CIE Topic 25 Equilibria REVISION

Summary

Highlights

• Chris explains equilibrium in chemical reactions: Le Chatelier's principles FTW! ⚖️
• Dive deep into acid-base reactions and how conjugate acids/bases form. ⚗️
• Get to grips with the pH scale, from acids to bases, and the math behind it. 📊
• Buffers are discussed in detail; they're key to maintaining stability in solutions. 🧪
• Learn about solubility product and what makes substances dissolve. 🌊

Key Takeaways

• Understanding equilibrium concepts is key to mastering chemistry. ⚖️
• The pH scale tells you everything about acidity and basicity. 🔬
• Buffers are real lifesavers—literally in your bloodstream! 🩸
• The common ion effect is brilliantly simple yet incredibly effective in reactions. ⚗️
• The partition coefficient shines a light on solubility in dual solvents. 🌊

Overview

Chapters

• 00:00 - 01:00: Introduction The introduction begins with Chris Harris from Allen Chemistry welcoming viewers to a video on equilibria, which is topic 25 in the Cambridge Internationals specification.
• 01:00 - 07:00: Overview of Equilibria The chapter titled 'Overview of Equilibria' seems to introduce the available resources for students studying the CIE chemistry syllabus, specifically for both year one and year two. It mentions a YouTube channel named 'Alloy Chemistry' which offers a complete set of videos targeted at helping students with revision. The channel encourages support through subscriptions and mentions that PowerPoint slides used in the videos are also available for purchase as additional study material.
• 07:00 - 16:00: Acids and Bases The chapter titled "Acids and Bases" begins by mentioning the availability of supplementary revision materials that students can purchase to support their studying. These notes are noted as being valuable for money and useful for marking up with additional personal notes. The chapter acknowledges the broadness of the topic, indicating that it encompasses a substantial amount of content and various options for exploration. The initial introduction suggests a detailed and comprehensive examination of the subject matter.
• 16:00 - 27:40: Calculating pH of Acids and Bases The chapter 'Calculating pH of Acids and Bases' discusses the interconnection between various CIE topics, specifically focusing on equilibrium. It highlights that the equilibrium topic in the second year builds on the understanding developed in the first year. For readers unfamiliar with equilibrium concepts, it is advised to review the year one material on equilibria.
• 27:40 - 36:40: Dissociation Constants and Weak Acids The chapter "Dissociation Constants and Weak Acids" begins by assuming some prior knowledge from the reader, specifically on the topic of equilibrium and reversible reactions. Key principles such as Le Chatelier's principle, which are typically covered in year one chemistry, are mentioned as foundational concepts. The discussion prepares to delve into conjugate pairs, but emphasizes a brief review of equilibrium concepts, highlighting the role of catalysts in these reactions.
• 36:40 - 59:00: Buffers and Their Calculations This chapter delves into buffers and their calculations, exploring their relevance in the context of equilibria influenced by changes in concentration, pressure, and temperature. It discusses the role of conjugate acid-base pairs, which are linked through the transfer of protons. The fundamental concept that acids are proton donors while bases are proton acceptors is emphasized, connecting the discussion to wider principles of acid-base chemistry.
• 59:00 - 69:00: Solubility Product and Calculation The chapter discusses the concept of solubility product and how to calculate it. It refers to principles from year one related to Bronsted-Lowry acids and bases, emphasizing the process of proton transfer between acid and base. It reviews that an acid donates a proton to a base, forming the conjugate acid (BH+) and conjugate base (A-). The generic equation for this reaction is presented and explained in the context of solubility calculations.
• 69:00 - 77:00: Partition Coefficient This chapter explains the concept of partition coefficient and its relation to acids and bases. It focuses on the behavior of acids (HA) in donating protons in the forward direction, and bases (A-) in accepting protons in the reverse direction. The chapter further clarifies the conjugate acid-base pairs, identifying HA as the conjugate acid and A- as the base in a particular reaction scenario.
• 77:00 - 79:00: Conclusion The chapter discusses the concept of conjugate acid-base pairs in chemistry. It explains that A- and its corresponding acid form a conjugate pair, while B and BH+ form another. BH+ is identified as the conjugate acid because it donates a proton to form B. Conversely, B is referred to as a base because it accepts protons from BH+. This process of proton donation and acceptance is crucial in identifying the roles of acids and bases in chemical reactions.

CIE Topic 25 Equilibria REVISION Transcription

• 00:00 - 00:30 [Music] hello my name is chris harris and i'm from allen chemistry and welcome to this video on equilibria so this is topic 25 for the cie that's the cambridge internationals specification so if you are studying the cambridge
• 00:30 - 01:00 international's um syllabus or the qualification then this is the perfect place to get your targeted revision material um i have got the full range of year one and year two videos for the cie specification all of it's on alloy chemistry youtube channel please hit the subscribe button to show your support for this project that'll be massively appreciated um all of these um videos as well there are actually slides powerpoint slides that can be purchased from my test shop if you click on the link in the
• 01:00 - 01:30 description box below you can get a hold of them there they're great value for money and can be used to kind of supplement your revision material as well so your notes about you can use them for your notes if you wish you can write notes against them so um yeah so they're available there go and have a look there for a little bit more information right so let's make a start on this um so obviously this is this is a quite a big topic actually um it's fairly big it's a lot of content i think there's a you know there's a lot of different options here so
• 01:30 - 02:00 obviously you'll find that the um all the cie topics they kind of merge together um a little bit there's a lot of overlap with some of these topics obviously this is a year two topic so um you'll find that there's an equilibrium topic for year one and obviously equilibrium topic for year two kind of builds what you've learned in the year one topic if you're not sure and anything um to do with equilibria i strongly suggest that you go and have a look at the year one video first and also called equilibria and before we
• 02:00 - 02:30 look at this one because i'm going to make some assumptions that you know some of that stuff as well okay right so let's just kind of pick up from there then and let's have a look at obviously just kind of briefly before we go into conjugate pairs um looking at equilibrium obviously equilibrium has to do with reversible reactions so there's information such as leicester tillia's principle which you will have seen from year one and the reversible general reversible reactions and the impact of catalysts and um
• 02:30 - 03:00 concentration and pressure and temperature what impact if any it has on equilibrius there's gonna be a lot of this kind of brought in here but just advancing it and one of the ones was about acids and bases and looking at conjugate pairs so conjugate pairs is basically um linked by the transferring of a proton this is related to acids and bases as i mentioned so any species that has gained a proton okay remember acids are proton donors and bases or proton acceptors so
• 03:00 - 03:30 any species which is gained a proton is the conjugate acid and a species that has lost a proton is a conjugate base so legacy you've seen this in year one about bronsted-lowry acids and bases so i've written a generic equation here showing an acid in the base so here is your acid as you can see there b is your base and you've got bh plus and a minus remember this acid will donate a proton to b to form bh plus so
• 03:30 - 04:00 that makes nasa's and obviously the base received the proton to make that class as a base so h a is an acid in the forward direction like i said donates the proton and a minus is a base in the reverse direction um as it accepts a proton from bh plus to form h a so we definitely that's a base now the conjugate pairs are is basically h a and a minus okay so the conjugate acid is obviously h a in
• 04:00 - 04:30 this case and the conjugate base is a minus for these two so these are what we call a conjugate pair we also have another conjugate pair here which is the b and the bh plus so bh plus would be the conjugate acid because it would have to donate this proton to b or well effectively the proton would have to be released from that to form b so that is definitely an acid and obviously b in this case would accept the protons from bh plus so it's definitely a base
• 04:30 - 05:00 so this these are basically conjugates and that's what we cast you might see that word kind of dotted around so obviously water reacts with acids to form h3 or plus which is your conjugate acid okay because that's going to donate the proton and it reacts with the base to form a conjugate base which is o h minus so you can see here we've got um some reactions here just as an example so you've got your base which is aqueous water and that'll form bh plus an oh
