This tutorial provides a lucid deconstruction of redox stoichiometry, turning the often-confusing logic of back-titrations into a streamlined, exam-ready process. It is a highly efficient resource that prioritizes clarity and practical application for serious students.
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AQA A-Level Chemistry Live Revise for the examsAñadido:
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>> If I called you tonight, would it feel the same? [music] >> Or have we both moved on to different names?
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If only for a moment. I've been up since three. scrolling back.
>> Oh my gosh. Right. Um Yeah. What's happening? What's happening? What's happening? I'm running late.
Yeah. Um here's a nasty question for you if you want to have a go at it. Um you'll need a calculator for this one.
So, um what is it? Jeez Louise.
This is a redux titration. Some people said they wanted to go for a redux titration. I don't know. Shrink my head.
Don't know what kind of sick people said they wanted to do a redux titration.
Someone said it last time. So, um, yep.
There we go. Let me have some water.
So, yeah, redox titrations are pretty rough. I will be going live tomorrow.
Um, probably a little bit more of a detailed live tomorrow. a bit more functional. We're running all over the place today. So, um yeah, got some questions to go through. Anyway, I'll be going live tomorrow and potentially Saturday and Sunday as well for Chem.
So, let's have a little look what's going on.
Um, so you might want to have a go at this one and then I'm going to go through how to do this.
Oh my days. [sighs] I still ain't got no glasses. So yeah.
Okay. What we saying?
[sighs] Yeah, it it isn't nice this question. I hope you guys are all good. Um, work out the moles and hope for the best. Yeah, yeah, yeah. Yeah, yeah, yeah, yeah, yeah, yeah. Um, practical exam question packs are good to do. Yeah, I mean for paper one, some people are predicting I mean that's a prediction. Make up a standard solution.
That one's nice and easy. You should know how to make up a standard solution by now. Use the before and after weighing method. Um, yeah. [laughter] Uh, make use the before and after weighing method. Tip your stuff into a beaker. dissolve it. Stir with a glass rod until it's fully dissolved with distilled water. Then tip your washings and your mass of whatever solid that you're using to make your standard solution. Pour that into a volutric flask. Um, and then you want to say add your washings. So, your glass rod you used in the beaker to stir it and everything. Add those distilled water washings into the volutric flask. And then make up to the mark. As you get close to the mark, use a dropping pipette to make sure and look at the meniscus and make sure it sits exactly on the graduation line or the mark on the volutric flask. And then you want to put a stopper in and invert it several times. And that would be how to make a standard solution. Some people have predicted that that's coming up for some reason. I've seen some YouTubers and people predict that coming up. And also your um chemical test tube reactions uh are what I think like I've seen three or four different chemistry YouTubers predicting test tube reactions. You know your halide ion ones and and etc. They predict those coming up in this paper one. But who knows? I'm not sure about predictions.
[sighs and gasps] Okay, let's have a go at this. So um yeah, this is this looks disgusting. I might have to make that bigger. Can you can Can everybody see that? Or do do I need to make that a bit bigger? Jeez Louise.
Let's go through how to do this one for you guys.
So, straight away, you're in the exam, you get a disgusting six marker like this. You want to set your watch by it.
Don't take more than eight minutes over this. Okay? So, extract the marks you can out of it, but don't waste your whole life on it. So they start by dissolving 10 grams. So they've given me mass. So straight away I'm like, "Oo, you're giving me mass. Nice." Of fees 7 H2O. So we got a hydrated one there.
And we got the MR of it. 2779. So straight away we're like, "Hey, you're you're giving me mass over MR time moles." So hopefully you settle your nerves and you're like, "Let me start by calculating the moles of that." So, we start off by going 10 over 277.9 and that is mass in grams over the MR. And we're going to start straight off by smashing out the moles.
Yeah, I'll go through and I'll give you a clue.
Um, and that is 0.0 3598. Listen, I'm going to use roundings cuz I can't be writing all those long numbers. I've just decided that's not going to go. It's not going to cut it.
So 0.0 I'm going to call it 36 moles.
Don't know if that's my face is on that.
Is my face all over that thing? Yeah.
Okay.
Yeah, we do love redux titrations. So we got 0.036 moles in that amount. Now let's bear in mind, let's make a note.
