good day good day everyone uh so we meet again right so today i want us to to cover assets and bases um which is quite an important chapter um maybe let me make a disclaimer to this uh assets and basis is quite a voluminous chapter it has quite a lot of detail in it so i will not in my attempt try to be as exhaustive in my approach to it um there's information or there's uh things that we must know from grade 11. i'll try to just go back a little bit to it um but what i want us to do is just to simply move on uh from uh towards the grade 12 stuff very swiftly and what i'm i'm going to do in my attempt is that i'm going to make sure that i give you everything that you need uh you know just to master the grade 12 syllabus however so what i will do is um i'll break the lessons on assets and basis into different parts so i will not cover everything during this lesson but i will also do my best to make sure that i provide another lesson on assets and basis all right so if you are ready uh i'm also ready i'm ready to go and by the way for those of you who have not subscribed where have you been we've been learning on this channel and um there's some more lessons you can benefit from okay just make sure to hit that subscribe button and if you need to get in touch with us my email address is available on my bio and you can get in touch with me otherwise just throw a comment let me know how this lesson goes okay right shall we go okay so first of all uh when we start with acids and bases um we're going to limit our scope to two theories that we're going to do okay the first one is um referred to as the arrhenius theory uh okay there i'm writing the name wrong there okay i will not attempt to give you the historic background you can read that for yourself okay so that's the arena's theory and there's the laurie bronstead theory of assets and bases okay so it's named after two people all right so um in this case so wanna know first of all right so when we talk about an acid according to the arrhenius theory okay which is where we're going to spend most of our time today right you want to know um what would be the definition of an acid according to arrhenius okay so arenas simply just said an acid is a substance that ionizes in water to give us hydronium ions okay so h plus or we can use h3o plus just please note uh for now and in future that h plus and h3o plus are exactly the same thing okay only difference is that here we've shown that the hydrogen ion okay the hydronium ion is actually embedded onto the h2o molecule okay so it doesn't just float around on its own uh it's actually on h2o so that's why we have that h3o plus okay right so an acid is a substance okay that dissolves in water and therefore um gives us this hydronium ion this h plus ion okay so that's according to the arrhenius theory so when we look at the base however what arena said is that a base is a substance that dissociates okay so this one ionizes this one dissociates uh if you ask me you know it's more or less the same thing but it's a substance that dissociates um in watering or in solution okay to give us hydroxide ions okay so um you can simply look at a an acid as a substance that kind of donates okay hydronium ions and a base as a substance that kind of donates okay uh hydroxide note the difference in the terminology hydronium okay that's h3o plus and hydroxide okay that's o h minus okay right so those are the substances that uh we are referring to when we are talking about acids and bases yeah let me try and focus my camera a little bit okay so essentially when we look at the arrhenius theory every time that we take an acid plus a base and i want you to please note this because we're going to talk about titration in just a little bit right anytime we take an acid plus a base okay what it will give us is that it gives us a salt okay now when we say salt we're not we don't mean table salt but it is part of what you also get by the way we get a salt okay and we also get h2o water okay right i'm going to show you just a couple of examples just now and by the way we're going to use this um quite a bit when we do titration okay which i'm going to show you in a in a few okay just remember that every time that we look at an acid it will donate the h plus okay a base will donate the o h minus and note this the h and the o h that's what gives us the h2o okay right just to talk a little bit of course we're going to uh we we said we're not going to really spend time a lot on the lowry bronsted theory but i would still want you to know um what it entails okay so larry bronsted theory uh an acid according to this theory and we're going to look at it in the um in in the videos to come right so he said okay oh rather they said okay right an acid is simply a substance now please i want you to note so according to the larger bronsted theory an acid note it's nothing different okay it's a substance that donates hydronium ions when in solution okay so an acid is an h plus donor hydronium iron donor when in solution okay or it's a substance that ionizes in solution to give us a high concentration or a concentration of hydronium ions right but now the difference if you note so between this and that there isn't really much of a difference okay but the difference really comes in when we now have to look at the base and what is the difference because here a base is simply a hydronium ion acceptor so the difference is the reaction here so one donates an h plus and another one accepts