• 05:00 - 05:30 minus and obviously the acid reaction with the water as well so it's just showing you the impact of a base and an acid in terms of accepting a proton and donating a proton so what you need to be aware of is what the conjugate bit is so where you pair them up and make sure you're aware of it a lot of this stuff here like say the acid-base reactions you know proton donor proton acceptance etc is all stuff that you've done year one make sure you look back at that if you're not showing what this is um obviously we're just going through um
• 05:30 - 06:00 some of the stuff for the year two content here okay so still kind of sticking with the acid base reactions because a lot of these reactions are equilibrium so that's why they kind of fit into this topic so when they react with each other obviously protons are exchanged as we've seen before so we can see in this generic example the h a donates the proton um to b and obviously producing your positive and negative ions so in terms of equilibrium so remember this this is obviously an equilibrium reaction and if
• 06:00 - 06:30 we add more h a or b so if we add more of this then equilibrium will shift to the right to use it up and vice versa if we add more of this it could have been shifted to the left to um to use that up so this is obviously based around licia tilly's principle and obviously that was done in year one so obviously the water um in this example here behaves as a base okay when um acid is added to it and the reason
• 06:30 - 07:00 why is because the water molecule is accepted a proton to form h3 or plus this is also known as a hydroxonium ion or hydronium and it's got two words or two different names should i say so strong acids um anything that's a strong acid equilibrium lies well over to the right so you're producing loads of h plus ions um weak acids don't lie too much over to the right they kind of lie more to the left so they're weak acids so this was too for example when
• 07:00 - 07:30 you looked at organic chemistry in year one you would have seen that carboxylic acids are weak acids the equilibrium shifts right to the left or lies while well to the left so you don't produce many of these ions here now to simplify this a little bit further you'll see this is obviously the the hydroxone mine is actually what is produced this is what makes something acidic um but for simplistic purposes we can just kind of cut the kind of h2o bit out of this and just put h plus you'll see it's kind of written in various ways h plus is fine and because that's the
• 07:30 - 08:00 ion that's obviously the important bit that makes something acidic in reality though hydroxonium is the is the true molecule that makes something acidic so just in case you kind of see some kind of confusion around that right we're going to look at um obviously looking at acids and bases here and looking at equilibrium so the natural thing to kind of move on to is ph because obviously that's strongly linked with with acids and bases so there's quite a bit of maths in this
• 08:00 - 08:30 topic here there's a lot of equations and different formulas there generally is anyway with chemistry but obviously this is physical chemistry so you're going to have a lot of um you know formulas in this but ph is basically a logarithmic scale that measures the concentration of h plus ions in solution okay so um you those you have done um maths again if you do maths you've probably heard of a logarithmic scale and if you haven't if you don't study maths for example um then don't worry there's a
• 08:30 - 09:00 calculator thankfully that will kind of massively help you here as you'll see in a moment um but the ph scale ranges from zero to fourteen zero being very acidic seven b neutral fourteen obviously being very basic you'll you'll know this already the equation though is the important bit so ph is minus log to the base ten of multiplied by the concentration of h plus ions so ph can be calculated when we know the
• 09:00 - 09:30 concentration of h plus ions that exist in solution so let's look at an example here we're going to calculate the ph of 0.03 moles per dm cubed of hydrochloric acid so here we are hydrochloric acid is a type of acid obviously it's a strong acid so our assumption is that it dissociates fully basically all of the hcl molecules dissociate to produce h plus signs and cl minus signs and so this means that the concentration of hcl
• 09:30 - 10:00 molecules in the first place equals the concentration of h plus ions okay so we assume that they are the same so we can use the equation um to work out the ph and the equation is ph equals minus log of h plus and the concentration of h plus is 0.03 so we got a ph of 1.52 so the concentration of hydrogen ions can also be calculated when we know the
• 10:00 - 10:30 ph as well so we can kind of go the opposite way so let's have a look so let's calculate the concentration of hydrogen or calculate the concentration of hydrogen ions of nitric acid with a ph of 1.7 so we're kind of working backwards here so to help i'm going to put a picture of a calculator there we are okay so normally when you're looking at log so say if you've got ph it's minus log h plus you would hit obviously minus and
• 10:30 - 11:00 then hit that button where it says log and then just put in the concentration of h plus but when we want to work out the um concentration of hydrogen ions we have to kind of well we have to rearrange the equation so the concentration of h plus signs is basically the um it's at the inverse log of so it's 10 to the minus ph so that's what that 10 means now in terms of what you hit in the calculators you hit the shift button at the top you hit the log button that
• 11:00 - 11:30 will give you the 10 bit and then this box here will start to flash and then what you do put what you put in there is minus the ph they've given you which is 1.7 now you should get 0.020 moles per dm cubed when you do that so this is the way in which we use the ph equation to work out the concentration of ions or we can work out the ph okay so obviously we've looked at how we work out the ph of acids let's just look
• 11:30 - 12:00 at some other types of acids so far we've looked at um obviously hcl which is what we call a monoaprotic acid in other words one molecule of hdl will produce one h plus ion and some acids can donate more than one proton though and they're called polyprotect uh polyprotic or poly basic depending on what we're looking at so here like say nitric acid is a monoprotic um or monobasic um substance depending on what it's obviously reacting with so here one mole of hno3
• 12:00 - 12:30 will produce one mole of h plus ions sulfuric acid is a diprotic or die basic um molecule so this is one mole of sulfuric acid will basically produce two moles of h plus ions because obviously it's h2so4 and phosphoric acid is triprotic or tribasic depending on what it's reacting with so one mole of phosphoric acid will produce three moles of h plus ions so
• 12:30 - 13:00 just be aware of these because obviously the concentrations of some of these these are all what classes strong acids but the concentration of the actual molecule will not be especially for dye and triprotic will not be the same as the concentration of the number of hydrogen ions they produce because one of these can actually produce three times the amount of h plus ions um as the number of molecules of this so it make sure you're aware of that when you do your calculations well we'll go through some examples later
• 13:00 - 13:30 okay so let's look at the calculation of the ph of some of these strong acids that we've seen there now when we calculate the ph of strong acids we've got to make sure that they dissociate fully that's the assumption that we make okay when we're looking at these so that's obviously that's quite important so for monoprotic acids obviously hydrochloric acid is an example nitric acid is an example of a monoprotic now these will dissociate to produce one h plus sign for every
• 13:30 - 14:00 acid molecule and so this means actually we can just say that the concentration of the acid is going to be the same as the concentration of h plus ions so for example if we're looking for the ph of not points say the ph of 0.25 moles per dam cubed of hydrochloric acid and then we can basically make the assumption that the concentration of h plus signs equals the concentration of acid and then we can put that into the ph equation and get a ph of 0.6
• 14:00 - 14:30 so diprotic acid so an example would be sulfuric acid so h2so4 so these produce two h plus ions for every acid molecule okay so this means that the concentration of the acid will equal two times the concentration of the h plus ions okay so we've got two lots of h plus ions so that's the ratio that we're looking at here so for example the
• 14:30 - 15:00 ph of 0.25 moles of gm cubed of sulfuric acid basically the concentration of the acid is equal to obviously because the one acid molecule of a diprotic acid will produce two um h plus signs so that's why i've got the two there so in other words 0.25 moles per dm cubed of acid will produce twice as many h plus signs as a concentration so it'll produce 0.5 moles of gm cubed of h plus ions so when we're working out the ph of a diprotic acid