All right. Let's let's keep it all good and above board. So we got the MR, we got the mass. We've already stuck a mark on the board. We're cooking on gas. We got 250 cm cubed. You know what they're going to do? They're going to take a sample. No one titrates 250 cm cubed.
The solution was left to stand exposed to air and some of the iron 2 ions. So, Fe2 plus ions became oxidized to Fe3+ ions. A 25 cm sample. A lol.
They've taken 25 cm sample.
Okay, so we got a 25 cm sample has been taken.
So I guess this is our original amount of iron and then we've got a 25 cm sample taken down and that required 23.70 cm of 0.01 one mole potassium dromate solution for complete reaction in the presence of SS dilute sulfuric acid that's just means it's acidified that helps you with your half equations to know when they say this and it's redox you know that there are some hydrogen ions I don't need to put brackets and concentration we don't need that but you know there's hydrogen ions knocking about okay so what we want to do here my second thing that we can do is I want to know that it's gone from 250 to that. So, we've divided by 10. At some point, I need to times by 10 to go back. So, you want to whenever you see that in a titration where they had 250 and then they went to 25, say to yourself, right, I've I'm only using a tenth of the moles. I'm using 0.036 whatever moles. So, I'm going to need to times back to get back to that. Anywh who um apparently some of these ions it sounds like they it's full of iron 2 plus ions. Why not? It's FeSo4. So it must be it was all Fe2+. Some of them were oxidized to three ions by the air.
They want to work out how much was oxidized by the air. So they're taking the excess Fe2+ ions that are left over that haven't been oxidi oxidized and they're neutralizing those with dromate in a redux reaction kind of a yeah um so it's not really neutralization but you know what I'm saying they're reacting those with that. So the the second bit okay so if you're like bafted at this point the second bit is go hello I've got a shaky hand and I can't actually do that. So I've been given conchk and volume for K2 CR whatever it is potassium dromate acidified. So K2 CR2 07 let's work out the moles of that. At least you're getting two marks right for the moles. So you're going to go 23.70.
We divide that by a th00and. You guys are cool with that. That's 0.02370.
Has anyone worked out what the moles of potassium dromate are that we used? Now this is the you know I'm such a flop shot. I'm I it's so easy for me. That's mole per decimeter. So that is volume and that's conk. And we're going to get moles. You know that N over CV equals moles. So that is going to give me what?
0.01 * 0.02370.
That's going to give me Oh my gosh.
2.37* 10us 4. Does everyone agree with that? Because I don't want to flop hard early on.
Everyone anyone got ahead right. Yes.
Yes. Yes. Nice.
You can start from the bottom of the question. A lot of people do that. Very good. You can start from the bottom, but you kind of also need to know the original moles at some point. And then what you're going to do, you're going to get your This has told me how this has actually told me how many moles. This is going to tell us how many moles of the Fe2+ was left over. So, in the question, they've kind of said, listen, they've said we had this um we kind of had a beaker of Fe2+ and some of them became Fe3+. Not all of them. Some of them became Fe2+. We reacted all of the remaining Fe2+ that were left with this.
So, if I know the moles of this, hopefully I can find out the moles of Fe2+ were left. We know it's going to be less than the original Fe2. We can't have more moles of that than the original. So, we know then we're going to subtract and work out the missing amount of Fe2+ and then we'll know how many had been oxidized by the air and we'll get our six marks. So the next thing to do now we know the moles of that and the moles of that. We got to work out the ratio between potassium dromate and the iron 2+ that we're in a reaction. So you might say sir sir I'm confused. How do I know it's not the iron 3+ that's getting down? Well let's let's draw the u the redux equation out for potassium dromate. Now it's a redux equation. Potassium ions are spectator ions. They don't change their oxidation state. So, we're going to go like this.
Who can tell me? You should all know the chromate ion charge 2 minus. We know that is going to be + 6, right? Cuz that is 7 * 2 minus is -14. The reason why it has an overall 2 minus. It's cuz it's got + 12. It's got two chromates. Each one is + 6. And you notice chromate ions they go to chromium ions which are 3+.
Now once you've got that that should be your starting point for the redux half equation for the chromate ions in the reaction. What we're going to do if I can get hold of my pen properly and it don't run around. It's always being a brute.
What we're going to do is we're going to look here. Whenever you are doing a redux titration, they will mention acid and you know you're going to use hydrogen ions. But not only that, when you have oxygen in a redux titration, what do you have to put on the right hand side, guys? Whenever you have oxygen.
