back onto the oxygen those two bonds or the two electrons in the bond collapse back to become a set of dots on the oxide anion very simple once you do many repetitions do some more homework with you i love practicing because you're becoming proficient and speedy here we have a set of conjugate pairs i like to hook together the conjugate pairs and label them knowing that this terminal carbon actually has three h's there and now obviously a negative charge meaning that carbon is missing its third or its fourth bond there so um that is only two hydrogens the negative there and then i see nitrogen with a negative one so i know there's a set of dots up there and what's happened the one with one less hydrogen and i can label those the first having three h's at the terminal end of that carbon makes it an acid as compared to only having two hydrogens at the terminal n that must be its conjugate base in these two conjugate pairs the nitrogen in the central atom there is a pair of dots over here it's protonated so the one with more hydrogen always acts as the acid so we've hooked together conjugate pairs remembering that we always originate the arrow in the electron rich area and we go to the electron deficient proton so the first arrow is reaching out for that terminal end hydrogen and when it plucks it off it forces the electron in that in the bond the shared pair collapses back on the carbon and so you don't see it but that's a set of dots making that a negative one formal charge three bonds and a set of dots instead of electron making it a negative one formal charge so we have two arrows and conjugate pairs oh my one more even i'm really doing your homework for you so here i have a carboxylic acid see that cool c o o h and i know there's some dots on here to finish out the valence electrons i know that an acetic acid or or any carboxylic acid will form a conjugate base when it loses a proton these two are conjugates and hydroxide and water are conjugate pairs so here i start the arrow think about polarity this is a very positive proton making this negative area on the oxygen electron rich we reach out for the proton so my first arrow kind of looked a little messy there i'll redraw the first arrow starts from the oxygen dots and reaches out to the proton forcing this bond to collapse back on to the oxygen a concerted mechanism where a proton has been transferred forming conjugate pairs whew those are a lot of good examples i think we've mastered bronsted-lowry acid base and proton transfer let's bring in a little numbering system specifically the pka chart next so the next steps we're going to examine how strong or how weak an acid is now in gen chem 2 you talked about strong versus weak and setting up an equilibrium i represent the term strong as the ability to 100 ionize in an acid reaction or a basic reaction ionize for example hcl is one of the seven strong acids that means when i place it into water it 100 ionizes to form its aqueous ions right and the strength of the acid lies in the amount of ions it can produce so a strong acid creates lots of ions you're favoring the right side of our dissociation equation whereas a weak acid let's just say for example acetic acid here's the ionizing proton because it's weak it only partially ionizes to create its aqueous ions that the equilibrium lies far to the left favoring the molecular form remember acid is measured by how many protons there are in water that's the whole fundamental scale of the ph system strong acids give me lots of protons in water weak acids don't give me as many protons in water so this represents a very weak acid favoring the left side of the arrow as a matter of fact about 99 of the dissociation lies to the left of that equation so comparing the term strong versus weak relates to the degree of ionization how successfully can the acid dissociate to form its aqueous ions so the strength of an acid or a base is also very helpful to predict how reactions will progress and as we get our feet wet in learning how acids compare in terms of proton transfer we're going to do so in two ways quantitatively using numbers a pka chart to be specific and then in the next section we'll talk about it qualitatively just comparing strength uh based on structure is the negative charge being able to be strengthened or stabilized and so we'll again next chapter qualitatively in this section we'll do it quantitatively pka charts out and ready so when we examine ka or pka it's using numerical data to compare how strong or how weak acids are in just general chemistry you know just a an h a to represent any acid and when i place it in water it dissociates it ionizes it breaks apart falls apart dissociates dissociates into aqueous ions of course this would be the conjugate base of h a and this would be the conjugate acid of h a the more ions we have the stronger the acid so that means if i compare how much do we have in terms of molar units of the aqueous ions set over the molar units of the reactants we get an equilibrium expression and i know you did these before in gen chem 2 especially an equilibrium expression you started your journey with equilibrium talking about kc the concentration kp the pressure of gases you then moved into ka for acids kb for bases you probably even spent time doing ksps the solubility solubility product of salts all of those are equilibrium expressions that simply represent the expression of the ratio of products over reactants where those brackets stand for molarity units which are concentration units and it's just an expression that gives us a number and it's the magnitude of the number how large or how small the number is that's of interest to a chemist the unit itself the k constant has no unit it's just a number and the magnitude gives us great information and comparing well let's suppose we have an equilibrium that's really favoring the forward direction making a whole lot of aqueous ions what would that do to the ka constant by the way do you see how keq incorporates the water concentration here and the ka eliminates the water because it's a pure liquid do you remember learning pure liquids and pure solids fall out of ka expressions or kb expressions k expressions in general so because it's a pure liquid it's a liquid it dropped