• 15:00 - 15:30 it's really important that we're quoting the concentration of h plus ions not just the acid so here obviously we've got twice as many h plus signs per acid molecule so it's going to be 0.5 that goes in there and obviously that gives you a ph of 0.3 there really just be really really careful with this a lot of the time some people look at that and just say oh it equals the same put the number in and they'll get it wrong so just make sure you're taking into account that obviously your triprotics as well so
• 15:30 - 16:00 phosphoric acid obviously it's multiplied by three instead so similar i think you get the idea so it's a similar similar process okay so let's have a look at calculating the ph of strong bases here so obviously similar to strong acids when you're calculating the ph of strong braces bases make sure you assume they dissociate fully so let's have a look at the example or we assume sorry to dissociate fully so here's an example of a strong base so
• 16:00 - 16:30 your for example sodium hydroxide that will dissociate to form n a plus and oh minus ions and most strong bases they actually dissociate to produce one o h minus sign for every base molecules for every one of these you get an o h minus ion and this means obviously the concentration of base equals the concentration of oh minus signs very similar to what we're saying with an acid now to calculate the ph of a base we still need the h plus signs we still need to constant calculate the
• 16:30 - 17:00 concentration of h plus signs and to get this we need to use a different equation called an ionic product of water expression so kw so kw is the concentration of h plus multiplied by the concentration of h minus ions so to work out the concentration of h plus which is this we need to know what kw is and we need to know what the concentration of which minus ions are now at a specific temperature so make sure obviously the temperature will be
• 17:00 - 17:30 given the value can change depending on the temperature so once we know the h plus value we can obviously work out ph using the equation that we've seen before so for example let's look at calculating the ph of 0.15 moles per gm cubed of sodium hydroxide solution at 298 kelvin so kw is 1 times by 10 to the minus 14 so that's fixed at room temperature so 298 kelvin so that makes it a bit easier
• 17:30 - 18:00 so what we have to do is substitute the figures into the kw expression so we've got um 1 times 10 to the minus 14 equals concentration of h plus times by 0.15 and then secondly we rearrange the expression to find the concentration of h plus so h plus sub c 1 times 10 minus 14 divided by 0.15 so we've just shifted that around to get h plus the subject
• 18:00 - 18:30 and then that tells us the concentration of h plus now we've got that we then need to put that into the ph equation to work out what the ph is so ph is minus log concentration of h plus ions and that gives a ph of 13.18 check this value obviously we're talking about sodium hydroxide it's a base so we'd expect it to be close to 14 in which case it is if you're getting a figure that's suggesting two or one something's gone wrong because we know
• 18:30 - 19:00 it's a base so this ph has got to be um towards 14 okay because that's obviously the on the ph scale okay so we've looked at strong acids and we've looked at strong bases and we looked at the formulas we can use for that but we're going to introduce a um a different constant here which is the acid dissociation constant ka now ka is used when we don't have a strong acid so in other words we're going to use a weak acid instead
• 19:00 - 19:30 there we are and now weak acids only dissociate slightly so remember we said that start strong acids will dissociate fully weak acids dissociate slightly so in aqueous solution we have to use this different constant to work out their ph values so for strong acids like you say we could assume that the concentration of h plus especially for monoprotic acids of course and we can assume that the concentration h plus ions equals the concentration of acid for strong acids
• 19:30 - 20:00 and we can't do this for weak acids so we use ka now weak acids exist in this equilibrium so very similar to strong acids except that one assumption is obviously that only a small amount of the weak acid here actually dissociates so in a strong acid all of this we assume all of this breaks down to form h plus n and minus but here we can't make that assumption so we can say that the concentration of h a at equilibrium is approximately the
• 20:00 - 20:30 same as the concentration of h here at the start of the reaction so right at the start when we had 100 of this is roughly about the same as what it is at equilibrium and obviously the equilibrium law can be applied here as we have an equilibrium reaction um so we can use the k expression to represent the reaction as well so there we are so ka equals obviously this is the units of this is moles decimates cubed but basically it's products divided by
• 20:30 - 21:00 reactants so it's the concentration of h plus concentration of a minus divided by the concentration of h a at the start remember we've assumed this is roughly the same as what it is at equilibrium which is basically what we're measuring so the second assumption we make is that the dissociation of the acid here is greater than the dissociation of water present in solution so we can assume that any h plus signs that are knocking around in solution have actually come from the acid and not from the water
• 21:00 - 21:30 that's actually been dissolved in so we can say that the concentration of h plus signs here are about the same as the concentration of a minus because obviously to get this you need to have this and vice versa so that's the other assumption that we make with um with this react with this obviously expression here so what we can do is simplify this k expression and put it to this so we'd say ka equals concentration of h plus squared divided
• 21:30 - 22:00 by the concentration of h a which is your acid okay so we can use the k expression to calculate the ph of a weak acid so let's have a look through this so let's calculate the ph of 0.03 moles per diem cubed of ethanoic acid at 298 kelvin so the ka for ethanoic acid is 298 kelvin is 1.76 times by 10 to the
• 22:00 - 22:30 minus 5 moles per decimal cubed so the first thing we need to do is write down our k expression so ka equals h plus squared because we assume that h plus and a minus is equal and divided by the concentration of ethanoic acids because that's your weak acid so we need to rearrange the equation to calculate um h plus squared because that's what we want to work out remember if we're not the ph we need to know how much of these the h plus we've got
• 22:30 - 23:00 so h plus squared equals ka times the concentration of ethanoic acid we put the figures in and we get the concentration of h plus squared to be 5.28 times by 10 to the minus seven now we need to work out what the concentration is of just h plus signs so we must square root this value here okay to get the concentration of h plus that gives us seven point two seven times by ten to the minus four
• 23:00 - 23:30 moles per decimeters cubed and then finally we can calculate the ph so ph is the minus log of h plus and we get a ph of 3.14 again check the answer this is a weak acid so we'd expect it to be around about three or four somewhere around there and this is exactly what we've got so that's that looks fine to me okay so we can use this k expression to
• 23:30 - 24:00 calculate the ka or concentration of a weak acid as well so let's have a look at this example so we're going to calculate the concentration in moles per decimal cubed of methanoic acid at 298 kelvin with a ph of 3.14 and the ka for methanolic acid at 298 kelvin is 1.77 times by 10 to the minus 4 moles per decimeters cubed so here we're going to kind of we're just calculating something different to what
• 24:00 - 24:30 we've mentioned before so the first thing we need to do is calculate the concentration of h plus signs using the ph equation so ph equals minus log to the base 10 of h plus ions rearrange that to get this remember that's the shift and log button to get that so i'm going to put in 10 to the minus 3.14 because that's what we've been given and that gives us a concentration of 7.24 times by 10 to the minus 4 concentration of h plus ions
• 24:30 - 25:00 so write down the k expression there it is there okay so we know this bit and actually we know this bit as well because we've been told what that is there that's the strength so we're going to put all this into the equation we're going to rearrange it because remember wanting to work out the concentration of um methyl sorry that's the ka so we want to work out this bit here the concentration sorry about that so concentration of methanol acid so we're going to rearrange that we're going to get h squared divided by ka
• 25:00 - 25:30 there's your h plus squared that's what we've just worked out before divided by 1.77 times by 10 to the minus 4 and this gives you the concentration of methanoic acid to be 2.96 times by 10 to the minus 3 moles per decimeters cubed so calculating ka um is basically just steps one and two so you just do exactly the same and then you just substitute your numbers that you don't actually need to rearrange it at all so ka is basically
