hello students we are at topic 14 on acid-base equilibrium this is a continuation from the chemical equilibrium and solubility products which you have learned in jc1 and if you recall some of the things i mentioned before when we are dealing with acid-base equilibrium we are still very much focused on a system at equilibrium so that means that uh the construction of the ice table uh for easy uh doing and then as a as a way of solving problems is highly useful yeah so for this topic i would advise you to continue using the ice table to help you soft although along the way you might realize that using the ice table might not be absolutely necessary because eventually the outcome seems to be the same yeah but it does helps you to focus your mind and then think about the variables and think about the assumptions which you have made when you are working through the problems okay so without further ado i will just jump straight in into the introduction yep so you probably need to recall what you have learned in topic for under reactions and stoichiometry on the bronsted-lowry theory of acids and bases okay so first of all we need to understand that an acid is a proton donor and a base is a proton acceptor based on the bronsted theory of acids and bases okay so when there is an acid-base reaction taking place it actually involved the transfer of a proton from the acid to the base okay solely by this definition so if we are going to look at this particular system where you have ethanoic acids so this is ethanoic acids and you dissolve it uh in water and most of you who have learned in secondary school that ethanoic acid is a weak acid so it will partially ionize in water to give you um ethanoids which is what we call the conjugate base and of course in this case h2o accepts a proton in this case h plus from uh atonomic acid uh to give h3o plus so we call h3o plus the conjugate acid okay so if you if you look at the terms here is actually arbitrary okay so what do i mean by arbitrary okay so if we are looking at a reaction going from left to right uh in this case my etonoic acid is actually the acid which give water the proton so we will call ethanoic acid the acid here of course and then water the base yeah but after ethanoic acid gives a h plus to water it actually becomes a base because the negatively charged uh species or the negatively charged center uh is actually located largely at the oxygen so if you remember the structure for eternal weight it will look something like this yeah uh of course there's a lone pair here so the lone pair can actually accepts a h plus from h3o plus uh to give you back ethanoic acid and that is why this particular system uh is in equilibrium and that is why this whole topic is known as acid-base equilibrium because we are interested in uh when we dissolve a certain concentration of ethanoic acid in water how much of it does it ionize or how much of it does it dissociates okay so we are going to use the term uh dissociation and ionization interchangeably because both terms are correct in describing the situations okay uh we will not go into details like for example in this case uh you will know that ethanol is what we call resonance stabilized uh based on what you learned in organic chemistry okay so because um for the atonoid and ion you know that the oxygen pi bond is what i've shown you in red over here and then of course the o minus you have a lone pair of electrons that can overlap continuously with the c double bond o okay so we will not uh discuss too much about this because this will be covered in the other organic chemistry topics yeah so for this particular topic we are solely interested in understanding um the amounts uh which ionize and then uh perform some calculation and uh of course the the bot of the topic later on is actually on buffer solutions and titration curves and then uh determining the ph of buffers etcetera yeah but before you learn all those uh stuff you need to uh be comfortable uh in identifying uh what exactly is the acid and what is exactly the base and then when an acid ionizes or give the base the h plus it becomes the conjugate base and then when the original base accepts the h plus it becomes what we call the conjugate acid okay then in this case if i may just erase this portion right you can actually view this reaction uh using arrow pushing so i'm going to just write it out over here so we have a ethanoic acids okay over here and then um the o will take on a partial negative charge the hydrogen partial positive charged the water here has the lone pairs of electrons okay over here yep and then um this oxygen will be partial negative charge and then the h partial positive charged okay i think from here you can understand that uh the transfer of the h plus to the o involved the lone pairs of electrons attacking the h and then in this case it caused the heterolytic cleavage of the o h bond yeah so the electron uh the bonding electron pair is now being pushed towards the oxygen center or you can say that it breaks towards the oxygen center and um in this case after the reaction so if i may just represent it with this arrow yeah so after the reaction um you are going to get something like this um with an o minus okay and then of course that's the lone pairs over here so this lone pair um it actually comes from the heterolytic cleavage of the oh bond yeah but for those of you who are sharp enough right you also notice that originally it already has two lone pairs yeah so after the lone pairs after the heterolytic cleavage right it doesn't really matter which lone pair we're talking about because they are um you can't really distinguish them okay and then um after that you're going to get a h3o plus so you're going to get an o h here and then an extra h and then with a plus charge at the oxygen yeah those of you who are familiar with formal charge yeah so this plus is a full charge as well as a formal charge at the oxygen okay then of course uh you will also notice that for this particular system um because the ethanol is a conjugate base yup so the lone pair will also abstract h plus back from h3o plus otherwise known as the conjugate acid and then now the o h bond will undergo heterolytic cleavage and break towards the oxygen and in this case it will neutralize the charge and then i will get back atonic acid and h2o yeah so where the position of equilibrium will lie towards largely depends on um which acid is stronger and which base is stronger okay so uh in short we are actually saying that um is atenoic acid a stronger acid or is h3o plus a strong acid okay and is h2o a stronger base or is uh ch3co2 minus a stronger base yeah and this whole topic along the way we are really uh trying to decide uh which is stronger and which is weaker and then um we need some numbers to guide us and that is why along the way you learn things like uh ka we are going to introduce you to a new equilibrium constant like ka for acids and then kb for bases etc etc okay um of course those of you who are already quite familiar i think you should be able to tell that uh h3o plus is obviously a stronger acid than atomic acid yeah and and that is why uh ethanol is actually a weak acid and the position of equilibrium actually lies towards the acid side i mean the ethanoic acid side okay and it has a relatively uh small ka value in the region of 10 to power -5 yeah and those of you in secondary school you might be familiar that the amount of ethanoic acid which ionizes is maybe around less than one percent okay yeah and i'm just going to move on from here uh yeah of course another term which you need to be familiar with is the is what we call the uh conjugate acid base pair yeah so once ethanoic acid loses a proton it will form attenuates so we will say that ethanoic acid and ethanol it forms a conjugate acid-base pair okay and at the same time h2o and h3o plus also forms a conjugate acid-base pair yeah so uh basically conjugate acid base pair differ by a h plus okay yeah uh this is something generic uh i don't think i need to go into details so do flat out if you don't really uh understand this because i think i've used the example above to illustrate uh this portion uh relatively well okay so i will just straight uh and go into exercise 1.1 so in exercise 1.1 you are supposed to identify the conjugate acid base pair in each of the following reactions yeah so over here for the first question we have nitric acid and ammonia reacting to give us nitrates and ammonium and in this case you notice that nitric acid loses a h plus to ammonia so in this case we will call uh nitric acid the acids and of course after nitric acid loses the h plus it forms nitrate we will call nitrate the conjugates based okay and in our case here ammonia accepts the proton so we will call our ammonia the base and then after it forms ammonium we'll call this the conjugates acids yep and then for question two the same thing um ammonia reacting with h2o and in the process form ammonium again so yup so ammonia accepts a h plus from h2o so we will call ammonia the base over here and because h2o loses a proton so we will call uh h2o um the acid okay and of course our ammonium is now the conjugate acids and o h minus is now the conjugates uh based yeah so do take note that uh whether something is a conjugate acid of a base or whether something is a conjugate base of an acid it really depends on the partner which you are referring to yeah so um remember that in any equations uh our focus is the forward reaction which is from left to right okay so if we are to reverse the reaction so for example for example the first question right if we have to reverse the reaction so let me just erase this portion for simplicity okay so if we are to reverse the reaction then um we will refer ammonium to be the acid here because it loses a h plus two nitrates yup okay so now uh if we reverse the reaction okay so do take note of it don't listen to it halfway huh okay so if we reverse the reaction yeah uh ammonium will now be called the acids and of course ammonia here will be called the conjugate based and then nitrate gains a proton from ammonium so nitrate will be called a base and nitric acid will be called the conjugate acid yeah so uh it's a perspective thing uh there is no absolute it depends on your reference point so that is something uh very important which you need to take note of okay then uh moving on to 1.2 where we'll be discussing strong and weak acids and bases so as what you may have already know in secondary school a strong bronzer acid is one which dissociates or ionizes completely in water to give h3o plus okay so this term dissociates and or ionizes is important terms which you need to use in your answer yeah but of course we underline the term completely because the complete dissociation or the complete ionization uh is a key criteria for deciding whether something is a strong or weak acid yeah and uh for the purpose of our discussion uh in uh most of our