out we get the ka expression if we have a strong acid we favor the forward direction of this equilibrium we have very high concentrations of the ions we have big number over smaller number and so you can see ka is very large bigger than one for strong acids on the opposite thought plane here let's suppose that we had a very weak acid which means that the forward direction hardly takes off and we're favoring the left side of the equilibrium equilibrium is lying to the left and this is what it would look like for a very weak acid weak acids do not dissociate with much success and so we end up with a smaller number of products over a larger number of reactants to give us a fraction which is a decimal point which means it's smaller than one the weaker the acid the smaller the ka the stronger the acid the larger the ka and we can see these ka values having you know a variety of magnitudes of number they can range from being very very small numbers one times ten to the negative fifty that's like 49 zeros then a one right point zero zero zero forty nine of them it's a very small number up to 10 billion 10 times 10 raised to the 10th power so there's such a large magnitude and these are very small very large numbers which are hard to work with so we prefer to use a pka scale instead of just the ka value itself now remember p is just a mathematical term it means negative log if i have a ph for example i take the negative log of the hydrogen ion if i have a poh i take the negative log of the hydroxide ion that's just what p means and so if i have a pka i'm just taking the negative log of the ka value and it comes up with a more manageable number so instead of having numbers 10 to the negative 10 or 10 to the positive 50 those are big and quite small you can just be looking at numbers from negative 10 to 50. you know that's much easier to manage but keep in mind it's a logarithmic scale so even a small movement of number is an incredibly large difference in value this is an example just a little snapshot of your pka chart and again in your text and in your e-learning i provided you one to print and it has a much more extensive list of acids in their conjugate bases but as you look over this table as you read on the left column these are acids and they're doing a nice job even of highlighting the proton for you in red the ionizing proton and then over here once it's lost its proton it forms its conjugate base and so the right column is its conjugate and in the middle you see its published pka value notice that the very strongest acid listed here is sulfuric acid h2so4 has a value of negative nine well that's a very small number indicating that it's a very strong acid because it's a p k a so a strong acid has small pka values and as you progress down here we have a protonated ketam ketone you see this carbonyl carbon c double bond o leading out to an extra proton making the formal charge on oxygen a plus one it has a value of negative 7.3 underneath that you see hydrochloric acid hcl here we have a hydronium ion h3o plus one here's your carboxylic acid see that cool carboxylic acid this is an anhydride this is phenol which means that alcohol is attached to an aromatic ring here's just plain old water here we have a primary alcohol that term primary means that that oh group is attached to a carbon that's simply attached to one other carbon and just giving you idea this would be a tertiary alcohol because the carbon attached to the o h group is attached to three other carbons and it has a difference in the pka value and again your chart provided either from your book from printing and e-learning it's going to be a provided test taking tool page uh get familiar with how to read this chart and do so quickly the worst thing that could happen is you you use your quiz minutes staring at your chart not understanding where to find things now let's emphasize uh for example notice sulfuric acid is negative nine and hcl is negative seven on a number line here's negative seven negative eight negative nine as you're getting smaller and smaller down the the number line sulfuric acid lies at negative nine and hcl adds at negative seven well the difference in a number line jump is only two units but remember this is a logarithmic scale that's the power of 10 so going from 7 negative 7 to negative 8 is 10 going one more is again times ten so from negative seven to negative nine that's two zeros 100 times stronger sulfuric acid is in terms of its acidity as compared to hcl now both are very strong but we can see just the slightest change in number gives us an incredibly different strength of the acid i'm going to be just a little bit sneaky and have about five six of these questions and the redundancy i know will be there but it's on purpose to make you use your pka chart and really get familiar with it so i'm going to ask you to look at two structures use your pka chart and determine which compound is more acidic it's not a tough question when i ask you more acidic i want the smaller answer the smaller the pka value the stronger the acid now that's an easy question to answer but you've got to go to work and find the pka values so first of all find acetic acid this is a carboxylic acid functional group over here we have acetone notice this is a ketone the functional group is called a ketone but we're looking at the carbon next door specifically so the carbon in the second position from the carbonyl carbon can you find the pka values of those two protons well here i have reported let's see if you found the same thing this carboxylic acid i found the value on my pka chart to be 4.75 notice this is the carboxylic acid this is its conjugate base then i went to work and found this structure matching the acetone and found its value is 19.2 now that i have the answers in terms of pka values it's quite simple the smaller the number the stronger the acid the only trip up is if you find the wrong structure or can't seem to find the you know what we're looking for in the chart and just remind yourself that that 19.2 minus 4.75 gives me roughly 14 and a half right and so what that means is that carboxylic acid is 10 raised to the 14th power times more acidic than acetone do you see that's a really large number it's one with 14 zeros after it