• 25:30 - 26:00 just the concentration of h plus divided by the concentration of your weak acid and that basically will tell you what the value of k is so pretty straightforward okay so um let's have a look at kw okay which is obviously the ionic product of water and we kind of looked at bits of this already but we're going to kind of go through it a little bit more detail so obviously water exists in equilibrium as well so when you look at a glass of water you don't just have
• 26:00 - 26:30 i'm sorry you don't have just have loads of water molecules in there you actually have um ions floating around in there so water dissociates into hydroxide ions and hydroxonium ions as we can see in this equation so that's normally how it exists in water now we can simplify it by doing h2o um with produce h plus signs and oh minus signs
• 26:30 - 27:00 now the equilibrium law um can be applied here um as we have an equilibrium reaction and you can see that obviously you've got product over um reactant as you can see there so that's the kc expression but water doesn't actually dissociate very well it's quite weakly dissociates actually so we don't produce that many h plus and h minus signs in comparison to h2o as i've mentioned before and water is it's really dissociated very well so we can assume that the concentration of water has a
• 27:00 - 27:30 constant value and so for that reason we can just kind of ignore that so we need another expression to represent this neat unique property and if we multiply the two constants case k c and the concentration of water we get a new constant and we call it kw and this is the ionic product of water okay and the units of this are mole squared dm to the minus six there it is okay so kw equals
• 27:30 - 28:00 concentration of h plus multiplied by the concentration of o h minus ions so this is the ionic product of water now we'd seen this before when we used it to work out the um the concentration of a base so we use the kw expression this is basically just saying where it's come from and obviously the reason why we call it water is obviously because we're looking at the equilibrium of water which produces both of these ions here so this is the expression that we've just seen
• 28:00 - 28:30 before so some important points related to the value of kw and so the value of kw is the same in the solution at a given temperature so it doesn't actually change obviously the temperature will change it but in terms of that it's always going to be fixed kw has a value of 1 times by 10 to the minus 14 mole squared dm to the minus 6. make sure you remember that it's really important and obviously like i mentioned before if the temperature changes the value of kw
• 28:30 - 29:00 changes as well so pure water um has got an equal split of h plus no h minus signs which is kind of obvious because you can't have h plus signs without which minor signs because they're both derived from the same water molecule so therefore we can assume that the concentration of h plus signs in for water is the same as the concentration of h minus ions and so when we're referring to pure water we can actually write the kw's
• 29:00 - 29:30 expression as h plus squared into something like that pkw and you might see this um you might see this as well um you know in your exam but pkw is calculated from kw and it's just calculated using the following um equations so for example pkw is the minus log of kw and so therefore k w is basically 10 to the minus of p k w now this might look familiar because
• 29:30 - 30:00 actually if you look at p h the little p effectively is not not direct link but the p is means minus log of something so this is the minus log of kw so ph is the minus log of h plus ions so that's what the little p just was you can kind of remember it like that the little p just means the minus log of something and this is just like the ph equation except we're putting kw in there instead of h
• 30:00 - 30:30 so pkw is used to display kw values on a much smaller scale so it's handy for when you're drawing a graph instead of getting a graph with huge variation in numbers putting anything in a log scale makes the kind of figures proportionate to each other so you still get the same shape graph but it means it condenses it down into numbers which are a lot more manageable and allow you to plot the results much more accurately so you know those who do maths will probably appreciate that's what um that's what a logarithmic scale does it tries to kind
• 30:30 - 31:00 of condense it down into into something which is more manageable and easier to plot okay so let's see how we can calculate pka so like say it's it's another pk is just another way of measuring the strength of an acid similar to ph and obviously the lower the value the stronger the acid there's no difference really between pk and ph in terms of what it's measuring so remember pk is the minus log of the value of ka
• 31:00 - 31:30 and so to calculate the pka value for an acid with a ka value of 7.52 times by 10 to the minus 3 again relatively straightforward for modest so pk is just the minus log of ka and so therefore the value of pk is 2.12 now to work out what k is we just do 10 to the minus of peak here so it's that shift log and then you put minus the pk
• 31:30 - 32:00 value in so let's have a look at an example so we're going to calculate the ph of 0.0250 moles per decimeters cubed of ethanolic acid at 298 kelvin which has a pka value of 4.75 at 298 kelvin sorry excuse me um so the first one um so calculate ka so the first thing is ka equals 10 to the minus pk so ka is times
• 32:00 - 32:30 is 10 so shift log and then minus 4.75 because that's what they've given you there and that gives you a k value of 1.78 times by 10 to the minus five moles per decimeters cubed and then secondly we need to calculate the value of h plus from the ka expression okay because remember we want to work out ph so ka equals the concentration of h plus squared divided by
• 32:30 - 33:00 the concentration of ethanoic acid so we need to rearrange this to get h plus squared so h plus squared again it's ka multiplied by your concentration of ethanoic acid put your numbers in there and that should give you that value remember this is squared so we need to take the square root of that to get the concentration of h plus and that gets us 6.67 times by 10 to the minus 4 moles per decimeters cubed and then finally we need to calculate
• 33:00 - 33:30 the ph so ph is the minus log of h plus signs obviously put your numbers in there and you should get a ph of 3.17 again check the value does that look at sensible well we've been told that we have ethanoic acid so that's a weak acid we'd expect it to be about ph 3 or 4 because it's weak acid so that makes sense that fits always check these numbers never just work out a number and just say well i don't really know what it means this is the ph of a weak acid so we'd expect
• 33:30 - 34:00 it to be there okay so we're going to move on to something a little bit tricky here um and this is buffers now buffers um is a chemical that resists the change in ph when small amounts of acid or base are added okay remember this is resisting they don't stop the change of ph they just resist it from changing okay so it's not preventing it from happening and there's two types of
• 34:00 - 34:30 buffers that we need to be aware of and this is acidic buffers and basic buffers and we're going to go through each one of them so let's look at acidic buffers so basically a buffer is something which effectively resists the change in ph if we add something to it so um they resist the change in ph acidic buffers resist the change in ph in order to keep the solution below ph seven and that's um by definition something which is acidic they're made from a weak acid and the
• 34:30 - 35:00 salt of its conjugate base so for example um an acidic buffer is using ethanoic acid so that's your weak acid and it's salt so it's salt would be something like sodium ethanoate this these two together in a beaker creates an acidic buffer okay you must have this you must have this kind of set up here it's got to be um it's salt in any buffer solution there's two
• 35:00 - 35:30 equilibrium equations at play okay and the two will actually coexist in the same beaker um this gets a bit complicated because obviously we're going to talk about a lot of different kind of equilibrium reactions here and hopefully i'm going to try and make this as clear as i possibly can because this can be really tricky so here we've got our weak acid here which is ethanoic acid there's the ion that it will form and h plus so this is it dissociating now we
• 35:30 - 36:00 know that weak acids dissociate weakly so we have effectively high concentration of this and a very low concentration of these two here so equilibriums line well over to the left in its own right it's salt on the other hand is obviously the size dissociates strongly if not all of it um so the equilibrium lies well over to the right so if we add this into solution it will dissociate to produce loads of this and loads of that so these
• 36:00 - 36:30 concentration comments below are going to be really really important when we're looking at trying to describe what a buffer does so what happens when we add an acid so that's h plus two this buffer well let's have a look so the h plus ions imagine these will these two will exist in tandem imagine you've got that reaction and that reaction happening in the same beaker okay you've got both of these happening so we add h plus signs um and his plus