discussion related to h2 acid-base equilibrium we are dealing with uh the acids dissolving in water so uh that's why you notice this particular portion where they said uh ionizes all or dissociates completely in water to give h3o plus and when we have h3o plus over here you might be wondering why is it not h plus but uh h3o plus well um you might be aware that h plus essentially is a proton so it's actually not very stable actually it's not stable at all yeah so uh of course we use h plus as a proxy for us to uh discuss um a representation of an acid yeah but um scientists have of course uh discovered over the years that um h plus is uh not uh the reality so it doesn't really represent the reality so in most cases when we are dealing with acid-base equilibrium we would like a h-plus to be donated or to be in this case stabilized by a water molecule so in the process it forms h3o plus yeah but scientists have actually uncovered that it is actually more complicated than just h2o plus yeah it could be h3o plus but stabilized by another water molecules or another two water molecules so you can have species like um h5 o2 plus or h7o3 plus and and and so and so on so um it's actually very complicated but um in most cases when we are performing calculation uh we will try to simplify the equation yup so uh if you are wondering whether you can continue to use h plus i think that should be fine in most cases yeah but just be aware that h3o plus uh might be something which you see uh more common especially in the topic on uh essay-based equilibrium yeah even in advanced chemistry texts at the university level or or even at the graduate level they tend to use h3o plus but of course um when they use that at that level uh they look into slightly different things like yeah um there's a difference between diluted solution and concentrated solution because the properties uh can be very different yeah so for acid-base equilibrium for the purpose of our discussion for h2 we also tend to just focus on dilute solution yeah because as the solution becomes concentrated then uh they display slightly different behavior uh but you will not encounter them um very much in h2 okay so uh yeah so in this case uh there are certain bullet points which you need to be aware of like for example um the conjugate base of a strong acid has a low tendency to accept a proton back yeah so uh in that case i think um it will be plain to notice that if i have a strong acid here which ionizes completely in water to give h3o plus i will not expect chloride to abstract uh h plus back from h0 plus to give me hcl yeah because hcl equals on its own right it doesn't exist uh i mean it doesn't exist as molecular hcl so meaning when we are looking at this um we are not looking at this form in in water so this is not the form that exists in water so what exists in water is actually chloride and h3o plus yeah so that is the reality yeah uh and that is why i focus on telling you that um the solution must be diluted because for concentrated solution you do get some molecular hcl floating around and then um maybe it will exist in the air space above the concentrated solution and things will get really complicated yeah so um yeah i mean without going to very complicated things yeah we let us just focus on dilute solutions okay so uh in dilute solution you will not get molecular hcl yeah uh please understand the term um whenever we try to say let's say molecular ammonia molecular hcl in aqueous solution yeah so molecular ammonia exists in um aqueous solution as well as molecular ethanoic acid uh because these are weak bases and weak uh acids so they do not ionize completely yeah yeah and that's why uh the bot of them exists as the molecular form but for hcl being a strong acid ionizers completely you will not see any molecules of hcl around and um it is the same for the rest of the strong acids like h and nitric acid hbi et cetera et cetera okay yeah so uh in contrast a weak bronsted uh acid will ionize partially or you can say it dissociates partially so the term partially and completely will be important for us to uh recognize that uh indeed you understand the difference between a strong and a weak acid yeah and of course the question is how partially is partially how completely is completely yeah normally in a standard uh h2 kind of questions which you will get in exam uh normally when we are talking about weak acid uh you are going to get a k value of the magnitude say 10 to the power of minus 3 or lower yeah so uh and that's why it is safe for you to say that uh these are relatively weak acid but sometimes you will hear your tutor saying oh this is a strong weak acid so what exactly is a strong weak acid yeah so um okay so weak acid in this case is the noun yeah so we are talking about weak acid that is uh relatively strong because the ka value is not very low but not very high yeah and um in that case you will know that when you are trying to make approximation later on we are going to do some calculation you realize that if you are trying to make some approximation you will not come across as very accurate okay yeah so uh be aware of the double-headed equilibrium arrow so it's like this uh i think in chemical equilibrium we emphasize this right uh never write your arrow like this huh because this represent resonance okay so uh this is incorrect so you should not be writing a double arrow heads okay yeah um moving on um a strong bronsted based uh in this case will ionize completely as well so uh in this case strong bronzer base they tend to be ionic compounds like sodium hydroxide solids or potassium hydroxide so when you dissolve them in water they do not exist as the solid anymore so so you are just going to get um a solution full of sodium plus and uh o h minus okay um on the other hand for weak bronsted bases they tend to be molecular form if you learn more organic chem you'll notice that they tend to be amine based yeah so uh ammonia will be among is not an amine so ammonia is ammonia but when i mean amine base i mean like for example one of the h in ammonia is being replaced by an alkyl group so let's say it's r group okay so maybe i use a very specific example methylamine yeah so methylamine is an example of an amine and because of the lone pairs on n here so uh the lone pairs of electrons have the tendency to abstract h plus from the solution so therefore methodology is a weak based or we can we call it a weak uh bronsted base okay so there are other examples which uh you are uh you're gonna deal with them um as you move on in your organic chemistry education okay so um remember that the strength of an acid is not the same as the concentration okay so you can have a very concentrated weak acid but it is still a weak acid because it does then ionize completely in water okay let's move on to exercise 1.2 where you are supposed to match each acid on the left with the correct descriptors on the right okay so we have a very dilute nitric acid so uh nitric acid as we discussed earlier on is a strong acid so therefore is a dilute but a strong acid okay so dilute and strong acid so we match this too um ethanoic acid at four moles per diem cube uh pretty concentrated yeah so ethanol s is a weak acid but uh this concentration is quite high so it's concentrated so it's a concentrated weak acid okay uh this is a very dilute weak acid so it's a dilute and weak acid okay so uh do take note that there's a there's a difference between uh something that is concentrated and it's uh strength uh it's very important for you to distinguish that okay so without further ado we'll move on to uh topic two um on calculations involving acids and bases as uh what you might have remember from the earlier chemical equilibrium and solubility product topic um any equilibrium calculation um you can use an ice table to actually help you in terms of visualization uh to to work through the problems yeah so i would really suggest uh using ice table to help you uh but of course after a while you'll realize that for acid-base equilibrium um the variable tends to be quite similar so uh you might not feel that the ice table is required yeah but of course before that we need to um define uh what we meant by ph and uh poh basically they the when you put a p in front of either the h or the och or even um k uh we are taking the negative log to the base ten so basically p h simply means negative log to a base 10 concentration of h plus in the solution okay and um so therefore if you do a little bit of manipulation you'll realize that the concentration of h plus is just 10 to the power of minus ph but you do not need to memorize this so don't memorize these are things that you can work through with your mathematics okay so uh same idea goes for poh so i will not elaborate okay 2.2 we will introduce you to uh an idea of the autoionization of water uh under standard conditions yeah so uh pure water does conduct electricity slightly uh because they auto ionize so auto ionize you can replace this work with self ionization so it's a very slight ionization so that means that um itself as the base and itself as the acid so they uh they transfer a proton to one another um or rather from one party to the other party so uh one will get h3o become h3o plus the other become o h minus so this is known as auto ionization okay um the simpler way of course uh is just to look at h2o uh which involved the heterolytic uh fission or the heterolytic breakage of the o h sigma bond and in our case um to give um h plus and o h minus yeah uh so in terms of organic chemistry uh because most of you if not all of you uh hopefully remember some stuff you learn from organic chemistry so this arrow pushing actually um goes from the o h sigma bond uh you push towards the o here yeah so um in the process you will get a h plus okay and then of course originally i have two lone pairs after you push to the o then um the o h minus at the o there will get an extra lone pair i will represent with yellow okay so uh it will be a negative charge okay so in this case the equilibrium constant as you have seen here uh this is indeed an equilibrium so we can actually uh write out an equilibrium constant kc uh where h plus oh minus divided by h2o and as you know h2o has a concentration of about 55.8 of 5.6 moles per mq being a constant uh it will not be reflected as well so uh if you bring this across you'll get a constant term yeah so uh this particular constant term we call it kw yeah so kw uh is known as the ionic products of water yeah and as per all equilibrium uh constants uh only dependent on temperature so it's a temperature dependent variable and as you might be aware the breaking of the oh bond i mean it's solely i won't say slowly but largely results in uh the formation of h plus and o h minus yeah so the auto ionization of water in this case uh you would expect it to be endothermic yeah so therefore um the higher the temperature you expect the position of equilibrium to lie more to