hello and welcome to your second last lesson of equilibrium so this is the time of year that the weather starts to get nice and this is a time of year that normally we'd be using the outdoor classroom behind the cafeteria so I figured I might as well bring ourselves to our own outdoor classroom so hopefully you're able to watch this lesson outside or at least do some of your schoolwork outside or after this go outside and thank you fo chemistry um but today we're gonna be looking at a continue our work with acids and bases so the next two lessons have been split up into two concepts today we're gonna look at the idea of weak acids and bases and then our next lesson we'll look at strong acids of bases um so I want to remind you where we left off last lesson we said that weak acids and bases what makes them weak is the idea that they will not totally dissociate so what that means is they won't totally break into hydrogen ions and hydroxide ions so they're actually going to form on equilibrium so acids and bases that are considered to be weak will form an equilibrium that's what we're gonna study today so we're gonna be looking at this concept of ka to start all that ka is is the equilibrium constant for acids the a standing for acid so I gave you a really great example here of acetic acid that's that white vinegar that you might put on your lunch acetic acid is considered to be a weak acid so we've got our acetic acid here notice that it's written out like one organic compound because it is an organic compound this is the functional group for a carboxylic acid right carbon double-bonded o single bonded o H so when acetic acid is mixed with water it is going to partially dissociate so what that means is it's going to partially break into our hydronium ion that we see here and it's conjugate base that we're gonna see over here notice that the conjugate base looks almost the exact same as acetic acid the only difference is it's missing a hydrogen that hydrogen has been added to that water molecule we do have this equilibrium arrow here though because this isn't gonna happen fully part of the acetic acid is gonna stay intact it just kind of be floating around in water and the other part of acetic acid is gonna break into these hydronium ions and this conjugate base here so the extent of ionization the extent of how much it's going to be broken down is represented by that equilibrium constant that ka so ka is actually called the ionization constant how we write ka is the exact same way we've been writing our other equilibrium constants we take the concentration of the products and we place it over the concentration of the reactants notice here that I actually only have one reactant in the bottom I only have acidic acid in the bottom that's because in this case we're always going to be looking at our weak acid combining with water water isn't a liquid so it wouldn't be included in that equilibrium expression here so why is ka useful it's gonna help us to measure the strength or the weakness of a weak acid I know that sounds really weird saying a strong weak acid but there's there's a range right we have weak acids and we can have some weak acids that are weaker than other weak acids so a stronger weak acid is going to have a larger ka and a weaker big acid is going to have a smaller ka so essentially the smaller the ka the weaker the acid these stronger the ka the stronger the weak acid remember strong acids are not gonna have a KA they're not gonna form an equilibrium so when we say stronger acids we're still referring to those weak acids not those strong acids that we're gonna cover in our next lesson another way to measure how strong something is is this idea of percent ionization so ka is one measure of acid strength the other measure of acid strength is percent ionization and it's again just another way to look at how that ion is broken down so it's gonna be my concentration of hydrogen at equilibrium divided by my initial concentration well let's think about this hydrogen ions are a product and the initial concentration is a reactant so really this is just a way of looking at ka it's almost the exact same as ka but we're multiplying it by a hundred so we're turning it into a percent often times a percent is easier to think about than a really small number like eight point six times ten to the negative six that means nothing to you but 0.1% that's something your brain can process a little bit easier so it's just two different ways of looking at the same information we're gonna do two examples today we're only gonna do one example referring to weak acids so let's go ahead and take a look so we talked about acetic acid being a weak acid another weak acid is this idea of propanoic acid also an organic compound so most carboxylic acids that we studied organic chemistry are considered to be weak acids so we've been told a couple different things we were told this idea of 0.1 molar that's the concentration of propanoic acid so we know the concentration of propanoic acid is 0.1 molar and we know that this solution so remember solution refers to when it's