• 36:30 - 37:00 ions will react with the ch3co or minus ions in solution okay so we know that's what will happen it'll go for this negative charge you got a little bit here but the majority of it's coming from the salt there's a high concentration of these from the salt as you can see and effectively more ch3coh is produced so the h plus will react with this there's loads of this knocking around in solution and and that will form this okay so you'll get more
• 37:00 - 37:30 of ch3cooh and so equilibrium effectively shifts to the left if we had to kind of put these side beside it's moving to the left hand side of this equation here remember these are working in tandem with each other okay another separate equations but they're happening in exactly the same beaker so we can interchange them like that so what happens when we add a base to this buffer which is a which minus well the which minus signs will react with the h plus ions in here so they'll
• 37:30 - 38:00 react with that and they could react with this but if you think about it sodium reacting with sodium ions with hydroxide ions will form sodium hydroxide which will dissociate very quickly and basically produce these two ions back again so that's not really that much good so it will react with this though and change the kind of chemistry of what you've got now the problem is there's not many of these in solution so when the o h minus signs react with the um
• 38:00 - 38:30 uh when the orange minus signs react with the h plus ions these have to reproduce so equilibrium will actually shift to the right so it will go this way so this will dissociate further to replace the h plus signs that have been lost to um obviously by the hydroxide ions and this is obviously on the base of alicia tillia's principle which you've seen in year one and so equilibrium shifts to the right to replace that
• 38:30 - 39:00 now you can see this is quite complicated there's a lot going on here you've got to be comfortable with all of this here because we're going to go into some calculations in a minute um which is going to be reliant upon you understanding this bit here okay so hopefully this should make sense if not go through it again and and make sure you've got it okay so acidic buffers can also be made from an excess weak acid and a strong base as
• 39:00 - 39:30 well so we make it in two ways so an example of an acidic buffer is using ethanoic acid so ch3coh that's a weak acid and we can react that with sodium hydroxide and basically all the of the base reacts with the acid in the following way so we've got um your weak acid there and there's your base that's your hydroxyl ions from the base and that'll form your um ethanoate ions which are here and water
• 39:30 - 40:00 now here we have an excess of ethanoic acid we've got loads of this so we'll still have some left over after all of the oh minus signs of reacted loads of this okay remember it's an excess weak acid so um in our beaker we have a mixture of weak acid so some of this we have its salt okay which is this bit here so we've got the ions from there and we've got water so effectively we have all the
• 40:00 - 40:30 kind of beginnings of a buffer and because there's still some weak acid in there it partially dissociates to form this as well so we've got this from the weak acid we have loads of this because that's been formed from the hydroxide ions and and obviously we've we've got some water in there as well so this can also creates a buffer solution as well it's just in a slightly different way so just be aware of what these acidic buffers are
• 40:30 - 41:00 okay so moving on to basic buffers so basic buffers they resist the change in ph just like acids do but they keep the solution try to keep the solution above ph seven um and they're made surprise surprise from a weak base and it's sold so let's have a look so a basic buffer um is ammonia for example ammonia which is nh3 so that's the weak base and its salt is ammonium chloride so nh4 plus cl
• 41:00 - 41:30 minus so if we mix these together we get a buffer and obviously in any buffer as we've seen before we have two equilibrium equations at play so the two equations remember they coexist in the same beak and that's how we can interchange between them so let's have a look here's your weak base and this obviously reacts with water from the ammonium ions and which minus ions because it's a weak base we have a high concentration of ammonia and we don't
• 41:30 - 42:00 have much ammonium ions or oh minus ions at all so equilibrium is lying well over to the left okay and then the salt and the salt will dissociate quite readily obviously it's dissolved in solution so you'll get high amounts of ammonium ions and you'll get high amounts of cl minus signs you won't get many ammonium chloride um molecules because these will dissociate fully so equilibrium lies well over to the right
• 42:00 - 42:30 on this occasion okay so let's go through the same concepts as what we've done with our acidic buffers so what happens when we add a base which is a which minus ions to this buffer so see if you never think where do you think the oh minus signs will react with out of all of these reactants are these products here well the o h minus signs are actually going to react with the ammonium ions in solution okay so it's the nh4 plus ions
• 42:30 - 43:00 so that's negative so that's going to go for something which is positive luckily we have a large amount of um ammonium ions we've got a high concentration of these produced predominantly obviously from the salt um and because um if we add more h minus signs these will react and use up the ammonium ions and and that will shift equilibrium and obviously we produce more nh3 and h2o is produced obviously what loads of
• 43:00 - 43:30 these these will react and we'll get produce more of these products here so equilibrium will shift to the left now what happens when we add acids to this well the h plus signs are going to go for the o h minus signs and we don't really have a large amount of these at all so what's going to happen um obviously we don't have a lot of them so they're going to have to be remade or replaced because they're being used up so equilibrium will actually shift to the right in this case okay so that moves to
• 43:30 - 44:00 the right the nh3 and h2o will be used up to replace the which miners that have been lost or reacted um to to form this okay let's look at some titration curves and buffers and now you would have seen titration curves in year one this is really looking at um the buffers the the buffer side of it first so obviously we can see evidence of buffer action in a
• 44:00 - 44:30 titration curve so if we see here remember you've got ph here and you've got a um a conical flask or something down here and then you've got a buret and then you're adding obviously acid in this case it's going to be a weak acid is going to be in the buret and then we're going to add um an a strong base into this into the solution here so obviously we add it and initially the ph changes um of the ph changes quickly as there's loads of oh minus ions from the strong
• 44:30 - 45:00 base to react with the h plus ions from the weak acid okay so that's right at the start initially it starts off pretty quickly then what happens is that over time the curve then becomes more level and what's happening here is we're creating that buffer solution okay so the buffer solution eventually we're adding a strong base okay to this weak acid and a buffer solution has been created so remember what it's doing is just trying to resist the change in ph okay
• 45:00 - 45:30 there comes a point though when actually um you start to run out of acid molecules because the acid molecules are reacting with the oh minus ions and you all keep adding them up adding them in it gets to a point where you run out of acid molecules um and it's at this point when you've used all the acid molecules up they've all reacted and then it just collapses so the o h minus signs are in excess of any acid molecule that's left
• 45:30 - 46:00 we get a very sharp rise in ph at this point and this at this point here this is called the equivalence point so this is where um effectively the buffer is broken and the concentration of acid to oh minus signs is the same so this is the equal and obviously when you add more which minor signs um obviously that then definitely makes it um a strong base so just making sure that you know that this bit here this bit is the evidence of buffer action at the start and at this
• 46:00 - 46:30 point the vertical bit of the graph this is where um obviously the buffer is eventually broken at that point okay so let's have a look at some uses of um buffers so obviously buffers have a lot of uses um that's obviously in household products or naturally occurring in living things and so a good example is blood um obviously buffers are actually everywhere to be honest um many and like say in food products as
• 46:30 - 47:00 well if you think of them normally actually they call them stabilizers if you see them on the back of say ready meals and things like that they're called stabilizers it's effectively a buffer that's what it is and it stops um your food from going off too quickly it tries to regulate the ph and make sure that it stays as fresh as it can be for as long as possible and but blood definitely has one so it's obviously vital to make sure your blood ph is maintained as close to 7.4 as possible and obviously our body