so this process we call we call by the way protolysis hydronium ion can also be used interchangeably with the proton okay the reason for that is because the hydrogen atom has got one electron and one proton okay in its nucleus and once you take away the electron it becomes h plus and the only thing that remains in the nucleus is um uh the the the proton so in this case we say well this guy is simply a proton donor okay so it's a proton donor that's an acid okay and whereas this uh a base is a proton acceptor okay so that's the fundamental difference between the two okay so according to the uh um the louder bronzer theory one donates and another one accepts um just to show you where this may be applicable i'll come back to this in just a little while so this is applicable by the way mostly when they give you say for argument's sake you've got hcl okay let's take that's an acid and i reacted with h2o okay so usually that's a reversible reaction okay so what i will get is h3o plus okay plus cl minus now please i want you to stay with me uh just a little bit okay so i want you to see where the importance what the importance of this is so if you look at this reaction you've got hydrochloric acid starting on one side as hydrochloric acid okay okay so it starts out as hcl and what does it become when it goes on the other side it becomes cl minus can you see that right so what happened to hcl to end up as cl minus so in this case what did it do it must have donated it must have given away its h plus to become cl minus so according to this it donated an h plus so therefore it becomes an acid okay right so if i look at h2o it started out as h2o what did it become when it it gets to the other side it becomes h3o plus so what happened to h2o to become h3o plus it became or it must have accepted an h plus right and therefore had now uh three hydrogen so in this case it's accepting an h plus so it is a base okay so um if you take the reverse reaction okay so now we start with h3o plus all right it started as h3o plus and what does it become when it gets to the other side it becomes h2o so it means what happened to h3o plus to get to the other side right uh it must have donated an h plus so in this case it is uh sorry um it is an acid okay so this guy is an acid it's an h plus donor cl minus starts out as cl minus again right and what does it become when it gets to the other side of course it becomes hcl right okay so what happened to cl minus to become hcl it must have accepted so therefore it becomes a base okay according to that definition it is a base now please i want you to note something quite important here normally under this theory we will then ask you uh the following okay identify the larry bronstead acid-base conjugate pair so that's where they normally ask you about conjugate pairs right i want you to note so the conjugate pair will always be between the acid on the one side of the equation of the reaction right and the you always pair it with the base on the other side i'm going to make it quite easy for you in just a few so hcl is an acid so i have to pair it with the base on the other side what's the base on the other side it's cl minus right so in this case it means i'm going to pair hcl with cl minus so that's the first conjugate pair right so that's hcl and cl minus okay all right and i want you to note here again acid on the one side i pair it with the base on the other side okay so in this case there's my acid base conjugate pair right so in this case i know another conjugate pair will be h3o plus and h2o so that's another conjugate pair by the way they like these questions uh particularly for the multiple choice section okay so uh just know how to do that right but uh let me just show you in a very simplified way uh how you can identify acid-base conjugate pairs so the acid-base conjugate pair will always be the substances that kind of look similar in the reaction and the only difference between them is just an h plus okay or at just an h look at this can you see that these look similar right what's the difference between them just an h can you see right um h3o and h2o these look similar right and what's the difference between them just another age can you see so in this case uh you can actually just identify them easily that way you know what i want to do perhaps let me just give you another one that would um kind of uh solidify you know your understanding of this principle okay let's take another uh substance okay and we're going to come back to this arena's theory uh in just a few so let's take for instance let's take uh nh3 alright and i reacted with h2o again right and what i get there is nh4 plus okay plus o h minus okay i hope i can fit that into the shot okay so nh3 plus h2o nh4 plus and oh minus right so i want to identify my acid-base conjugate pairs right so um if you follow the second method that i showed you the easy one okay well it should be simple it you all you'll just simply say is that okay substances that looks similar all right in this reaction it's nh3 and nh4 plus and that you you would be correct in saying that okay so nh3 and nh4 plus sorry about that so that's nh4 plus so that's my acid-base conjugate pair right okay and then which one is another one um so and um which one is a an acid-base conjugate pair you're right it's h2o and o h minus okay so you'll notice that because i've got those two they're exactly the same h and o h but the difference is that this one has got two hydrogens whereas