mixed with water right to make a solution we have to mix it with some sort of liquid so the solution has a pH of two point nine six calculate the percent ionization of water so essentially what's happening is we have our propanoic acid our propanoic acid is in solution so it's therefore mixed with water we didn't list water in this formula as a reactant because of the liquid it's not gonna because I'm included in the equilibria expression but it's not wrong to list water so you can absolutely have waters of reactant here it's just not gonna play a part so this propanoic acid is mixed with water and it's breaking down into hydroxide or hydronium ions right that's gonna end up being in all reality h3o plus and this is its conjugate base here again the conjugate base is only different by an H so we've gone ahead with this propanoic acid we've pulled off an H and we've got that conjugate base and that H is are now gonna be disassociated that H is gonna be over here so the idea is is that this is going to form some sort of equilibrium some of my propanoic acid is gonna break apart some of it's gonna stay the same our job is to figure out how much our job is to figure out the percent ionization using this other piece of information using this pH so how we are going to go about and solve this it's gonna be the exact same way we were solving our other equilibrium questions anytime we're dealing with an acid-base equilibrium we're talking about an ice so we are gonna start using an ice table notice that I listed my reactants and products at the top I have one reactant I have two products so I've got three columns I start with a concentration I was given my starting concentration of propanoic acid that was point one molar so I've got my point one molar here and at the start of my reaction the very second that I mixed my propanoic acid with water I have zero conjugate base and zero hydronium ions because they haven't had a chance to break apart so initially I've got point one and I've got zero and I've got zero my change is just gonna be using my coefficients my balanced chemical equation we are gonna talk about the really nice thing with acid-base reactions they are always balanced they are always at a one-to-one here because all we're doing is we're just pulling off a hydrogen ion so just by pulling off this hydrogen ion it has not changed the number of hydrogen atoms it has not changed the number of oxygen atoms it has not changed the number of carbon atoms so I've still got a one here I've got a 1 here and a 1 here that's always gonna be true so we're not dealing with different coefficients this time we're not dealing with exponents this time everything is a 1 so for my c row here it's just gonna be negative 1x why is it negative for propanoic acid because i'm losing concentration of propanoic acid we're gonna see that concentration go down as my conjugate bases forms I'm adding concentration of my conjugate base here and I'm adding concentration of my hydrogen ion my last step is to fill in my econ my equilibrium call in here and all I do is I can I combine my first and my second row so I'm gonna take point 1 minus X so my first one is gonna be point 1 minus X and then this is just gonna be a change of X and this is just gonna be a change of X because 0 plus X is just gonna be X so we filled in an ice table the next thing that you want to do is we need some sort of way to solve for X before we were solving for x using ka in this case we don't have ka but let's to go back to a question we have another piece of information and other piece of information is our pH so in this case we're gonna use pH in order to solve for X so I want to remind you um that pH is related to our concentration of hydrogen ions so we're gonna calculate the concentration of hydrogen or hydronium ions using the pH we're gonna do so using this formula so my concentration of hydronium ions is going to be equal to 10 to the negative pH my pH was two point ninety six so my concentration of hydronium ions is equal to 10 to the negative two point nine six plug that in your calculator and you're gonna get a small number one point one times ten to the negative three moles per liter so that negative three tells me that this is a small number how is this useful so let's think about how knowing my concentration of hydronium or H+ ions is useful well if I think about it um a ch3 my hydronium is formed at equilibrium when propanoic acid is broken down so h3 is formed by releasing H+ ions from propanoic acid so essentially h3o my concentration of hydronium ions is this concentration here in my ice table right if I think about this I'm gonna scroll it here for a second move this over here so you can see that if I think about this at equilibrium my concentration of H+ ions I know what that is right this doesn't happen you don't have a concentration of H+ ions until equilibrium so that one point one times ten to the negative three is equal to my equilibrium concentration well this here before was just X so I now know what X is so I can go ahead and I can fill in X here and I can also go ahead and