Okay. So the reason we'd know this is a redux titration is because this titration is not acid and base. They titrated like aquous Fe2+ ions. They'll start mentioning ions. So Fe2+ ions to Fe3+ and they'll either mention like potassium manganate what is it KMO4 or they'll mention CR2 uh K2 CR2O7 is very typical of these redux titrations. So they'll start mentioning ions and stuff and you'll realize it isn't straightforward as like acid and base. So what we're doing here with this redux titration, you're going to need water. Excellent. I can see people are putting water. How many waters do I need, guys? If I got seven oxygens on the left, how many waters? So you standard need to know it always turns into CR3+. Yeah. Every time you do that half equation. Yeah. So I need seven waters. Excellente. So, we're going to go 7 H2O.
Now, there's some skills to Redux equations, and some of it is just a bit of good luck, winging a prayer stuff.
So, if I have 7 H2O, they told me that it was acidified with excess acid. So, I know I have hydrogen ions knocking about, which is just as well. Otherwise, I couldn't form the water. So, they'll always mention that. That's another clue you're dealing with the redux. So, I'm going to have 14 H+ over here.
Why? Cuz I've got 7 H2O there.
Now, what I want to do is check the charges. Well, oh, I got CR2. So, I need two chromate ions here. Hopefully, everyone's with me. Now, let's add up the total charges. 2 * 3+ is 6 plus on the right hand side I've got uh that's not a plus that's that was the oxidation state over here I've got 12 plus so one of one of the ways you can work out how many electrons we need to add to the left hand side is that in redox half equations this is half of the full equation it's the chromate side we need to balance the ionic charges on either side of the arrows so once you've written your equation out. Check the ions balance and then add electrons to balance it out. So, how many electrons do I need to add on my left hand side?
6 E minus. Thank you. So, we're going to add 6 E minus here. Now, I haven't added it very well. That should be in a straight line, not underneath. You can see I'm running out of space. You could have worked that out as well from the oxidation state. Some people are like, "Sir, I don't like doing it like that."
You could work it out as well. We're going right that 6+ is going to go to a 3+ oxidation state. Balancing the numbers, I need two of them. I'm going to need six electrons to take two six pluses down to two three pluses. We're going to need six electrons. So, you could do it via the oxidation state as well. But you can also do it by just saying, look, the overall charges need to balance. And now what we have, we have 2 minus 14 plus and six electrons.
We have a plus six charge over here. We have a plus six charge over here. We are balanced. Whoopdedoo. So that half equation is done. Now you got the easiest half equation in the world to do. You got the iron 2 plus half equation. And just think to yourself, this is a reduction. Chromate has been reduced. This is gaining. This is reduced. Now for those of you that might be thinking, sir, sir, sir, is it um are we talking about like Fe3+ here? Right.
Fe3+ could do this.
Um, right. So, the Fe3+ could become 2+.
No, it can't do that, can it? So, we should work out that this is not what's going on.
If this is uh reduction, we want oxidation over here. And it that can't happen. That Fe3+ can't give away electrons. So hopefully you can work out it has to be Fe2+ that can give away an electron. So we had to be starting with Fe2+ going to Fe3+ because we needed a supply of electrons. And this comes to the bit where this half equation relates down here. This is just a normal moles coefficient situation and amount of substance. They just add in this. Do I need to combine these? I can do it if you want me to combine them, but you should be able to work out. Now, we got six electrons here. What do I need to do to Fe2+ to balance it?
What do I need to do to this half equation to balance it?
Multiply the whole thing by six. We need six electrons here as well. Balance the two half equations. And to get six electrons, I need to oxidize six Fe2+ ions to six Fe3 plus ions. So I now know the ratio is 6:1.
Now if you want, we could combine this.
In some questions, they will say write the overall equation. All you do there when you write the overall equation is take the reactants from here with the reactant here and then the products all there once you've balanced them and write them all out. And then you do any further cancelling because occasionally you will need to do a bit of further cancelling of hydrogens and and bits and bobs. But I don't really need to do that. Once I get to this, once I get to that, I can see that my ratio is 6:1.