the right and uh therefore you expect the value of kw to increase as well okay yeah so uh there is a need for you to to do some very simple manipulation um again uh please do not commit to memory because if you try to commit to memory then you realize that the learning of this topic is no longer interesting yeah so um you need to be very comfortable in converting uh from kw to pkw um in fact uh throughout um this whole acid-base equilibrium so for example from ka to pka expression okay so uh you need to work a little bit on your mathematics as well so uh basically you are taking a negative log base then um on both sides and then because this is multiplication so uh it become a plus and then um because we define negative log base 10 of h plus s p h so this is essentially p h plus p o h okay yeah i know this is straightforward but uh you know sometimes when it comes to exams uh uh under exam pressure or when you're doing our tutorials you might forget yeah so uh these are something which you need to take note of as well okay yeah then uh moving on to 2.3 um relationship between h plus o h miners and um ph yup um if you have heard me earlier on okay so um some of the things i'm gonna mention here i think will be uh obvious to some of you okay so uh as i said to you earlier on and it also appears in your data booklet um at 25 degrees celsius standard condition kw is 10 to the power minus 14. okay so this value is also given in the data booklet so you need to know it's actually on the second page okay so therefore when you express it pkw is equals to ph plus poh and this is equals to 14 yeah and uh that is why for solution that is neutral at 25 degrees celsius we say that is ph 7 okay but this is not true when you change the temperature okay so when i change the temperature the neutrality ph value can change okay so maybe when we are close to boiling point of water maybe the neutrality ph is about 6.5 but that doesn't mean that water is acidic okay the reason is when we increase the temperature we are encouraging the ionization of h2o but when we are encouraging the ionization of h2o right in the process we are forming the same amount of h plus and o h minus so therefore the concentration of h plus and oh minus balances of each other so we are still dealing with a neutral solution yeah we are only dealing with an acidic solution if there is an imbalance of either party so if we have more h plus than o h minus we get the acidic solution if we have more o h minus then h plus we get an alkaline solution right so and so forth okay so that is something you you you need to be aware of and confident with yeah and uh this is also something which student tends to make mistake in exams because sometimes we'll change the condition we'll tell you that maybe at say 80 degrees celsius the pkw is 13 for example and then therefore ph is 6.5 so some students will think that the solution is acidic which is not true okay so remember that similar to kc kp and ksp okay similar to k s p k c and k p um equilibrium constants these are all temperature dependent variables okay so k w k a later on you're going to meet ka and kb or this they will change the value will change according to temperature okay so uh i think we have a table over here which uh i think showcased to you the different values of the concentration of h plus so i will not go into details uh if you want you can calculate the pkw yourself okay so we have an exercise uh 2.1 over here and um we reference the above data on kw state whether the following statements are true or false okay so uh ionic dissociation uh is an endothermic process yes this is obviously true uh uh how do we tell number one is biologic uh what do i mean by logic uh earlier on i mentioned that the only way you can get h plus and o h minus is by the heterolytic cleavage of the o h bond so this is largely bond breaking so if it's largely bond breaking i'm going to expect the reaction to be endothermic but of course you can argue that uh water on its own uh gets stabilized by hydrogen bonds yeah so uh technically you should also consider the interaction between the water molecules and water molecules as well as the water molecules and h plus and the water molecules and o h minus uh before you can be con before you are confident to say that uh this is an endothermic process yeah uh so we are making an assumption that the interactions more or less don't change too much and that's why we can we are confident that it's an endothermic process but of course uh talk is cheap so we need to look at data so you notice that as the temperature increases the value of kw increases as well yeah so therefore from this particular evidence uh the ionic uh ionization or the self ionization of uh water the self dissociation of water is an endothermic process okay then um for two and three as water is heated up uh concentration of h plus and h minus increases you'll notice that these two statements did not say the other party so uh i hope you can understand that when concentration of h plus increased concentration of o h minus will increase accordingly as well so uh 2 and 3 are tied to each other remember that they are tied to each other it cannot be h plus increases more than oh minus and suddenly the solution becomes acidic so this is incorrect okay part four in pure water the concentration of h plus is higher than o h minus at 60 degrees celsius it is nonsense because pure water h plus concentration should be the same as o h minus concentration okay um part five water remains a neutral liquid at 60 degrees celsius so of course this is correct i think we emphasized that quite a bit just now because the concentration of h plus and oh minus are the same and then for part six the ph of pure water at temperature greater than 25 degrees celsius is seven this is incorrect it should get lower because uh of the increase in ionization of h2o so you get a higher h plus and oh minus concentration you take a negative log of them you get a value below seven okay uh when in doubt you can just punch your calculator and verify okay then moving on to 2.5 over here right um we have the ph of strong acid and strong base solution okay this is not very inspiring because uh basically the concentration of h plus is the same as the concentration of the strong acids in a dilute solution i always like to emphasize to students is in a dilute solution because for a concentrated solution things can get tricky yeah but i think you can take that off the equation for h2 because i think in h2 we will not tend to go into uh complicated stuff like asking you to calculate um uh things involving concentrated strong acid or even concentrated weak acid when i mean concentrated here i really mean very concentrated yeah because that really involves a totally different ballgame okay yeah then if the concentration of acid or base is 10 to the power minus 7 or lower i think this is important we need to incorporate the autoionization of water in our calculations okay so you have to take them into consideration in your calculations okay uh then some of you might be asking then why is it that uh if the concentration of acid is uh bigger than 10 to the power minus 7 we don't need to include the autoionization of the water molecules uh that is because you just imagine that let's say you add um 10 to power minus 7 to say 10 to the power minus 3 the numbers is actually very close to 10 to the power -3 okay so it becomes the 10 to the power minus 7 becomes relatively insignificant that's the reasons okay yeah then i think we'll just start off with uh exercise 2.2 uh to warm up a little bit so we have a we have an aqueous solution containing um one point zero times the power five moles per mq of hi which is a strong acid okay so your job is to calculate the concentration of h plus in moles per diem cube and then hence the ph of the solution okay so uh as what you have seen earlier on concentration of i will be the same as the concentration of h plus because h i is a strong acid which ionizes us completely and therefore the concentration will be 10 to the power minus 5 moles per diem cube okay and uh ph is simply just minus log base 10 of 10 to the power minus 5 okay and we will get uh five point zero zero okay so we will get we are getting ph five over here okay so it's a straightforward calculation and then um for question two um an aqueous solution containing zero point zero one moles per diem cube of uh beryllium hydroxide uh take note that this is a dye acidic base so one equivalent of baron hydroxide gives you two equivalent of o h minus okay and then um i and then i mix with 0.05 moles per mq of koh okay and then of course uh my my job is to calculate the ph of this uh solution okay so uh over here uh we need to look at the total concentration of o h minus so we need to look at our o h minus total yeah so that would be the contribution from barium hydroxide which is a diacidic base so we need to multiply 0.010 by 2 okay and then we need to add this to um 0.050 okay uh you think so from the from the way the question is phrased uh they are not mixing two solutions together so the solution itself already contains zero point per dime of various hydroxide and 0.05 0 moles per mq or koh so there is a need for it to to do anything with the volume term yeah so if you add the two up uh you are going to get the total oh minus concentration to be 0.070 moles per dmq over here okay and then um therefore uh we are we can easily calculate poh which is simply uh minus log base 10 concentration of h ah minus okay so uh if you punch your calculator you should get around 1.15 okay and then we can easily calculate ph over here that is 14 minus poh and then we should get 12.8 okay yep then um we have question three uh equal solution contains one point what sorry 1 times 10 to the power minus 8 moles per mq of nitric acid yeah so uh in this case the student might simply just say that the concentration of h plus is equals to 10 to the power minus 8 moles per diem cube yeah and then after that uh the student will simply calculate p h and then just minus log base there and eight and then the students will just happily submit this as the answer but this is actually not correct yeah because for uh for a aqueous solution at 25 degrees celsius um ph cannot exceed seven because that would mean that there's an imbalance of o h minus against h plus yeah so uh then in that case uh why is this incorrect yeah because um when you are performing this calculation uh because the concentration of h plus from the strong acid is less than 10 to the power minus seven you have to incorporate the auto ionization of water in your summation okay of concentration h plus so in this case um there is a missing part over here so um we have to add uh 10 to the power -7 here and it gives us 1.1 times 10 to the power minus 7 moles per diem cube and then when we calculate ph punch the calculator we should get 6.96 okay yeah those students uh at this stage right if you