I can fill in X here so we now know my equilibrium concentration of propanoic acid is point one minus one point one times ten to the negative three and these are just my X values now that I know what X is I can go ahead and I can solve for ka we're still going to be following the same steps that we were before so before I can solve for ka I need to write myself an equilibrium expression so ka is going to be equal my products / my reactants so I have my two products here I have my conjugate base and I have my hydronium ion and I'm dividing that by my concentration of my reactant that was my propanoic acid notice that I don't have any exponents because everything is a 1 so I don't have to worry about squaring values now for my I stay by actually know what these numbers are so I'm gonna go ahead and I'm gonna plug them in both my conjugate base and my hydronium ion are the same they're 1.1 times 10 to the negative 3 so if we're gonna be multiplying those two things together it's the same as squaring it so you can use that squared button on your calculator so make things easier or you can just type in 1.1 times 10 to the negative 3 twice it doesn't matter we're gonna divide that by our concentration of our reactant our reactant was zero point one minus one point one times 10 to the negative 3 that's gonna give me this KA value here so that's how we would figure out ka here since KA is considered to be small right this is a pretty small number we would consider that so phenolic acid is a weak weak base it's got a small K ASA propanoic acid is a weak weak base we're not quite done yet note that the question asked us for percent ionization so if we calculate ka that's great question might ask you for KA we're gonna look at percent ionization luckily for us we've already done the hard work so now all we have to do is we have to take the concentration of ionized propanoic acid that's the same as our concentration of our base and we have to divide it by our initial concentration of propanoic acid and we're gonna multiply that by a hundred so this first value here this concentration of our conjugate base is actually from our ice table right we know what the concentration of this is gonna be I see what you got there we go we know what the equilibrium concentration of this is gonna be it's right down here we calculated it in our ice table we know what the initial concentration of propanoic acid is gonna be it's right here it was given us to us in the question so we're gonna take this number and we're gonna take this number and we're gonna go ahead and we're gonna plug that into our formula right here so we've taken our concentration of our conjugate base from our ice table we're dividing it by our initial concentration of our acid we're multiplying it by a hundred and it's giving us a percent iron eyes ation of 1.1 percent so that's telling us that only 1.1 percent of our acid is breaking apart only 1.1 percent of our acid is ionizing the rest of it is staying intact as a whole acid so that's ka that's that's really it what we're gonna do now we're gonna do one more example and we're gonna look at Kb the good news is that KB is calculated the exact same way in the exact same steps except for one step at the end so I want to remind you that here we work out we were calculating the concentration of hydroxide ions uh sorry hydronium ions h3o plus or H+ in a base we're not gonna be looking for our concentration of H+ we're gonna be looking for our concentration of O H so let's go ahead and look at measuring the strength of weak bases to measure the strength of weak bases we're gonna be using the value K be K standing for equilibrium and be standing for base it's the exact same as ka but for bases so we're gonna call it the base ionization constant same idea as before stronger bases are gonna have a larger KB weaker bases are gonna have a smaller KB so it's a chance for us to measure weak bases and order them in what's stronger and what's weaker let's go ahead and take a look at an example so we're gonna be looking at something that's used to manufacture dyes we're gonna be looking at an elite so we've given you the formula for aniline here notice that you don't see a specific Oh H um it is considered a weak base but it's a weak base that has NH 2 in it often times things that have any 2 in it will also act as a base we were told when aniline is dissolved in water it's gonna become a weak base when the solution containing of 0.05 3 7 moles per liter of aniline is prepared the pH was determined to be eight point six eight what is the KB so we even print pretty much the exact same information as last time we were given our initial concentration of our base we were given our pH and we're being asked for KB so in order to solve this we're gonna set this up the exact same way we're gonna start with an ice table so what I've given you here is the equilibrium reaction of what happens when an allene brokes breaks down and when it's dissolved in water so when we take this and we dissolve it in water we're gonna get an aqueous solution we didn't include water here because