this one has got only one again that looks similar the difference is that the one has got um three hydrogens and the one has got four so the difference between them is just an h plus okay so those are acid-based conjugate pests but let's see let's go back to this method okay so so that you can identify which one is the acid which one is the base and the other way around right now you always start so this is nh3 what does it become when it goes to the other side it becomes nh4 plus what happened to nh3 to end up as nh4 plus and you you'd imagine it must have gained it must have accepted okay a proton h plus in order to become nh4 so in this case what does it become it is a base so nh3 is a base why because it accepted an h plus and became n um and became nh4 plus all right and then um h2o it started as h2o and what does it become when it gets to the other side it becomes o h minus can you see so in this case i start with h2o and it becomes oh minus what happened to h2o to become h2oh minus it must have donated an h plus k it had two now it has one so in this case uh if it's a proton donor therefore it must be an acid okay so this guy is an acid all right and then um going back again we start with nh4 plus okay what does it become when it gets to the other side it becomes nh3 right so in this case nh4 started as uh nh4 and it becomes nh3 so what did this guy do he must have donated the h plus to become nh3 so in this case we know that guy is the acid okay and um oh minus starts as o h minus but what does it become when it gets to the other side it becomes h2o so what happened to this guy to become h2o he must have accepted so as a result it is a base just to confirm what we said there we said which ones are our acid-base conjugate pairs you always take the acid on the one side based on the other side acid on the one side based on the other side right so in this case what do we have we've got a base here we must pair it with that acid there on the other side see so we've got an acid here we must pair it with the base on the other side and that's how we end up with nh3 as a base okay uh nh4 plus is our acid so that's our pair okay and then we have h2o and oh minus as another pair all right so i hope that kind of makes sense to you right so as i said i'm not going to stay too much on this side but i at least appreciate the fact that um you now understand how to do those acid-base conjugate pairs right so what i want us to do is i want us to stay a little bit on this side and we're going to talk about titration in a few right and uh talking about titration remember um that is where acids and bases actually that's the reason why we have acid and base reactions because um the only solution to acids which if you remember in grade 11 we know that acids are quite corrosive right and because they are corrosive nothing else can neutralize an acid except a base and that's why we are studying this section okay so that's why we want to know how much acid we need in order to neutralize a base now let's take a few examples here so say for argument's sake i take hydrochloric acid okay all right um and i react it with a base sodium hydroxide so remember in this case these guys give us uh a salt so an acid plus a base gives me a salt so i know i'm going to have h2o there right so it's going to give me a salt it's h2o um and whatever is left i need to knit together so my salt there would be nacl okay and then plus h2o right now ladies and gents the most important thing when it comes to these reactions is always to make sure that it is balanced now at this point in grade 12 you probably will be given the reactions okay you probably will be given a balanced reaction why because what we expect of you is just simply to be able to use them okay so we expect you to be able to use it in a sense that you'll see in just a few when we do titration so look at this we've got hcl and we've got sodium hydroxide it gives us a salt nacl and it gives us um water right now if you look at this we want to check is it balanced so let's start so hydrogen and another hydrogen there we've got two on the left we've got two on the right hand side okay and then note we've got chlorine one chlorine so one chlorine there one sodium one sodium one oxygen and one oxygen so um it seems like we are fine on this one now please i want you to note once you know that it is balanced okay right so what does that mean it means for every one hydrochloric acid that's a coefficient there there's a one there there's one sodium hydroxide that we need and as a result it just simply means that for everyone there's one that's the ratios that you need so you can just forget about the right hand side all we need is this you'll see how we apply that in just a few okay right let's take another reaction um by the way we don't expect you to memorize or remember these reactions uh per se you know um i often get questions from learners yeah but how do i know what the product is please for now please don't focus on that for the exam we probably will give you uh if we need you to apply that we probably will give you uh these reactions and just expect you to know how to apply um the process or the calculations on titration so uh take another example let's take another acid so h2so4 okay that's sulfuric acid okay sorry to just jump in there and just give you the assets um um probably by now uh from grade 11 you would have known which ones are the acids and which ones are the at other bases right so we've got