are short enough uh you might recognize that a simple addition like this is actually conceptually incorrect okay so this is kind of uh conceptually uh incorrect but it is a good proxy okay it's a good proxy why is it conceptually incorrect because the ionization of our water uh in this case um 10 power minus 7 right um and the the presence of the strong acid which is nitric acid right will affect the autoionization of water yeah okay because um in okay when we are looking at the auto ionization of water uh there is no excess of h plus or oh minus that will shift the position of equilibrium am i right to say that yeah so because of that um we can simply just take ten to power minus seven and then um and then calculate the ph but when the system already have uh nitric acid we are kind of dealing with um i mean a common ion but we are not talking about um salts salty i hope so i hope you see the difference we're not talking about salts not dissolving but yeah so we have a h plus in the system so it will affect the autoionization of water yeah so it will actually suppress us the auto ionization of water by a little bit so you are expecting less contribution from h2o ionization yeah so the actual value if you if you incorporate that in your calculation you should be getting about 6.98 but of course to do that uh you need to learn a concept known as the systematic treatment of equilibrium but that is not um in the h2 curriculum so you don't have to worry about that okay yeah so i repeat um whatever we are doing here is just a proxy because it's conceptually incorrect to consider the two uh system independently because whatever h plus in the system will affect one another okay uh if you are confused just think back about the ksp which you have done last year okay uh if you're not so confused i think it's okay you once you can read out on the systematic treatment of equilibrium and that would be very interesting okay so with that um let us move on to exercise 2.3 okay then um exercise 2.3 we have an aqueous solution contains 0.10 moles per calcium hydroxide you are told that it's a strong base okay and then you're supposed to calculate the concentration of which minus the most per diem cube enhance the ph at 25 degrees celsius even if they don't tell you 25 degrees celsius you can assume that okay because uh it's kind of no choice okay so again we will use the same logic because calcium hydroxide is actually a diacidic based yeah so therefore the concentration of o h minus is simply 0.10 multiplied by 2. so i'm going to get 0.20 moles per m cubed so this is simply the concentration um of o h minus uh being contributed by 0.1 moles per inch of calcium hydroxide okay and then um when we calculate poh i'm just going to take negative log base 10 of this and i'm going to get 0.699 okay and of course uh ph is simply 40 minus that so i'm going to get 13.3 okay so this would be relatively straightforward okay next uh we look at question two of exercise 2.3 over here we have solution x uh i condesh hcl ph 2.2 i have another solution y same volume as x okay but my y contains nitric acid ph 1.6 i mix the two solutions together calculate the ph of the uh final mixture okay uh what most students will do is uh maybe they will take 2.2 1.6 and then simply divide by two and then actually that is incorrect uh yeah uh because technically you have to calculate the h plus contribution or h plus concentration due to each of the uh solutions okay so um i'm going to calculate the total concentration of h plus so total okay this will be the contribution from hcl which essentially is 10 to the power minus 2.2 so for those of you who are a little bit confused by this how come i can just take 10 to power minus 2.2 right so remember that h sorry sorry remember that ph is minus log base 10 concentration of h plus 2.2 so essentially i can shift the minus to the right so i get minus 2.2 uh then i because it's log base 10 so h plus concentration is 10 to the power of minus 2.2 yeah so concentration of h plus due to hcl concentration of h plus due to nitric acid that is 1.6 and then uh because i mix the two solutions together and each of the solution has the same volume so uh that means that when volume doubles concentration of half okay and then from here i'm going to get uh punch my calculator 0.0157 uh moles per diem cube okay and then uh i will calculate my ph based on this concentration which is minus log base 10 um concentration of h plus okay and then i'm going to get 1.80 okay yeah so i hope uh you'll find the calculation are not so not as tough as you think but of course we haven't gone to calculating um weak acids okay so for 2.5 we're just going to get to the mid of the chapter so we have uh we are going to define two variables acid dissociation constant ka and uh base dissociation constant kb yeah actually it's a little bit weird to to say uh base dissociation constant because the base actually do not dissociate unlike um i mean weak base don't dissociate weak base ionizers the reason why we based don't dissociate is because your weak bases are largely in molecular form so when you put them into water uh they cause the h2o to ionize in that sense like in in the process produces o h minus the base itself uh forms the conjugate acids so your base do not break up in that sense so uh it's a bit weird uh to call it base dissociation constant but it has some legacy issues so uh we just keep it but i think it's good to be aware why is it called base dissociation constant okay then uh of course you have seen this uh in our starting portion uh when we are discussing ethanoic acids okay then uh we have h a uh partially ionizes in water to give a minus and h3o plus so over here we are going to write a ka expression yeah so that will be the equilibrium concentration of um h plus or h3o plus for that matter a minus divided by the equilibrium concentration of h a yeah then um you'll notice that the bigger the value of ka right that means that uh the greater the extent of ionization yeah and of course the bigger the value of ka the smaller will be the value of the pka okay so it has this uh inverse relationship again no need to memorize uh once you are in doubt just punch calculator and i think the answers will be quite obvious so uh don't worry about all these little things do not memorize unnecessary things okay in fact this is the single most important equation you need to know uh throughout your studies for acid-base equilibrium uh and this equation you actually don't have to memorize because it actually comes from here and most of you can actually write this out so uh in all honesty acid-base equilibrium should be one of the easiest topic in chemistry but uh i don't know students somehow feels that is the most difficult okay that is why when you study this topic it is important for you to focus on the fundamental understanding of the phenomenon rather than uh blindly memorizing equations okay if you want to memorize actually the only thing you might want to know is the relationship between pka and ka or the relationship between p something and you know and that's something and then of course uh how to write k expression of course that is not very difficult okay so uh i'm going to move on uh base ionization constant base dissociating constant okay so base ionized partially in the process uh gives you the conjugate acid and o h minus so again you can write a kb expression out you can define pkb as well and the greater the kb value the smaller the pkb so same logic so these are just some kb values of the various uh bases okay the relationship between uh ka kb and uh kw okay so there is a little bit of derivation here uh and sometimes the the students might not be um uh very confident about what exactly is going on so again uh in the end they will end up memorizing equations okay so uh fret knots that's not supposed to be their way okay so uh if i'm going to zoom into this right remember earlier on that we say that we define this as the acids okay and uh of course when this is the acid ethanoid becomes the conjugate base okay so from here we can actually write out this particular ka expression okay but the conjugate base itself right uh being the base of a weak acid itself is also a weak base and therefore itself can also ionized uh in water and in the process produces ethanolic acid and o h minus yeah so over here we can also define a kb expression yeah so remember that whatever weak acid that ionizes uh the base it produced can also ionize back because the base it produced is also a weak base yeah so what we're going to do here is we are attempting to combine the two equations okay so we are multiplying k and kb together and you notice that when you multiply the two together right uh the numerator here which is the ethernet will cancel out the denominator here and then um the denominator which is atomic acid will cancel out the numerator for kb and then eventually you're going to end up with kw equals to concentration of h plus plus uh multiplied by concentration of o h minus and then after that when you expand this out you'll notice that ka times kb equals to kw okay how do we get this because ka and kb equals to kw from here okay right so uh that's how this thing uh is obtained and then after when you take the negative log base then of uh i mean negative log base then of the concept of k a and k b and then kw on both sides you end up with p k a plus p kb equals to pkw okay yup so uh sometimes uh when students reach here they get super confused because there seems to be more equations to memorize but uh honestly you do not need to really memorize this because again as i've shown you above the the logic is that uh for the same system uh which means for the same conjugate acid base pair you can allow the pk and pkb adding them up together which eventually will give you pkw uh because essentially each of them is responsible for their own h and o h minus okay i mean i mean that that's not exactly the correct way of saying but that that is one way way which you can look at things like yeah but uh as we have shown here mathematically uh this is how it works okay then so this is a very important expression because from this particular expression you'll notice that if i give you the ka of ethanoic acids you can easily tell me the kb of its conjugate base which is at the nodes so vice versa if i give you the kb of adenoids you can easily calculate and tell me the kb of sorry the ka of ethanoic acid okay so um i know there's no example here so maybe i will uh just illustrate with one simple example which is um an example i often use to discuss students uh so for example ethernoic acids okay ethanoic acids uh pka is 4.76 okay uh i think if you are familiar with weak acids this is one of them that you need to know off the top of your head so pk is 4.76 yeah then um what about eternal weights what is the pkb so if you truly understand this you you'll notice that it's simply 14 minus 4.76 so you are going to