two three four five six seven eight nine 10 11 12 13 14. isn't that an incredibly large number and so when we say it's more acidic it's incredibly more acidic based on the numbering system of a logarithmic scale the pka chart represent zeros just like the ph scale does try another draw these structures take a moment look up their pkas and then identify the more acidic compound i'll wait a minute you do it pause the video okay so that means you're back and you've got an answer let's see if it matches mine using my pka chart i found the alcohol group that's a direct attachment to the benzene ring called phenol has a value of a pka 9.9 i then went to the pka chart and found that for water and it's 15.7 the stronger acid has the lower number i'll do another one with you from your homework hydronium hcl pause the video come up with an answer you can verify come back to me when ready okay welcome back i'm just gonna say yay i'm celebrating with colors all right so here's what we found h3o plus has a pka of negative 1.74 hcl has a value of negative seven the more negative uh the more small the number is negative seven makes it a stronger acid easy peasy oh my goodness another one you don't have no homework left i can see this is sulfuric acid this is the protonated ketone find the kas come back when ready pause the video and come back are you ready i found negative 7.3 and i found negative 9 on the pka chart respectively the stronger the acid the smaller the number remember negative numbers going on the number line they get smaller as you proceed down right so negative 9 is a smaller number than negative 7.3 cool we're doing it super proud of you you know what we can also use the pka chart for is quantifying basicity how basic a species is in comparison to another so notice what we have here i said we have conjugate acids written in the first column and their conjugate base written in the far column notice the pattern says the stronger the the acid the smaller the number but strong acids have very weak conjugate base acid strength increases as you read up the left margin the base strength increases as you go down the column the strongest acid has the weakest conjugate base so really what we're asking ourselves when we say who's the weakest base you just flip that and say who is the strongest conjugate acid and use the pka values from there so for example here i have a functional group known as an anhydride a and anhydride notice here that there are indeed hydrogens and therefore we have a set of electrons as well so here's one two a set of dots and one hydrogen and that's giving that carbon a formal charge of minus one remember this is a base so you're looking at the left side of the pka chart to find it find the structure that matches that picture but it's gonna did i say left oh linda i want the right side without looking at one i know i messed up the left side gives you acid the right column gives you its conjugate base and here of course we see another lone pair of electrons on this carbon giving it a minus charge this is the carbon that's directly next door to a ketone go to work and find which one is the stronger base using your pka when i ask use the stronger base i'm really asking you which acid is the weaker acid which has a conjugate acid that is weaker than the other one so you're using your chart don't cheat yourself find those values and then we can compare when you found the base on the right side the conjugate acid on the left side you saw a value of 9.0 for the conjugate acid and on the other side you found the conjugate base and acid given a 19.2 all righty so those are the structures you found using your pka chart and what do they mean so this tells me the stronger acid has the lowest number the strongest pka value right is the lowest number strong acid weak base the bigger pka value meant that was a weaker acid therefore this was the stronger base the bigger pka tells me that that conjugate base is stronger than the conjugate base on the other side that's our answer here's some more homework practice identify the stronger base in each of the following cases use the right side of the pkh art those are the conjugate pairs the stronger base will be below right because we're going this way in terms of stronger bases a stronger base and up here is a weaker base so you just really want to see where they lie on the pka chart so which one is below the other on the pka chart and that's your answer the stronger the base will be further down the column on the right side of your pka here we're looking at the terminal end of an alkyne the triple bond is the yne functional group the terminal hydrogen is an ionizable proton it can act as an acid it has a value of 25 for its pka and i say that the pka is going to the conjugate acid of that base the amide ion the amide has a pka published as 38. the conjugate acid is 38. remember this goes to the acid not the base the weaker acid has a bigger number therefore it has the stronger base so the amide is the stronger base because it has a weaker acid you ready for another these are really just sneaky ways of making you practice using your pka chart again this is the uh i n e and this is an a n e the alkane and alkyne terminal hydrogens and when you look on your pka chart the very bottom of the bottom there is the alkane the conjugate acid here the alkane has a pka of 50 and the terminal and hydrogen of the ion has a pka of 25. so clearly you see that this is the weaker acid and therefore its conjugate must be the stronger base we getting it oh yeah oh another one now look at these are bases so don't fall for it look on the left side the bases are on the right side of your ka chart the right side on the pka chart go to work and find their conjugate acids pka and we'll compare this protonated ketone has a pka of negative 7.3 that's the conjugate acid of this base this is a water molecule its conjugate acid is hydronium the hydronium has a value of negative 1.74 so this is the weaker acid therefore this is the stronger base i'm going to emphasize the skill again if you're reading down this is the conjugate base side you know that the conjugate acids increase as you go up the column the conjugate base strength increases as you go down so if you're asked for the