• 47:00 - 47:30 systems really need it to be 7.4 if it doesn't your cells won't function in the normal way and you you suddenly feel quite ill actually so um our body systems rely on this so a buffer is actually present in the blood and carbon dioxide plays a massive role in um kind of creating this buffer solution so we have um our carbonic acid and hydrogen carbonate system exists so you can see here we've got um your hco3 will
• 47:30 - 48:00 produce the h plus ions and hc or three minus so that dissociates quite weakly and then you've also got this um this equilibrium as well that exists here and effectively um your carbonic acid which is h2s h2so4 h2co3 is actually controlled by respiration in your cells so we breathe out um calm when we breathe out the carbon dioxide
• 48:00 - 48:30 and the level of carbonic acid reduces okay so carbonic acid remembers h2co3 so that reduces that amount there as equilibrium shifts right to try and replace some of these um um well that's their h2sco3 so when we breathe that out obviously the concentration of that reduces so this shifts right to replace some of the carbon dioxide that's been um that's been obviously breathed out so hydrogen carbonate hco3 minus and
• 48:30 - 49:00 this bit here is controlled by your kidneys and obviously a lot of the excess of this is removed and by your kidneys as well so what this is trying to do is buffer some of the changes in ph and you might think well how does your blood ph change when you think about it there's a lot of um different molecules moving in in and out of your bloodstream whether that's oxygen and whether it's carbon dioxide it could be glucose molecules you've got um you know hormones that's rushing around so
• 49:00 - 49:30 there's a lot of things kind of like being introduced in the bloodstream and removed and these things can have an impact on ph effectively um so you've got to make sure that this buffer system works quite well to ensure that your cells work in the normal way okay so this is where we're going to look at some calculations and here we're going to calculate the ph of a buffer so to calculate the ph of a buffer we need to know the ka value and the
• 49:30 - 50:00 concentration of the weak acid and its salt as well so let's have a look at this example so we're going to calculate the ph for buffer that contains 2.35 times by 10 to the minus 2 moles of methanoic acid and 1.84 times by 10 to the minus 2 moles of sodium methanoid in a one decimeter cubed solution so the value of ka at 25 degrees celsius for methanoic acid is 1.78 times by 10
• 50:00 - 50:30 to the minus 4 moles per decimeters cubed as you can see on there so let's write up the equation and the ka expression first so here's the equation so this is methanoic acid and that will produce h plus ions and your um ions your hydrogen carbonate i'm sorry your um methanol italians here your ka expression is h plus um and obviously hc or minus your methanoid ions so these are your
• 50:30 - 51:00 products divided by your reactants which are at the bottom there so in buffers um what you've got to notice here is that unlike in the previous examples where we wanted to work out the strength of an acid uh of the the strength of that sorry the ph of a weak acid we assumed that that these and these were the same and we could just put h plus squared when we're referring to buffers we cannot do that and we've got to keep these constituents separate okay so we can't use this equation here it's really
• 51:00 - 51:30 important whenever it's talking about buffers keep them separate like that okay we can't assume that there so we must use as well equilibrium concentrations here not initial concentrations okay so we've got to look at this at equilibrium once the reaction is set away it'll establish equilibrium and the values that we use here must be at equilibrium okay so we can assume that um salts um dissociate fully
• 51:30 - 52:00 okay and weak acids dissociate poorly so the concentration of the salt that we're using is basically equal to the concentration of a minus which is this bit here okay and the concentration of the acid at the start is equal to the concentration of the acid at equilibrium so when it's taught about the concentration of methanoic acid okay which is um obviously
• 52:00 - 52:30 this bit here then we can assume that the concentration there is going to be the same at equilibrium so there's a few assumptions that we're making here when we're doing these calculations okay then once we've done that we need to rearrange the expression to get h plus or the concentration of h plus on its own so the concentration of h plus is ka times by the concentration of methanoic acid divided by the concentration of methanoid ions
• 52:30 - 53:00 now here we've got the number of moles in the question okay so you can see here we've got moles of this and moles of that now that's no good we need to have them in concentrations because this is what the expression is telling us to do with this means concentration of um these of the acid here and not the number of moles so to work out concentration we need to do concentration as moles divided by volume remember this is in year one chemistry and in this case
• 53:00 - 53:30 um in this example it is moles divided by one decimeters cubed okay because the volume that we're using here is decimeters cubed so in this case we can say that actually the number of moles here if you divide that number by one you're just going to get that number again so we can in this example we can say that the number of moles is the same as the concentration of each of them as well so the concentration sorry the value of ka is 1.78 times 10 to the minus 4 we've been given that the
• 53:30 - 54:00 concentration of methanoic acid is 2.35 times over 10 to the minus 2 which is there and the concentration of methanoid ions is 1.84 times by 10 to the minus 2 because we've been told it there okay so we put all that in and we get the concentration of h plus to be 2.27 times by 10 to the minus 4. okay and then the fourth bit we need to do is obviously calculate the ph of this
• 54:00 - 54:30 buffer and we do ph the minus log of h plus put the numbers in and we get a ph of 3.64 okay so that's what we're doing here you've got to be aware of these assumptions we're going to use these assumptions for the um next um for well for any type of ph or calculate the ph for buffer you've got to use these assumptions that's really important okay so let's kind of move on to kind of the final parts of this i suppose of the final section of this topic where we're
• 54:30 - 55:00 going to look at solubility of products some common ions and and stability as well so looking at um stability of solubility of stability in solution so we're going to look at solubility product first okay so solubility is the extent to which a solid dissolves in solution seems to make sense doesn't it so when ionic solids when they're added to water hydrated ions are present in aqueous solutions so you would have seen that um
• 55:00 - 55:30 in previous topics so this would have been in a topic 23 when we looked at um hydrated ions if you're not sure on that go back and have a look at topic 23. so the degree of solubility obviously depends on the type of solids but also the temperature and pressure that um that exists as well so there'll come a point where the water can no longer hold any more solid and this is the point of saturation and any additional solid adds to this point won't dissolve so you've
• 55:30 - 56:00 seen this if you add say salt to water you can't just infinitely chuck as much salt in as you want into that water initially it'll start to dissolve it will dissolve and then eventually come a point where you add more salt and it just simply will not dissolve any further the additional salt you're adding so that is saturation point now the maximum amount of solid that will dissolve in a solvent is called solubility okay so that's the word we use solubility solubility that has the units of grams
• 56:00 - 56:30 per decimeters cubes it's the amount of salt you add per volume of liquid that you're adding it to at which what's the maximum amount you can add to that it is temperature dependent obviously so if you take salt water and you warm that water up it will hold more salt it can dissolve more salt at that point so you'll see all these things quartered with a specific temperature so to get solubility in moles per decimeter cubed all we do is we divide
• 56:30 - 57:00 the grams per decimeters cubed by the molar mass of the solid that we're adding in and we'll we'll see an example here so the solubility of barium sulfate at 298 kelvin is not point 0.25 grams per decimeter cubed so let's give the solubility in moles pesticide so molar mass of barium sulfate okay is bas04 using the periodic table we get 233.4 grams per mole
• 57:00 - 57:30 solubility is 0.0025 that's in grams of decimals cubed divided by the molecular mass and that will give 1.07 times 10 to the minus five moles of decimates cubed so that's the solubility of barium sulfate at that temperature at 298 kelvin in a saturated solution okay um we have the the amount of barium two plus uh yeah amount of barium two plus ions
• 57:30 - 58:00 equals or the concentration sorry of bearing two plus equals the concentration of sulfate ions okay so that's the same so therefore we can say that the concentration of these ions is also going to be one point zero seven times by ten to the minus five moles per decimeters cubed fairly straightforward not too taxing okay so just make sure you know that word solubility okay so we're going to look at a constant it wouldn't be uh it wouldn't be equilibrium without another constant