get 9.24 okay now um is ethanoic acid a stronger acid i mean is ethanol acid able to perform its role as an acid more prominently or at the known weights being able to perform its role as the base more prominently this is a difficult i mean i wasn't difficult but this question might not give you you might not have a very obvious answer yeah but if you simply compare the pka and pkp you'll notice that the pka is a lower value than the pkb yeah so that is why you you realize that um the position of euclidean actually lies a bit more to the right when we are looking at the ionization for ethanoic acid so therefore as an acid uh p ah i mean as an acid atomic acid uh is a stronger acid than its corresponding conjugate base as a base yeah so that's what the value is trying to tell you again um these are not things which you probably want to memorize uh i mean along the way uh um as you i mean as you grow you'll be more familiar and then uh you'll notice that it's not as difficult as it seems like but right now i thought i just want to introduce to you some of these ideas okay and then 2.6 we're looking at the extent of ionization or the degree of dissociation so basically it's just the amounts which ionized divided by the emission yeah so uh we give it an alpha term so if alpha is close to one that is a strong acid or base because essentially uh the concentration of acid which ionizes or dissociates is the same as the initial concentration so weak acid has alpha much lesser than one why are we telling you this sometimes the question might require you to calculate the fraction of ionization and that is why we're introducing this to you next we are moving on to section 2.7 where we are going to calculate the ph of weak acid or a weak based solution okay so uh because the ionizers or dissociates uh partially so the solution concentration for the h plus and o h minus will be lower than the emission concentration of the solution okay so over here i will just use a relatively generic way of uh representing it so suppose that we have a weak acid ha in aqueous solution and then in water which is the aqueous medium it will ionize partially to give me h plus equals and um a minus aqueous as you have seen above this is defined as ka okay so uh we can use the same method um we use for uh chemical equilibrium so we can create an ice table where um the initial concentration so it would be good to put the units here as what we mentioned in chemical equilibrium because if you put the short form ice you may not know whether they are referring to initial concentration or emission number of moles of course for acid-base equilibrium usually we are looking at initial concentration okay uh we rarely work in terms of moles yeah okay so um over here we can have uh the initial concentration of c naught okay and of course uh we don't deal with water because this concentration is a constant um being way higher than that of uh the rest of species okay and then h plus and a minus um zero initially okay so the change um again we let the change be minus x okay similar to chemical equilibrium then we have plus x plus x here so uh the equilibrium concentration of h a will be c naught minus x and then the equilibrium concentration of h plus will be x moles per mq and a minus x moles by the m cube so similar to chemical equi equilibrium we define ka to be the concentration of h plus a minus uh together so that will be x squared over uh the initial concentration i mean the equilibrium concentration of h a so that'll be c naught minus x usually to simplify our calculation we will make an assumption here we will assume that the ionization is negligible as compared to the initial concentration so that will mean that we will assume [Music] c naught to be way greater than x yeah so because of that we will be able to get ka equals to x squared over c naught and then we'll be able to get x square equals to c naught multiplied by k a and then we can get x to be equals to the square root of c naught multiplied by k a this is the expression um that a lot of guidebook and even textbook wrote for students and students thought that they actually need to memorize it yeah so again my suggestion is i prefer the students to work from first principle so create an ice table it's not it doesn't take you too much time and then um formulate the actual k expression and then uh know what kind of assumption you're making and then from there um you will be able to calculate x uh easily and then as you know x is the equilibrium concentration of h plus you take the negative log to the base 10 of x and then you will get the ph okay so uh in general this is the way uh you should go about working through the problems okay so um again uh you have exercise 2.4 here which is a work example for you okay so although the answer is given to you so let me just uh run you through so again um uh first you need to find a ph and degree of uh dissociation so this is the alpha which we mentioned um earlier on at the end of section 2.6 okay so this is the alpha here yep and then you are given the k of ethanoic acid to be 1.8 times 10 to the power minus 5. so uh you can see from the ice table that uh the units is not written so as i mentioned it will be good to write it in so please write in the units okay and then um yup so you formulate the ice table to be like this and then you can write the k expression so uh again this is the assumption we make here so uh when you work from first principle uh you will be able to get x equals to 1.34 times 10 power minus 3 and then our ph to be 2.87 the reason why we ask you all to make the assumption that x is much lesser than the initial concentration for h2 is because somewhere in your curriculum it was mentioned that the solving of quadratic equation is not required for the solving related to acid-base equilibrium so let me just show you this relevant parts in our syllabus documents yeah so under your syllabus document asset-based equilibrium is under topic 10. if you scroll down under learning objective 10c it's written over here that the calculation of h plus concentration and ph values of strong acid weak monobasic monoprotic acid strong bases and weight mono basic acids and then there's a bracket component here it reads calculations involving with acids bases will not require the solving and quadratic equation yeah so because of this it will be good if you can make an assumption um that x is much lesser than the initial concentration and then after that you can solve it in an easier fashion but of course if you choose to do quadratic equation we cannot stop you yeah it's just that uh it might not be advisable because uh you will be penalizing yourself in exam yeah then some of you might be asking um is this assumption always true uh if you remember what i mentioned earlier on it is not always true uh the reason is because if your ka is of a relatively large magnitude say 10 to the power minus 2 or even in the low 10 to power minus or rather in the height and the power minus 3 right you might get um a relatively different answer if you if you make an assumption yeah but for the purpose of h2 we will just ignore that but as i said um if you wish to solve the quadratic equation uh you can go ahead but again uh penalizing yourself in the process okay yeah then um yeah whatever i mentioned i think it's uh given in this particular note over here and uh if you remember from the question you're supposed to calculate the degree of uh dissociation so that would be x so uh we will take x value one point three four times ten to the power minus three divided by zero point one multiplied by a hundred percent yeah so that's about one point three four percent dissociated yeah so i i guess the assumption is still quite valid yeah so uh moving on we'll move on to uh exercise 2.5 where you are supposed to calculate the ph and the degree of dissociation of a solution of a certain concentration of equals ammonia so again instead of ka we're looking at kb for now so um we'll work using the same method so we have aqueous ammonia um ionizers partially in the presence of h2o okay uh in the process it gives us um the conjugate acid ammonium and then um o h minus so again i stable so uh i have the habit of writing the concentration term okay over over here so that um you know what you're dealing with and then uh initial concentration is given to be 0.125 so that's 0.125 so uh the rest will be dash and then um the external ionization we are not sure so we let it be y so minus y plus y plus y we are defining kb over here so it's good to write this at the top of the equilibrium arrow and then the equilibrium concentration of ammonia would be 0.125 minus y and then uh ammonium will be y and y so again over here we will define kb uh to be y squared over 0.125 minus y yeah and over here we are going to make the assumptions so uh we are going to assume that um the concentration of um the initial concentration of ammonia is way greater than the extent of ionization then so therefore my kb is equals to y squared over 0.125 okay and then i'm supposed to calculate the ph so i'm given a kb value that's 1.78 times 10 to power minus 5. so 1.78 times 10 to the power of minus 5 multiplied by 0.125 i'll take the square root and this is equals to y okay so now i'll just punch calculator i should be able to get y to be 1.49 times 10 to the power of minus 3 moles per diem cube okay but take note that y is actually the concentration of h minus so if you take the negative log base 10 you are calculating poh not ph so just be careful so uh minus log base 10 uh 1.49 times 10 to the power -3 uh we will be able to get 2.83 okay so poh 2.83 ph would be 14 minus 2.83 and we'll get 11.2 okay and then to calculate the extent or the degree of uh dissociation we'll take y divided by 0.125 okay so uh i'll just do the right working over here so degree of dissociation uh would be equals to zero point oh sorry not zero point uh would be where is it one point four nine one point four nine times ten to power minus three uh divided by zero point one two five uh multiplied by a hundred percent so uh i'm going to get one point one nine percent as the degree of dissociation okay so uh the way to work out uh exercise 2.5 is very similar to uh exercise 2.4 so um if at this point in time you have some issues uh please go back and review them because it is highly necessary that you are comfortable in working uh through the basics of uh at least finding out the initial concentration of a weak base and a weak acid okay so that is exercise 2.4 and 2.5 okay and then we have a world example uh exercise 2.6 so uh over here we have a stimulant caffeine um which is a weak base so you need to read the question very carefully so that you know whether they are dealing with a weak base or weak acids okay so we have a weak base over here which ionizes again partially to give us the conjugate acid and oh minus and then we were told that a certain concentration of caffeine has a