strongest base compare their locations could be that easy you want the one at the lower end so when we said water was a stronger base that means it's lying below the ketone on that conjugate base side so all i have to do is find the one that's more toward the bottom to find the stronger base to find the stronger acid find the one highest up on the left side in comparison we can also use a ka value and this is kind of the last skill for our lesson today using a ka value to predict the equilibrium which side of the equilibrium is going to be favored so ka values that are used to predict if the left side of the equilibrium is being dominant or is it the right side that's being dominant and it really is as easy as identifying the strong acid compared to the weak acid in conjugate pairs so for exist example we have a bronsted-lowry acid-base proton transfer let's kind of just bring in the chemistry here what's happening we have a base that's very electron rich on that oxygen the water molecule we know based on polarity is electron deficient so it's positive on this hydrogen we see that the hydrogen is being abstracted plucked off removed transferred from the acid to the base forcing this bond to collapse and we form conjugate pairs the arrow originated at the base and it formed its conjugate acid water acted as the acid and formed its conjugate base once you label the conjugate pairs i want you to identify the two acids one will be on the left the other will be on the right identify the two acids and when you do that i want you to then find their pka values on the pka chart the published value of water is 15.7 on the pka chart the published value of pka for a tertiary alcohol tertiary means that carbon attached to the alcohol functional group is attached to three other carbons and it has a published pka value of 18. which one is the stronger acid well the stronger acid has the smaller pka it's up higher on the chart and always always always the equilibrium will favor the weaker side the larger pka indicates that this is the weaker acid compared to the one on the left so you can see the equilibrium pushes or favors the right side of the dynamic equilibrium the equilibrium will always favor the weaker acid and the weaker base identify the two acids and compare their ka values so for example another example we have a water molecule acting as an acid the conjugate acid that forms from the base is a straight chain alkane the base accepted a proton to form its conjugate acid the acid donated a proton to form its conjugate base when you identify the two conjugate pairs those that only differ by the presence or absence of a hydrogen identify the two acids and look up their kas the larger the ka the weaker the acid and you can clearly see this reaction would be pushing in the forward direction so much so i mean the magnitude is so different a 15.7 compared to a 50 and if you subtract that you get about 34. well that's one with 34 zeros after it that's a really big number so that really means that there's not much success at all in heading back to the left side of the equilibrium not much of an equilibrium it's more like an irreversible reaction because the magnitude of the different pka values is so significantly large here's some homework problems for you and i really should have started with having them covered and it's going to ask you to determine the position of the equilibrium now the secret lies in finding the kas so i said step one is uh locate the acids right find the conjugate pairs and label who's acting as an acid find the acids one will be on the left one will be on the right between this ketone and this protonated ketone those are called conjugate pairs the one with more hydrogen will act as the acid between hydronium and water hydronium has more protons and therefore it acts as the acid we want to just locate one on each side who's who are the acids in this dynamic equilibrium and once you locate them tell me their pka values what are they and that's where i need the chart and so i went to the pka chart and i found hydronium and it has a published pka value of negative 1.74 the protonated ketone has a published pka of negative 7.3 determine the position of the equilibrium says we equilibrium favors the weaker side the weaker acid has the larger number negative seven one uh seven i'll say that again negative one point seven four is a bigger number so you can see the dynamic equilibrium is favoring the left side of the equation this is the weaker side this must be the stronger acid side we favor the weaker side in the position of the equilibrium here's your other homework question in this section here we have a anhydride as a functional group two carbonyl carbons separated by a carbon there then all of a sudden remember this would have two hydrogens a proton has been abstracted so you can see the mechanism says these electrons rich on the hydroxide are reaching out to abstract that proton when it does so that bond collapses and you see the negative charge on that carbon we have a carb and ion form because it has five assigned electrons we see a formal charge of negative one remember the hydrogen's still there one of the two hydrogens and the other one went over here we had a proton transfer so here these two are conjugate pairs this is the more protonated it has more hydrogen on it compared to hydroxide so it's acting as the acid here are two conjugate pairs the one with more hydrogens notice this has two hydrogens whereas this only has the one this must be acting as the acid find the pka values for the acids we just highlighted going to your chart we see water has a published pka of 15.7 and the anhydride has a published ka of 9.0 the larger number is the weaker acid meaning the equilibrium is favoring the formation of the weaker acid so this equilibrium is lying on the right side favoring the forward direction in the dynamic process favors the weaker side and we're going to stop our video here there's a lot of work we've done good work team we've got uh the first three of our nine sections of chapter three the next lesson i'd like to group four five six and video three will group seven eight nine bring closure so that's my goal anyway good work today take a process break i'm going to need one too you