h2so4 and um which is an acid sulfuric acid and let's react it say for argument's sake with potassium hydroxide okay which is our base in this case so we know an acid plus a base will give us a salt so in this case our salt we know we'll have h2o right oh and an h from there so what we get on the other side is potassium sulfate so k2so4 plus h2o now i want you to please note in this case um this is not balanced okay i'll just make an example for you there okay so uh you'll see you've got two uh hydrogens plus one there which is three you only have two on the other side right um i'm not going to waste a lot of time on the balancing of this reaction okay right so i'm just going to show you quickly uh what we're going to do so what i normally do i just check the salt here so i've got two potassium there so let me just to put a two there right and what it's done now my potassiums are enough okay and the only thing now that is still missing are my hydrogens can you see i've got uh two plus another two now on the left-hand side and i've got only two there so i've got four on the left and i've got only two on the right so what i'll do to balance this is just put a two there and so that makes this uh quite balanced okay so now i know that a reaction between uh sulfuric acid and potassium hydroxide all right for every one of this i get two of those and you can forget about the right hand side well as soon as you're balanced okay the one thing that we need okay uh will be this here these coefficients here that have actually written on the left-hand side okay right now um so what i want to quickly do is skip over to talking about titration look you can look at as many examples as you want um in terms of these kind of reactions but what i want us to do is just quickly jump into the application how are we going to use this uh in theory okay right so let's jump into just one question that i want to do with you and then i think um we would have achieved what you want to uh for today so let's take this uh simple uh um question so what we want to do is to talk a little bit about uh titration so first of all um i won't really be exhaustive in the procedure around titration but uh just to talk about what exactly are we trying to achieve when we are titrating or when we are yeah when we are doing the experiment on titration what we're trying to do is to determine the concentration of an unknown substance acid or base right so say for argument's sake you pull out a bottle of vinegar you know from the cupboard and you don't know you know that vinegar contains acetic acid or what you call ethanoic acid right but you don't know what the amount or what the concentration of acetic acid is in that bottle of vinegar so what you do is you're going to prepare a standard solution what is a standard solution it's a solution with the known concentration so you'll take something else that you know perhaps perhaps something like you know bicarbonate of soda that's a known base right and then you prepare a you know just a solution of it you would know how much amount you put of the um bicarbonate of soda and as a result now what you can do is try to get the two to react with one another until you get to what we call the uh the end point okay this is when the acid and the base would completely have neutralized each other and how would you know that point um you know you put an indicator and when the indicator changes color then you know that you've reached that uh end point right this is when the um you know the the concentration or rather we say that the number of moles of the acid okay and the base right would have completely neutralized each other right so i want us to look at a typical example of such a reaction right so we are titrating two things this time we are titrating a solution of sodium hydroxide okay we know the volume of it which is 25 cubic centimeters we know the concentration of it okay which is 0.5 moles per cubic decimeters right and we're trying to neutralize hydrochloric acid okay of a volume of 50 cubic centimeters right and um we want to find out if 25 cubic centimeters of sodium hydroxide will neutralize all of this 50 cubic centimeters then how much acid what is the concentration of our asset so this is what i want us to do so when we get these type of questions okay this would be a typical grade 11 question okay and we can find it in grade 12 obviously it will have other things included there right so then what we do is that they gave us 25 cubic centimeters of 0.5 solution of sodium hydroxide is used to neutralize 50 cubic centimeters of hydrochloric acid solution with an unknown concentration right so we know the two things that are involved in the titration we know that we are reacting hydrochloric acid okay it is reacting with sodium hydroxide so the first thing that you're going to have is i want to know how do these react with one another as i said in grade 12 we'll expect um by then i mean we would give you these reactions okay so in this case what did we have we said we get sodium chloride plus h2o isn't it so you wouldn't have to figure this out on your own uh by the way even in grade 11 we would have given you the reaction all right now the thing that we need to always do is check the reaction is it balanced so let's check for every one so that's one hydrogen one hydrogen that's two on the right hand side that's two on the left one chlorine there's one chlorine there okay one sodium one sodium one oxygen and one oxygen so those