certain ph and then uh right now there are two parts to this question we're supposed to write down the kb expression and then after that uh calculate calculate its value at 25 degrees celsius okay so this shouldn't be too difficult so again the kb expression uh we have written it out and then we are given the ph of this solution which is 11.5 so um if you are sharp enough uh i mean if i may just uh make life a little bit easier by just roughly writing down the ice table yeah but so i'll just write it on the right hand side okay so normally we will let the the weak base be something like um a okay then uh plus h2o and then we equilibrate to give us a h plus plus okay maybe not a because a seems to imply that it's an acid so let let me represent it with b then okay so b h plus plus o h minus so again um because the working history will jump into the action so i thought i'll create an ice table for you to see uh in a clearer fashion okay so we have uh initial concentration 0.020 so dash dash dash and then we're looking at minus x here plus x plus x and then 0.020 minus x and then we have x and x do take note that you are aware of the ph of the solution okay uh it's unlike earlier on when you are supposed to calculate kb or rather uh you are supposed to calculate um the ph so yeah you are given the ph of the solution so from the ph you can determine poh which is what they are doing here and then from poh um uh you can actually determine the concentration of o h minus which is essentially the x yeah so over here they determine the concentration of o h minus and then uh they work out uh the kb expression over here which essentially comes from uh what i've written above so essentially the kb expression is um x squared over uh 0.020 minus x uh you'll notice that the x has already been calculated out for you okay uh that's what i mentioned earlier on because it's given to us indirectly yeah so we're just going to substitute in and then uh we should be able to get a kb value over here so uh yeah it's given to be this uh for those of you with a sharp pair of eyes right you'll notice that uh over here the minus x is included yeah you do you do not make an assumption and you just ignore it the reason is uh i mean it's just a very lame reason because you're not solving quadratic equation here so it's actually totally fine for you to include that in because it's not too difficult for calculation anyway okay so over here you shouldn't make this assumption okay then of course to calculate the degree of uh dissociation again uh it's not very inspiring i think uh most of you should be able to do that yeah okay so we'll just straight away uh move on to exercise 2.7 we have a ph of a weak acid uh h a concentration of this and then uh ph was found to be 3.0 calculate the ka and pka of the weak acids yup so this wouldn't be too difficult okay so uh we're gonna use the same logic okay so we have h a um plus h2o uh ionized to give us uh h3o plus plus a minus okay so we have the i stable here my apologies that i'm not going to write down the units because i need to work it out fast okay yeah and also to save space so initial concentration zero point one zero zero um is amount uh dash and then okay technically this is minus x plus x plus x okay i'm just going to do it a little bit differently so i want to see whether you can catch what i'm doing okay so for those of you who are already very sharp and you kind of like know what's going on uh you might be aware that actually x is already given to you indirectly because the ph is uh 3.0 yeah so for those of you who are really sharp you actually do not need to put in minus x plus x plus x you can just straight away put in the value so essentially uh this is minus 10 to the power minus 3.0 okay 10 to the power minus 3 10 to the power minus 3. so this would be zero point one zero zero uh minus ten to power minus three point zero so ten to power minus three point zero ten to the power minus three point zero okay so uh if you are sharp enough you can do that uh relatively quickly then so from here ka will be just 10 to the power minus 3.0 uh squared over 0.100 minus 10 to the power minus 3.0 okay punch your calculator you should be able to get 1.01 times 10 to power minus 5 moles per mq yeah and then from here pka can be obtained quite easily basically pka is just the negative log base 10 of the ka value so that's 1.01 times 10 to the power -5 and then will be able to get 5.00 okay so i hope um the calculation uh isn't too daunting um and um as long as you are systematic uh with your approach uh you should be able to get the answers uh relatively uh easily yeah uh yeah usually the students um based on your seniors uh i mean experience uh the main problem comes later on um not in front yeah and after a while they get confused with what they learn in front and that's where the problem comes okay so right now we have the strength of our assets and bases that will be under section 2.8 yeah so uh we have a few bullet points here so uh strength of acids measure how readily donates the protons then um depending on structure whether is it uh stable or not stable is there any factor that helps to stabilize it yeah so the greater the ka value the stronger the acids okay the lower the pka the stronger the acids as well yeah so uh as what we mentioned earlier on the strength and the concentration is not exactly related there are different concepts okay yeah so uh over here we have a questions uh ph and degree of uh or degree of dissociation reliable indicators of the strength okay so uh i would say uh if you are again if based on what you understand about uh strong acids uh so strong acid ionizes completely so therefore uh extent of dissociation or degree of dissociation would be a more reliable indicator okay so if if again uh i'll open this up for discussion uh in the tutorial so you don't really need to uh we don't really need to file an answer over here okay so moving on with exercise 2.8 um you you you have questions here in exercise 2.4 the ph and the degree of dissociation of a solution or eternal acid of concentration 0.1 were found to be so um early on um 2.4 let's see um what is the ph um 2.87 okay so that so the ph is given to be 2.87 okay so um we'll write it down over here so this would be 2.87 uh degree of dissociation was found to be um let's see uh one point three four percent okay uh or you can put it at zero point zero one three four or one point three four percent okay so this is one point three four percent okay then um you are supposed to uh number one um find the ph okay and uh be the degree of dissociation atonoid acid um of a lower concentration uh which is 0.001 moles per mq how have the ph and degree of dissociation affected by dilution okay so uh this is a dilution of about a hundred times okay so uh if you if we try to work it out over here so uh maybe i'll leave a bit of space um so again um ethanoic acid so h a um equilibrate to give us a h plus plus uh okay maybe i should uh put the actual one so that um i mean the the more correct form not the actual the more correct form um from what you have learned for what you're playing h2 so that um there's some consistency so we have an i stable over here okay so uh again the ka value so what is the ka value of ethanoic acids uh 1.8 times 10 to power minus 5. okay so we'll write that in so the ka is um the ka is equals to 1.8 uh times 10 to the power -5 then um one two three so we have zero point zero zero one then dash dash dash um and then um minus uh y uh 0.001 minus y and then this is plus y plus y and then y y okay then uh we have a k expression over here so k a is equals to um y squared over 0.001 minus y yeah again um we'll make the assumption that um the extent of uh ionization is much lesser than 0.001 so from here 1.8 times 10 to the power minus 5 will be multiplied by 0.001 okay and then this will be equals to y squared okay and then we'll take the calculator and punch the value 1.8 times 10 to the power minus 5 multiplied by 10 to the power minus 3 and then we take the square root okay so we'll get y to be 1.342 times 10 to the power of one two three four ten to the power minus four uh moles per diem cube and then from here we can calculate the ph okay so we'll try to determine the ph minus log uh base 10 of this will get 3.87 okay so this uh 3.87 to be uh for the ph then um what we're going to do now is we're going to determine alpha okay so alpha of course alpha is simply a y which is a 1.342 times 10 to the power of minus 4 divided by the initial amounts which is uh 0.001 0.001 and then multiply by 100 i think there's no more space so you'll get what i mean so therefore i'm just going to punch the calculator and determine the value minus 4 divided by 0.001 times 100 so i'm going to get 13.4 wow 13.4 percent okay so you realize that um as i perform a hundred times dilution okay so why is it hundred times if you look at the initial concentration it's given to be 0.1 over here okay so uh i'm doing a hundred times dilution and then and then you'll notice that the ph goes from 2.87 to 3.87 so it increases but the extent of ionization actually increases by 10 times so we get 13.4 percent okay so the more dilute the solution is you expect the extent of ionization uh to be greater okay so uh in this case uh which is a a better indication of um strength of the acids okay because the ph and degree of ionization actually change with concentration okay i don't know whether you'll notice that earlier on uh we have a question of course uh i mentioned that it's open for discussion but of course from this part uh it seems to give us an indication of which is a more reliable indicator yeah of course uh whenever we want to compare across the different assets uh i think what is fairer would be we need to ensure that their concentration are the same before we can discuss because if the concentration is different then uh i think the the discussion will not make a lot of sense okay so that's something which you might want to be aware of okay so therefore you could use ph you could use extent of dissociation to help you [Music] decide on uh which is a stronger a weaker acid but that's provided that the initial concentration for all the acids which you want to compare with or all this weak base which you want to compare with are similar okay maybe i wouldn't say the same but similar okay then the diagram over here will show you how the ph alpha and ka change of course uh ka doesn't change as you know because k is a temperature dependent variable ph will approach 7 okay because as the acid become more and more diluted the amounts of cage plus contribution become lesser and lesser so it will approach the auto ionization of water and then the extent of ionization will also increase uh approaches one so for a very weak acid you can tell that um uh its extent of ionization is close to maybe hundred percent yeah so therefore um maybe we did mention a little bit about this but uh ka would be the