• 58:00 - 58:30 with it so we're going to look at a constant called the solubility product okay so this is the ksp so obviously anything to do this when we're dissolving things in water and equilibrium is established between the ions that have been dissolved and the undissolved solid okay so when sparingly soluble solids are dissolved in solution so when you do have a sparingly soluble solid so for example a solid that doesn't really dissolve very well you're going to have some undissolved solid i.e salt and some that has dissolved
• 58:30 - 59:00 so this is the case when you have obviously the example we looked at before is where you have a um something that dissolves fully we can make that assumption that it all dissolves and therefore we can work out the solubility when you've got something that partially dissolves we can't make that assumption and this is where we're using ksp instead so a bit like when we looked at ph and we said if you had a strong acid we could assume that the concentration of h plus ions equals the concentration of
• 59:00 - 59:30 the acid that we started with and therefore just took it in the ph equation and then if we had a weak acid we have to use ka because we can't make that assumption this is similar except we're looking at solubility instead so if you have something that's sparingly soluble we can't just assume that the amount of solid or the concentration of solid is the same as obviously the the concentration of the ions that's been formed so if we've got a sparingly soluble solid we need to use ksp okay
• 59:30 - 60:00 now um a specially soluble solid of a of aabb which is this here so this is just an example of a solid dissolved saturation the following equilibrium is established so effectively there's our solid salt we add it to solution and we form these two ions in the solution here so we've got a positive and a negative ion and obviously depending on obviously the number of these um will depend on the number that goes in front so we can apply the equilibrium constant
• 60:00 - 60:30 kc okay to show how this looks remember we don't include solids in this at all we're just looking at aqueous solutions okay because we're looking at talking about solubility here so ksp is the equilibrium constant for a sparingly soluble product in a saturated solution okay so it's already saturated and we've got something which is sparingly solid but doesn't dissolve very well so it has the same formula as kc as i mentioned before
• 60:30 - 61:00 so ksp which is here is basically the concentration of a aqueous um multiplied by the concentration of b aqueous we're not including the solid here okay we're ignoring that and that's why i've just put these two here so ksp okay which the solubility product has a fixed value for a specific solution and temperature so it's always the same so let's have a look um at some further
• 61:00 - 61:30 examples to do with solubility products so ksp should always include state symbols in the expression so using barium sulfate as an example which is here this is how we write an expression in terms of barium to plus ions and sulfate ions so these are the ones that are formed when we add barium sulfate to water um and we're going to write see how we write this expression so ksp which is ba 2 plus and so4 2 minus so that was kind
• 61:30 - 62:00 of derived from the kc expression or this okay which is ksp solubility of bas042 okay squared now notice we're putting squared here okay because the concentration of each ion that the saturated solution or in the saturated solution equals the solubility of baso4 and we can square it so remember we've got these two being formed here with both of these so that and that we assume
• 62:00 - 62:30 is the same and we say the solubility is basically that squared okay so that's how we write it out so just like for kc we need to work out the units okay and obviously these will vary according to the reaction that's obviously been conducted here so work out the units of the products let's have a look at a different example here so let's work out the units the product of um bi2s3
• 62:30 - 63:00 okay so we're going to use the ksp equation here so we've got two lots of bismuth three plus and three lots of sulfur two minus okay so you would have seen this in year one chemistry it's all about how to write the units but ksp is effectively moles per decimal is cubed squared multiplied by moles per decimal cubed cubed so because we've got two and three there so that effectively means it's moles per decimeters cubed to the power of five and then basically if we expand that
• 63:00 - 63:30 bracket out we get mole to the 5 dm to the minus 15. okay so we can use ksp and this can be calculated by using your solubility values so let's have a look at some and calculation examples here so calculate the solubility product of lithium carbonate that's li2co3 when the solubility of this is 13.3 grams per decimeters cubed at 20 degrees
• 63:30 - 64:00 celsius so the first thing we need to do is write out the ksp expression okay so remember we're only including aqueous ions here okay so the reaction is c li2 3 s solid uh we'll give 2 li plus and c or 3 2 minus so the solubility product is lithium plus with a 2 next to it because we have two of them okay and then you've got or three two minus and we've only got one of them so it's just that so
• 64:00 - 64:30 that's the first thing is write out your expression and then we need to calculate solubility of li2co3 in moles per decimeters cubed we've been given it in grams per decimal squared we need to convert that into moles of dm cubed so the solubility of moles of gm cubed is the solubility in grams distributes cubed divided by the mr so that's 13.3 74. obviously this is calculated from the periodic table and then the solubility of moles decimates cubed is 0.180
• 64:30 - 65:00 so we know that one mole of li2co3 which is this here okay dissociates to produce two moles of l i and we can see that there and one mole of c or three two minus signs at equilibrium okay so this is the equilibrium set up here so therefore we can say that the concentration okay because remember we worked out the concentration of lico3 li2co3 the concentration of li plus is
• 65:00 - 65:30 going to be two times that so it's two times 0.180 and obviously the concentration of co3 is just going to be the same as the concentration of li 2 co3 which is 0.180 so then we now need to use these figures here to work out the solubility product constants ksp and so we put the figures in there that we've just worked out up here and we get 0.023
• 65:30 - 66:00 units really important so here obviously we can use the go back to the original expression here so you've got moles per decimeters cubed squared as you can see there multiply by moles per decimal skewed because of that one um that will give you overall mole dmq to the minus three cubed so that's mole three dm to the minus nine make sure you work your units out here you get a mark for doing that so just make sure you do really really important
• 66:00 - 66:30 okay let's look at the common iron effect and this is really straightforward dead easy um so the common iron effect is basically where we now we add an eye on to a saturated solution so kind of still in this kind of arena and that is the same as one of the ions that's already present in there hence common so common iron okay so let's look at example where we've got sodium chloride okay and we're going to add sodium chloride which is salt to water until we form a saturated solution
• 66:30 - 67:00 okay so we've seen these examples already so let's have a look at this reaction so the reaction is in equilibrium and it's in solution so we've got sodium chloride solid in equilibrium in water to produce ni plus and cl minus ions so we're producing our ions there so let's say we add more cl minus ions we're not adding more salt just the cl minus signs to this saturated solution what do you think will happen with respect to equilibrium
• 67:00 - 67:30 so we've got this equilibrium here okay it's saturated we're going to add more of these ions here so it's a common ion what are we going to what do you think that might happen well listed india's principle okay equilibrium will shift to the left to use up the chloride ions that we're adding into that use up the excess so what we'll see is actually because this is at saturation point we'll see a precipitate form so we'll start and see nacl or sodium chloride actually being
• 67:30 - 68:00 formed um and as a result we'll have a lower concentration of na plus ions so this will start to decrease and we'll get more of this being formed but because it's a saturation point it will form a precipitate so we've added a common ion okay which is the cl minus ion to a saturated solution so one of the ions in solution is obviously the same as the one that's been added and that's why we call it a common ion so obviously the common ion effect can
• 68:00 - 68:30 also be applied to sparingly soluble substances in solution too you've just got to have a saturation point and then you will see a precipitate has actually been being formed at this point and obviously in this case it's going to be your salt see so relatively straightforward not too difficult okay so this is where we're going to look at another coefficient and this one's not too bad actually it's relatively straightforward it's more descriptive this rather than um you know more than two maths heavier supports um