are actually um quite balanced so what do we need we only need to now find out uh in this case so we know for every one there's one isn't it okay so now we know how they react with one another we know they are stoichiometric ratios right so one is two one it is completely balanced so we know what actually is going on now ladies and gents all that we simply do in titration when we know that the co the solutions involved completely neutralize each other okay we have to have that as a condition when we know that they completely neutralize each other right so what we are simply going to do is now use our formula and the formula is as simple as follows it says ca va over cb vb is equals to na over nb okay and this by the way comes from um you know the the the solution or rather the the formula for concentration concentration is number uh number of moles divided by volume so when you rearrange that number of moles would be concentration times volume right and by the way not that it matters which one is the a which one is the b okay some use this to to designate that a is the acid and b is the base it really doesn't matter right if i decide to call this a and i decide to call this b right let's see what information do we have about a what do we know about hydrochloric acid remember we don't know what the concentration is so it means that ca is actually unknown can you see that all right and what about the volume of it did they give me the volume of the acid yes they said it's 50 cubic centimeters so it means i know what va is the volume of the acid is 50 cubic centimeters okay right now let's talk about the b okay uh sodium hydroxide what did they actually provide me with right they told me the concentration of the base okay the concentration of my base there okay 0.5 moles per cubic decimeters okay moles per cubic decimeters all right sorry i should put a dot there minus three okay and the volume of the base okay my b they said they used 25 cubic centimeters okay so as a result what do we have in this case all that we're simply going to do is say right let's put that in that equation there now please i want you to note ladies and gents you are under no obligation when you use this formula you don't need to convert we know that volume we're supposed to use cubic decimeters isn't it right but you can actually forgo that when you use that uh this that formula there why because both of them are in the same unit for as long as they're in the same unit you don't need to actually convert them okay right but if one of them of course is in cubic centimeters and the other cubic decimeters you might need to convert the other okay so now we're going to say all right what's my ca the concentration of the acid is unknown so i'm going to leave that as ca multiplied by the volume of the acid that's 50 we said we didn't need to convert that right and then concentration of the base that's 0.5 multiplied by the volume of the base okay i'm given that as 25 that's cubic centimeters as i said to you because both of them are in cubic centimeters i didn't need to convert now where do i get my n a and n p ratio remember we said these ones are the a this one is the a this one is the b so it means that what is my ratio for n a it's this ratio over there so my ratio is 1. you see why it was important for me to get the balance formula right and what about my nb right what's my ratio there it's also one okay so as a result n a over b and b it's 1 over 1 right so all that's going to happen now is that i'm just simply going to cross multiply i'm going to say c a times 50 times 1 so i'm cross multiplying okay ca times 50 times 1 that's going to be 50 ca all right that's the concentration of my acid so 0.5 multiplied by 25 that should give us 12.5 or times 1 and that gives us 12.5 and all that we simply do you're going to divide both sides by 50 there okay that cancels that and we say 12.5 divided by 50 and what do we get we get a concentration of 0.25 moles per cubic decimeters okay right so we get a concentration of 0.25 moles per cubic decimeters right i hope that makes sense right so that's how we're going to um tackle the titration question okay right um i think i want us to take another one the last one and i think we're going to end it there and then we're going to pick up the lesson obviously talking about the larry bronsted theory next time let's take another question right let's take this question um says in a titration reaction a solution with the concentration of 0.028 moles per cubic decimeters of potassium hydroxide that's koh is used to neutralize 30 cubic centimeters of 0.06 moles per cubic centimeters solution of sulfuric acid right so now we need to know how much of the potassium hydroxide solution and cubic decimeters would be required to neutralize the acid okay so note by the way we need to note that we are talking about a titration here so we are talking about a situation where we are neutralizing the acid and the base right so we said the first thing that you're going to do note what are the substances involved in this reaction okay so we note okay it's going to be potassium hydroxide it's reacting with sulfuric acid right so let's write down the reaction so we know it's potassium hydroxide so that's koh plus h2 so4 right okay and we know what we're going to get on the other side it's a potassium sulfate so that's k2 so4 right and i know i often get this question and people say well how do you know that it's going to be k2 here right please