best indicator of the strength because uh ka tells us uh the extent um where the position or equilibrium is lie towards okay then um we have a exercise 2.9 to end of section two okay so each of the following reactions contain two bronsted acids and two bronzer bases such as brief reasoning which is a stronger acid okay so from here you get a kc value that is very very low so that means that the uh position of equilibrium lies to the left so if the position of equilibrium lies to the left that means that o h minus is a stronger base as compared to ammonia and that also means that um the conjugate acids uh in this case are ammonium is a stronger acid as compared to water okay so from the position of equilibrium whether it lies to the left or right you can tell whether uh which between between the two pairs i mean between the pair uh from the reactant and the products you can tell which is a stronger base and which is a stronger acid okay and then of course for b uh we have this particular species you probably will not be familiar with it as of now but um i can draw it out for you so it's actually phenoxide so a phenol ring for o minus um reacting with ethanoic acids okay and um in the cross erection i get uh phenol so phenol looks like this and then at the no weight you'll notice that the value of kc is in the magnitude of 10 to the power -5 so that means that p-o-e position equilibrium uh lies to the right so if the position of equilibrium lies to the right that is actually trying to tell you that uh phenoxide is a stronger base as compared to ethanoids and ethanoic acid is a stronger acid as compared to phenol then some of you might argue how come it can't be the other way wrong uh just imagine if it's the other way around then i'm good i'm not going to get a very high kc value i'm going to get a kc value that is much lesser than one okay so just think about it and then see whether it makes sense okay so therefore in shorts there is a a a principle which um you might want to uh be aware of in chemistry so when when you are dealing with reactions involving uh prostate acid and bases right yeah the fall reaction will tend to be uh will tend towards the formation of a weaker uh conjugate base and a weaker conjugate acid yeah there will never be a reaction where you start off with a weak base and weak acid and you end up with a strong acid and strong base yeah because if that's the case the thermal dynamics will actually drive towards the backward reaction yeah okay so if you're still confused uh think about it i think exercise 2.9 is a very good exercise for you to to think about some of these cases yeah okay yeah so uh with that we'll end off section two okay we are going to look at section three on salt hydrolysis next okay so um what exactly is the salt so similar to what you learned in secondary school if there's a reaction between an acid and a base you will produce a salt okay so in this case potassium hydroxide and hydrochloric acid reacts to give you potassium chlorate yeah and you might be aware that the reaction between potassium hydroxide and hydrochloric acid essentially is an acid-based neutralization between o h minus and h plus okay so therefore k plus and cl minus they are just what we call spectator ions swimming around the solutions they do not take parts in the reaction okay and you probably learn a little bit that if a reaction involves a strong base and a strong acid the end point of the reaction the ph is actually neutral okay so that actually tells you that um k plus and chloride they themselves they do not undergo reaction with water so they merely form ion dipole interaction with water molecules yeah and then on the other hand uh if the cations or the anions react directly with water that is when salt hydrolysis takes place yeah so this is the purpose of this particular section on salt hydrolysis yeah so we want to reinforce or we want to emphasize that many many sorts are actually a result of the production of either strong base weak acids or strong acid weak based yeah so because of that the relevant sorts either the conjugate acid or the conjugate base uh they themselves are weak based on acid themselves so they are able to undergo a reaction with water to produce oh minus that h3o plus yeah and later on you'll realize that some cat ions will also undergo hydrolysis because the cations have very high charge density yeah so the cations themselves they will be able to polarize the oh bond of h2o to a very great extent and in the process uh produces a surplus amount of h plus yeah so that aspect would be quite important in your inorganic qualitative analysis and of course transition method chemistry which will come out later on okay so in general um ions um where the salts are derived from strong acid and bases they do not undergo uh hydrolysis so the final solution will be neutral at 25 degrees celsius yeah then um the sorts of weak acids and bases uh as well as some of the cations with high charge density as i mentioned earlier on they will undergo substantial hydrolysis uh to produce uh solutions uh with ph greater than or less than seven okay so for the next section uh i think we often talk about uh ethanoic acids so again we are just going to talk about ethanoic acid uh salts now so um sodium atomic is the salt formed from the wheat acid ethanol acid and the strong base sodium hydroxide yeah and ethanolic acid being a weak acid so its corresponding conjugate base uh itself will be a weak base but as a weak base it is a stronger base than water so in that case it will undergo partial hydrolysis based on this particular equation and in the course of this partial hydrolysis it's able to produce a small amount excess o h minus and this more about excess oh minus will result in the alkalinity of the solution okay so therefore uh even when let's say you perform an acid-base titration with sodium hydroxide and ethanoic acid um although ethanoic acid may be completely neutralized the final solution is still alkaline yeah so the alkaline nature of the solution is due to the fact that atomic undergoes partial hydrolysis in water given by this equation producing a subplot about oh minus okay sodium plus itself does not undergo hydrolysis because of its low charge density yeah so for sodium atomic i think it's quite clear cut the solution uh i mean if you if you buy sodium atomic itself the salt um and you dissolve in water the solution will be alkaline okay so uh for a strong acid and weak base is also the same thing so uh that applies to many ammonium salts because ammonium salts are produced from ammonia and strong acids yeah so um ammonium salts are acidic in nature because um ammonium cation is a stronger acid than water although it's a weak acid but it is a stronger acid than water so strong and weight are relative so that is certain these are concepts which you need to come to terms with when you are dealing with uh acid-base equilibrium okay so um when i pass you or when you dissolve ammonium chloride in water you are going to get an acidic solution okay so uh that is because the ammonium cation will undergo uh partial hydrolysis with water and in the process you generate a surplus amount h3o plus okay however um if i have a salt derived from a weak as in a wheat base such as ammonium adenoid yeah because eternoid ionizes or hydrolyzes ammonium also ionizes or hydrolyzes you don't know what is the outcome of the final ph yeah so the final ph of a solution depends on how strong the acid is as compared to that of the base so if ammonium let's say if the okay so in this case i think uh we have all the three cases here if the k of the cation in this case ammonium is uh bigger than the kb of the anion then the overall solution will be acidic because now um the cation is able to ionize to a great extent or slightly greater extent uh if it's the same then it's neutral and then the other way around if the anion is a stronger base then solution will be basic okay next we are going to move on to sorts containing aqueous metal cation with high charge density so this is an important portion for transition metal chemistry as well as your qualitative analysis so you probably want to be aware that some of the cations with high charge density such as al3 plus and off the top of my head would be chromium three plus okay uh and in three plus yeah so these cations um they will partially ionize or partially hydrolyzed in water and in the process uh they will generate h3o plus yeah so if you are interested you can go to your data booklets uh you you can just roughly look at the data of course the data booklet doesn't tell you which cation will hydrolyze but you can use that to kind of like make a prediction remember it's high charge density so it's not just the um cationic radius so um if you if you remember what i said um it would be l3 plus so uh the the radius the cationic radius is about 0.050 and then uh we're going to move on to the transition metal remember i said iron3 plus yeah so that's also about 0.055 and i think i mentioned um what is the other one chromium three plus yeah so yeah so these are usually cations with plus three charged and they have a cationic radius that is probably around 0.06 or less okay so you'll notice that for example iron3 plus with a radius of 0.055 nanometers um when you dissolve it in water the solution is actually acidic but iron2 plus the solution is not very acidic in fact it's very close to neutrality ph yeah so the charge density again uh if you recall from what you learned in chemical bonding is dependent on the charge uh over the chi on it radius okay and then we go back to this uh over here basically they are just trying to explain to you why it undergoes hydrolysis okay so this is something you'll probably haven't learned in h2 unless you have red ahead and look at transition method so if we have a l3 plus most of the cations um except for group one chaos they tend to be surrounded by water molecules so um it's not always six water molecules but at least for h2 most of the cases you maps will be six water molecules okay and because of the high charge density of the al3 plus center right it will polarize the old h bond of every single h2o molecules yeah so over here we just show you one h2o but uh truth to be told all the h2oh will be polarized and because of that the h tends to be lose quite easily to a h2o yeah so uh because of that then um usually um for quite a number of ar3 plus with six water molecules as the ligand we call this the ligand rate they will tend to lose the h plus to h2o to produce um a surplus amount of h3o plus uh even when you simply just dissolve in water yeah and uh that is why um the sorts of these uh cations pens i mean the salt of these cations tends to be acidic in nature because of the excess among h3o plus okay so this is basically just a schematic i mean to show you uh what is happening because most of you have learned organic chemistry so uh when you look at this particular