it's not going to be included in the equilibrium constant it's not wrong to include water you're just not gonna see any values associated with it we're gonna see here our difference is we're gonna get a conjugate the odd acid this time so what's happening is in this case we're actually going to see I remember a base is something that's gonna accept an H+ so our base here has accepted an h+ by accepting an h+ it's taken that h+ from a water molecule so it's formed o h hydroxide ions in concentration so this is gonna be our reaction it would have been given to you in this case we have our initial concentration of aniline here and we know when this reaction starts we have no conjugate acid and we have no hydroxide ions forms for this c column same idea as our acid reactions everything here is a 1 this is balanced and I don't see any coefficients so for my C I'm gonna be losing reactant and I'm gonna be gaining product for mighty might equilibrium Colin I just want to take my eye and my C and I want to combine it together so I'm gonna have 0.05 3/7 minus X I'm gonna have plus X and I'm gonna have plus X my next step is I need to use pH in order to figure out what my concentration of hydroxide ions are but here's my issue is that I don't have pH is gonna help me to figure out my concentration of h+ I don't have h plus I have a CH so I don't want to look at pH I want to look at Poh right I have an H here so pH not useful Poh is useful so if only there was a way to convert from pH to Poh we no there is it's my magic number 14 so I'm going to calculate the concentration of O H using pH my first step is I have to convert pH to Poh and to do that Poh is equal to 14 minus my pH so I've got my 14 here I've got my pH here Poh is going to be equal to five point three two why am i converting a Poh because God because pH cannot give me my concentration of O H but Poh can the dogs outside come on Minnie Minnie oops sorry oh that's okay uh so I'm not going to use my concentration of Rocio H here to calculate high concentration of O H ions so my concentration of O H ions is going to be equal to ten to my negative Poh we know what my Poh is we just calculated it it's right here so my concentration of O H ions is going to be equal to ten to the negative five point three to my concentration of O H ions is going to be equal to five point seven nine times ten to the negative six so what is this number why is it useful it's my concentration of O H ions but if I look back at my ice table my concentration of O H ions is X so essentially by solving for my concentration of hydroxide o H ions I've just solved for X so I'm gonna take this value here this x value this one here and I'm gonna plug it back into my ice table so I now know what X is here I now know what X is here and I'm able to solve for my equilibrium constant here because this is a number so ah one thing I want to point out here we're gonna save ourselves some time we're gonna save our table sell some time punching stuff into the calculator um what I'm noticing is a very very very very small change I'm taking 0.05 and I'm subtracting 0.0 0.0 0.2 and apply the hundreds rule here you don't have to you can still solve without it but it's a short cut to not have to worry about punching extra numbers into our calculator so to calculate 4 KB what I'm gonna do is I'm gonna set up my KB expression that's my ratio of products to reactants so I put my two products on top and I put my one reactant at the bottom here I know what these numbers are so I'm gonna go ahead and I'm gonna plug them in notice that here I did adopt the hundreds rule so rather than having 0.05 3/7 minus this number here I've just gone ahead and I've gotten rid of that X in the denominator you could go ahead and keep that X there it's just extra stuff to punch into your calculator so I've taken my X values since my conjugate acid and my OS are the same I'm gonna go ahead and take this number and I'm going to square it in my calculator you could punch them in individually and just multiply by the them by each other at the same thing and if I divide by my initial concentration I'm gonna find I get a KB value here I get an extremely small KB value since my KB value is so small we can go ahead and assume that this is a weak weak base so notice that ka and KB are calculated the exact same way the only thing is there's print extra step nobody there's one extra step and the one extra step is just to convert pH to Poh they're just one more step in order to calculate K be compared to ka so you have a couple of homework questions here one is a KA question one is a KP question and hint hint this is a fabulous question if you want to ask about most of the concepts in this unit right so this is a really good question where we're talking about pH we're talking about equilibrium we're setting up equilibrium constants we're using ice tables so this is a really good like long answer question in 10th summative so I have a great day and we'll be back next week and next week we're gonna be looking at just a little bit of redox as you work on your summative he's doing fine