just remember that potassium is in group number one so usually it only has one valent electron right whereas the sulfate ion okay this guy here has got a uh um a two minus right uh so it's it's got um an oxidation number rather of two minus so in this case um obviously you need two of these uh to react with one of that but that's another study for another day right so in this case it's ktso4 plus h2o remember we said an acid plus a base will always give us a salt and water right so that's our salt that's our water by the way i want you to please note we said we are probably going to give you these uh for grade 12 we don't require you to remember it we're going to give that to you and what you just need to make sure of is that it's balanced and by the way most of the time it's given to you in in balanced form right so remember that that's two potassium that's only one there so i'm going to put a two there and that balances out my potassiums sulfate should be okay there and my hydrogens now i've got two plus another two which is four so i'm going to put that two there now ladies and gents i want you to note once you've done this you make sure that it's a balanced reaction or by the way it will be given to you in balanced form all that you are concerned with are those stoichiometric coefficients there's a two there okay obviously here there should be a one right so now let's look at it once again okay if you want to maintain which one is the acid which one is the base that's really completely up to you and otherwise it really doesn't make any difference which one you call a which one you call b right if i decide this is going to be my a this is going to be my b there's absolutely nothing wrong with that okay right now i want you to note what do we know about potassium hydroxide okay do we know the concentration of it yes they gave us the concentration okay we know that's 0.028 okay right i'll just avoid writing the si units there if you don't mind so that i don't run out of space okay and then what else do i know about potassium hydroxide well they said um right so so they didn't tell me the volume of it in fact that's exactly what we're looking for so we don't know what volume of potassium hydroxide would be required to neutralize the amount of acid that we are given so in this case we've got sulfuric acid so we said okay the concentration okay that's our b now this is now going to be 0.06 okay and then uh what about the volume of the base okay so we know the volume thereof is going to be 30 cubic centimeters okay right and then all that we're simply going to do ladies and gents is use a titration reaction okay formula rather so we're going to simply say okay we know c a v a over c b v b okay and this is going to be n a over n b right now notice okay we said our c a value that's 0.028 okay that's our concentration multiplied by i'm going to leave these in cubic centimeters okay so the volume of my acid is unknown okay uh sorry the volume of my base rather is unknown okay so concentration of the um acid that's cp that's 0.06 multiplied by the volume of it that's 30. so it means that when i provide you with the answer that answer will actually that volume that is there should be in cubic decimeters in cubic centimeters rather because the other one was also in cubic centimeters now right we said what is the most important thing for my a what is the coefficient there my coefficient is two so there it is there okay my nb right what's my coefficient there that's going to be one okay so that is pretty much it okay so all that we're going to do now is just the simple mathematical gymnastics right so what we're going to uh let's just calculate we said cross multiply we're going to cross multiply there so that's 0.028 va times 1 so that will give us 0.028 va okay and then what about our b value so that's 0.06 times 30 times 30 times 2 okay so that gives me 3.6 okay and then all that i'm going to do is just divide both sides by two 0.028 to get rid of that but what i do on the left i need to do on this side uh two eight so it means that my volume value so three comma six divided by zero 0.028 and i get a value of 128.57 cubic centimeters okay but if you remember my question my question said i want that volume in cubic decimeters okay so how do i convert from cubic decimeters to cubic uh cubic centimeters to cubic decimeters so all that i'd simply do means that the volume that i want it will be 128.57 okay how do i convert to cubic decimeters you simply divide by a thousand okay right so if i divide that by a thousand okay so all that you get is zero point one uh two all right so i don't know how many you want to carry let me just say 0.129 cubic decimeters okay right so that's my volume value okay right so ladies and gents when it comes to titration this is it okay uh this is what you need to note and by the way uh for those in grade 11 um this is what you would have to know and master all right so in our next video what we're going to do is we're going to look at how we can now apply um the other theory remember we said we've got the arenas theory which is what we did today and we're going to now apply the lary bronsted theory and i'm going to show you how we calculate ph okay the ph of an acid or base but otherwise for now ladies and gents let's keep it here all right and looking forward to having another great lesson with you all right and please if you haven't subscribed don't forget to do the right thing and i'll see you next time sharp