diagram you can probably appreciate that the lone pairs on oxygen over here uh will undergo an acid-base neutralization uh and grab on the h and then um this particular oh group will become o h minus yeah so uh as the reaction proceeds um the charge reduces from plus three to plus two okay yeah uh you will encounter more of this in your in the qualitative analysis part of your practicals yeah so not to worry too much okay and then uh yeah so these are the cations which we which i mentioned earlier on as well like chromium three plus and i am three plus these are probably the few which um uh they are a lot more relevant for h2 okay so i think it would be good for you to know uh these three ah there are a lot others but you don't usually encounter them even if you are required to to use them then there will be information given to you okay so uh with that let us look at exercise 3.1 you are asked to state whether the following solutions are acidic alkaline or neutral okay so acidic alkali or neutral and then supposed to write equations including state symbols for any hydrolysis reaction okay so for a you have potassium bromides okay so for potassium bromides i think you can guess that it is the sorts of potassium hydroxides and hydrobromic acids yeah so uh therefore it's the sort of a strong acid and strong base i expect it to be neutral so there will there won't be any hydrolysis okay and then um sodium carbonates so for sodium carbonates um of course sodium itself it will not undergo substantial hydrolysis as what we mentioned earlier on because it's a group one cation yeah but um carbonates uh you might want to be aware that carbonate is actually the conjugate base of a weak acid yeah what we call a bicarbonate hco3 minus yeah and bicarbonate is the conjugate base of another weak acid known as carbonic acid which essentially is not very stable in water yeah so uh this will tend to be um i mean it tends to form co2 and diffuse away yeah so that is why in secondary school if you remember uh carbonates uh usually red acid to give you co2 and that is the reason why because carbonate is a base so you wrap h plus to produce carbonic acid so react with two h plus to produce carbonic acid and along the way um the equilibrium will shift towards the production of co2 and h2o so that is why you observe effervescence okay so therefore carbonate is actually a weak base so um since it's a weak base uh we will need to write an equation for a hydrolysis reaction that is taking place yup so i'm just going to use a purple ink so we'll have carbonates reacting with h2o and then equilibrates so it hydrolyzes to give you oh minus and hco3 minus now remember when you are writing hydrolysis reaction right uh you are just supposed to increase uh the it for in this case since you are writing for a base so you are supposed to increase the h of the base by one do not go all the way to h2co3 okay so remember this yeah uh because there's no logic if it goes all the way uh i think you have to think about it uh why is that the case yeah so unless uh i i spam or i add a lot of h plus in if not you will not go all the way to co2 okay then um for c i think um it's probably an example which you are a little bit familiar with uh chloride as you know is the conjugate base of hydrochloric acid so therefore it will not undergo hydrolysis but um this particular cation is actually the conjugate acid of methylamine yeah so it's similar to ammonia so ammonia is a base it's a weak base so methodology is also a weak base yeah so again you expect um methyl ammonium so uh this particular conjugate acids to undergo uh partial hydrolysis in water and in the cause of the reaction it gives you a little bit of h3o plus and the con i mean the conjugate based in this case uh methylamine okay then um for d we have ammonium uh hydrosulfite or hydrogen sulfide yeah so uh nah4 plus so we have um nh4 plus and then we have hs minus we are given the kb of ammonia and ka1 of h2s okay then you will be wondering um how useful are these data and then um what are they uh why are we given the kb of ammonia and the ka1 of h2s uh you are given that because uh from here right you can derive the ka of ammonium okay how do i derive the ka of ammonium so essentially the ka is simply the kw which is 10 the power minus 14 uh divided by the kb which is 1.8 times 10 to the power -5 okay so if we take a if we take a calculator and punch we should be able to get an answer out okay let me try that just a moment okay yeah so you should be able to get uh 5.6 uh times 10 to the power of minus 10 um or 5.56 maybe um we leave it as uh three assets 5.56 times the power minus 10 uh moles per mq okay then uh we can do the same um we are given ka1 of h2s so uh in case you are not very familiar so h2s is what we call a a dibasic acid so it will ionize twice so the first ionization i will be able to get h plus plus hs minus the second ionization um i will be able to get h plus plus s2 minus so this is ka2 yeah so in order to find out uh the kb of hs minus i will need to find the correspondence ka 1 which is associated with h2s okay so in this case i'm going to use the value of ka1 to find the kb of hs minus okay so uh with that then i'm going to find kb which essentially is 10 to the power minus 14 uh divided by 9 times 10 to the power minus 8 okay so um just a quick calculation um for me to get the answer okay so i i i'm getting 1.11 times 10 to the power of minus 7 moles per mq okay but what is most important is please take a look at these magnitudes the magnitude of the ka of ammonium is in the region of 10 to the power minus 10 but the kb of hs minus is a region of 10 to the power minus 7 yeah so therefore uh hs minus is a stronger base than ammonium s and acids and because of that i expect the hydrolysis of hs minus to be slightly more substantial as compared to the hydrolysis of ammonium so i expect the overall solution to be alkaline because um there should be a slight excess of o h minus s compared to h plus okay yeah so because of that solution would be alkali yeah in case i didn't mention earlier on uh this is neutral uh this will be basic or alkaline uh this will be acidic and then this will be alkaline and then ion3 chloride i think we briefly talked about that just now because uh chloride is the conjugate base of hydrochloric acid so chloride will undergo hydrolysis but iron 3 plus is the cation with a high charge density given here so it will undergo a relatively substantial hydrolysis to give me an acidic solution okay so uh because of that i'm going to write a balance equation out so um i'm going to expect um this is known as hexa a qua iron three complex okay hexa equal ion 3 complex with a h2o molecule then you undergo hydrolysis to give me um fe then h2o will reduce by one okay overall charge will reduce by one as well and then i will have h3o plus okay yup i forgot to write for um d uh so maybe you can write on your own uh so the main reaction is probably hs minus uh plus h2o uh h2o will be a liquid um and i'm going to get uh h2s uh plus o h minus so technically you should be writing for ammonium as well but uh this is probably more substantial uh yeah so there should be a slight excess of o h minus and then for 3.2 um for 3.2 okay so before that sorry before that um we need to look at the ph of stock solutions okay so again we need to consider the strength of the acid and base and then write an equation and then see which one produces a surplus amount of h3o plus and oh minus and then after that calculate the concentration of of h3o plus or h minus so i think um it works best with an example yeah so in this case we have um sodium ethanoids so again sodium plus does not undergo uh substantial hydrolysis so we will ignore sodium plus okay so um we have um atonoids so ch3 uh ch3 co2 minus uh equals plus h2o uh equilibrate to give me um adenoid acids uh plus o h minus okay and then um i stable okay given the concentration to be 0.10 and then um dash dash dash and then uh minus x plus x plus x so 0.10 minus x x x i am defining kb over here i'm not given kb i'm given ka so in this case from ka i need to find kb so that will be uh 10 to the power minus 14 uh divided by 1.8 times 10 to the power minus 5 okay and this will give me 5.56 times 10 to the power of minus 10 moles per dam cube okay so since uh i have already written a kb expression that would be x square over 0.10 minus x we assume x to be much smaller than 0.10 remember this is an assumption assumed okay so therefore kb will be 5.56 times 10 to power minus 10 multiplied by 0.10 equals to x so that will be square roots so x will be equals to 7.45 times 10 to the power minus 6 moles per diem cube okay so from here we can calculate poh minus log base 10 7.45 times 10 power minus 6. i will be able to get 5.13 so uh ph will then be equals to 14 um for 10 uh minus 5.13 i will be able to get 8.87 yeah so the overall ph is above 7. okay so um we are going to do the same thing for ammonium chloride okay so uh ammonium uh chloride is the conjugate sorry chloride is a conjugate base of a strong acid so it will not undergo substantial hydrolysis so uh ammonium will be the one which will undergo hydrolysis so i'm going to get ammonia aqueous plus h3o plus aqueous okay so again i stable um ammonium concentration is given to be 0.450 dash dash dash okay so uh we let it be y so minus y plus y plus y and then equilibrium concentration of ammonia will be 0.450 minus y and then y y okay kb of ammonia is given to be 1.78 times 10 to the power -5 but for this particular equation we are defining ka because we are looking at the hydrolysis of the conjugate acids okay so we need to find ka so therefore ka is equals to 10 to the power minus 14 uh divided by 1.78 times 10 to the power -5 and from here we are going to get um a value let me just calculate to see what i'm going to get 10 to the power minus 14 divided by 1.78 times 10 to the power of minus 5 okay so i'm going to get uh 5.62 times 10 to the power minus 10 moles per diem cube okay and from here i'm going to substitute this in write a k expression so k a will then be equals to y squared over 0.450 uh minus y and because we assume that y is we assume that y is much lesser than 0.450 uh therefore i'm going to get 5.6 times 10 to the power of minus 10 equals to y squared over 0.450 okay and then i shift the things around so y is equals to square roots of 5.62 5.62 times 10 to the power -10 times 0.450 okay so from here i'm going to get 1.5959 times 10 to the power of minus 5 moles per diem cube okay i'm going to calculate our ph which is minus log base 10 of 1.59 times 10 to the power -5 and i will be able to get 4.80 which makes sense because this is the conjugate acid of a wheat base so you undergo a substantial hydrolysis or sulfur hydrolysis in water and i'm going to end up with a acidic solution okay so with that we are done with section three and do watch out for the next video for section four onwards which is on buffer solutions