• 68:30 - 69:00 so this is called um so this is called the partition coefficient or kpc um so the partition coefficient is the equilibrium constant between the substance ability to dissolve in two different solvents okay hmm someone's interested so let's say we've got this beaker here okay and we have a beaker with an aqueous layer and we've got an ether okay which is just another solvent it's just it's just not water now these two
• 69:00 - 69:30 liquids are immiscible okay so they don't mix with each other now we obviously get the two layers and the less dense um whichever one is has a lower density depending on what you react in will sit on top of that wall will sit on top basically so in this case the ether is has a lower density than water and so therefore the ether will sit on top of the aqueous layer in this example could be the other way around though but in this example that's what we've got so let's suppose we're going to add a substance so add chemical a and
• 69:30 - 70:00 obviously this is the purple dots okay um that is soluble in both solvents okay so it'll dissolve in both but actually it's more soluble in the ether so you get more you get a higher concentration of chemical a in the ether layer rather than the aqueous layer so if we stir the mixture okay so we put the mixture in we give it a good stir what will happen is some of chemical a okay will cross over into the aqueous layer and vice versa at the same time at
• 70:00 - 70:30 the same rate and we reach in equilibrium at that point okay so we're mixing up we're mixing up and it'll start to reach an equilibrium as this suspension in each solvent is in equilibrium we can write an equilibrium constant see another constant um all because of equilibria how fascinating so um so you've got kpc it's basically it's dead straight forward this one concentration of a in the ether over the concentration of a in the water and that's with the kpc okay
• 70:30 - 71:00 so we can calculate the partition coefficient okay which is this here kpc when a hundred centimeters cube solution containing one gram of chemical there was shaken with 20 centimeters cubed of ether um 0.9 grams of a transferred into the ether layer okay so all of it so one gram okay is in the was in the water in a hundred centimeters skewed we added 20
• 71:00 - 71:30 centimetres huge of ether and shook it and 0.9 grams of the original chemical we added then moved into that ether layer so basically we'll just write the concentration of ether and this is in grams per centimeters cubed is 0.9 divided by 20 and that gives us that and then we work out the concentration of a and water which is going to be 0.1 grams divided by 100 and remember all this is in equilibrium so if you had initially had one gram 0.9 has moved into the
• 71:30 - 72:00 ether so that leaves you with 0.1 in the aqueous layer divide that by 100 and that gives us not point naught naught 1 grams per centimeters cubed and then your coefficient partition coefficient um is obviously 0.05 0.45 sorry 0.45 divided by 0.01 equals 45 and that's your partition coefficient now we can obviously rule these obviously work out the units
• 72:00 - 72:30 they're basically both equal so there's no difference there so the units cancel out so there are never any partition never any units for partition coefficients you're just gonna get a number like that okay so what affects the um partition coefficient what impacts the number the value of it so the value of that is governed by the relative polarities of the solute and the solvents okay so the solvents is the aqueous and the ether
• 72:30 - 73:00 layer for example and the solute is the chemical that we add into it that then dissolves in either both or one of the layers so let's use a slightly different example here so we're going to use an example a specific example i suppose of bromine okay and that's been added to a beaker of water which is the aqueous layer now this time and and a solvent um called ccl4 okay so that's tetrachloromethane effectively um so in
• 73:00 - 73:30 this example water is actually less dense and so it will actually sit on the top layer okay with the ccl4 sitting at the bottom so bromine is brown as we know so i've kind of put the kind of dots on there to represent bromine molecules now you can probably see obviously from the diagram that bromine is definitely more soluble in ccl4 than it is in water so we have a higher concentration of bromine molecules in um tetrachloromethane than we do in the aqueous layer
• 73:30 - 74:00 so the intermolecular forces between bromine okay so between bromine molecules themselves is van der waals forces so these are weak forces remember if you go back to year one chemistry when you're looking at bonding and intermolecular forces van der vaals forces the weakest forces um so in between bromine molecules ignore the solvents here at the stage just between bromine molecules and its found valves now the same forces um exist between ccl4 molecules so the
• 74:00 - 74:30 solvent ccl4 the molecules within the solvent will interact via van der waals forces as well so they're non-polar overall now when we mix the solution okay so the intermolecular forces are broken between the ccl4 molecules between them and the br2 molecules um now these are replaced by forces similar in energy between ccl4
• 74:30 - 75:00 and br2 and so the molecules um so the molecules of the energy barrier is quite low so in other words imagine if you kind of imagine yourself take yourself off into your kind of a molecule and and you're watching this happening so in the ccl4 um solvent you've got ccl4 molecules kind of interacting via van der vaals between themselves you've got bromine molecules interacting via van der falls between
• 75:00 - 75:30 themselves as well and then when they kind of mix when you mix them together they will form br2 and ccl4 molecules will form van der waals forces between themselves as well there's no kind of other forces being produced so we're breaking the van der vaals and forming van der vaals so there's not much of an energy barrier there at all however there are additional um stronger forces there's two stronger forces between water molecules so the
• 75:30 - 76:00 aqueous layer the water molecules will actually so if we can move our eyes to the aqueous layer they have van der waals forces but they also have permanent dipole dipole and hydrogen bonding forces so you've got two stronger intermolecular forces that exist between water molecules now if um you're adding bromine into that you have to break these forces between the water molecules first and then bromine
• 76:00 - 76:30 will effectively um form a a van der waals force between the bromine water bromine molecule and the water molecule which is weaker than the forces that water has so effectively water is interacting with itself it's got a really strong bond between the water molecules bromine rocks up and says hey i want to try and get involved with this um the water molecule will look and think well the strongest force i can form with you with the bromine molecule is van der waals that's it so why should i
• 76:30 - 77:00 really do that when i can form a stronger bond with my fellow water molecules so effectively it's not a very favorable water doesn't really see that as a good deal at all in that sense obviously don't speak like that in the exam but it's trying to kind of visualize what's going on here and clearly it's not an energetically favorable process at all so in essence the relative solubility of the solute and the two solvents will govern the size of the partition
• 77:00 - 77:30 coefficients and therefore the strength of the intermolecular forces okay so see what what governs the size of this is obviously the intermolecular forces um between the solute itself i.e between bromine molecules the two solvents okay so this is obviously between um water molecules and water molecules and ccl4 molecules and ccl4 molecules that's kind of bonding within themselves and obviously between the solute the
• 77:30 - 78:00 bromine molecule that's coming in and the solvents and the respective solvents so obviously um the the key thing is that you're more likely to get solubility if the bonds that are broken are about the same as the bonds that have been formed when you add your solute in there okay so these are things which can affect the partition coefficient so hopefully that makes a bit of sense okay so there's a lot there okay there's
• 78:00 - 78:30 a lot it's quite a big topic um like i say this is just topic 25 the full range of year one and year two revision slides are available on alloy chemistry click on the link i'm sorry just don't click the link i'll tell you about that in a moment um if you um you know have a look on the anime chemistry youtube channel the full range is there please uh hit the subscribe button that's what i'm gonna say and just to show you support for this project that'll be absolutely fantastic the link to actually buy these powerpoints if you wish to purchase them
• 78:30 - 79:00 um is in the description box below and you'll be able to get a hold in there like i said they're great value for money and you can kind of make revision notes from them as well if you wish um right that's it then bye bye