Now we know our ratio. I know that this many moles over here, this is my moles of CR27 2 minus. So how many moles of Fe2+ got oxidized by the dromate. How many moles got oxidized? Well, let's find out how many Fe2+ were left afterwards. So, how many got how many got oxidized cuz they didn't get oxidized by the air. So, remember some of them got oxidized by the air and then we're finding out the ones that didn't get oxidized by the air. And we should be able to find out the ones that got oxidized by the air by taking it away from the original. You get the C, right? So, let's go. 2.37 * 10 2us 4 and we want to times that by six. Is anybody there?
Oh, Sugar Ray Leonard calculator will work for me.
Oh my god. I think it's 0.01422.
Anyone on that? Yeah. Anyone? Anyone get that? 0 point. So this is the amount of Fe2 plus ions that got oxidized. 0.0 Jeez, I can't see properly. I'm like blind as a bat. 0.1422.
Is that right?
Yes. Okay. So we probably picked up three or four marks now. Now what we need to do is get it back to it says 25 cm sample of the partially oxidized was required. Right. What did we do? We left the 250. We want to find out calculate the percentage of iron ions that had been oxidized by the air. So we want to calculate the percentage. I don't even know if I need to go back to the total amount, but we will do. We'll go back to the total amount and do it out of 0.0.36.
So this is how many iron ions got oxidized.
What do we want to do? Times that by 10 to find out the moles of Fe2+ that got oxidized by dromate. and then take away from the original. I think that's what we're going to do. I'm going to times this by 10. I'm not sure we have to because if I if I kept it like this, I don't need to times by 10. So, it's up to you. Do you want to times by 10? You don't have to cuz you could just do it out of that out the 25%. The percentage that got oxidized out of 25 cm is going to be the same as the percentage that got oxidized out of 250.
I think I might be just making it harder by tsing by 10, but I'm going to go back to the original 250. So, I'm going to times by 10 and get 0.0.14 22. That is the amount. This is the Fe2+ that was left over after the air oxidation. So, we had 0.0.36 to start with. We're going to take away 0.0.1422 014 22 and then we're going to find out how much actually got oxidized by the air instead. So, anyone on that?
Let's let's do it and see if it works out. All right.
Hopefully I've not flopped.
So, apparently roughly roughly 0. Oh, no. 0.0.
Oh [ __ ] I've run out of room. Let's write that properly. Let me change the color. Let's go purple. Get the get the imperial colors out. The emperor emperors used to like wearing purple apparently. 0.02178.
I think we can just call that 0.0218.
[sighs] Hopefully I've not flopped. That's how many moles were oxidized by air. So that's how much we didn't have down here in the reaction. So just to recap, we had 0.0.36 to start with. Some of those got used up by the air in this Fe2 plus went to Fe3+ by the air. We wanted to find out how much got used up by the air. So we took the leftovers and we titrated it against the chromate. Um, once we worked out the chromate, how many moles of chromate got used up. They completely got rid of the rest of the Fe2+, we then had to work out the ratio and we realized it was 6:1 as in I wrote six to one the wrong way round. We needed six of them to one of them. So that meant that we had to times cuz when I said six to one, I meant we would need six lots of that to go with one lot of that. So we had to times that by six and that gave us 0.01422.
We then times by 10 to get back how many would have been left over in the original uh how many were left over in the original amount. So how much were we used up and oxidized by the air was 0.0218.
Now we got to work out the percentage.
So we're just going to go 0.0 218 divided by 0.0 36.
Hold on. Hold on. Hold on. That don't sound right at all. Times 100. What do we reckon? What do anyone get anything?
This is a level chem. Yeah, it's brutal.
Yeah, you're right. You're right. Well done. That is the right answer. Yes, that's the right answer. That's the textbook answer. Now, that's the answer if you use 0 point. If you don't use 0.02178, 0 2178. So, you're going to get a slightly different answer if you keep all your numbers in and I rounded. Yeah, that's the right answer. Okay, this was disgusting.
I can't remember who recommended it. We can do another one if you really want, but that would make me pass out. So, maybe we try something else. Um, let's try something else. We don't want to put all our eggs in one basket.
Listen, I'm not saying like I just Googled. You can try this yourself.
And by the way, I didn't get any of the questions. Just before I come live, I quickly Googled what's coming up in the exam. I asked AI. Has anyone asked AI yet? Has anyone got down and asked AI what's coming up?
Has anyone done an AI what's coming up?
Stick around. I'm going to have to science the [ __ ] out of this.
Oh, memorize which equations you can memorize. Generally, I it's always going to be in the same ratio, but you kind of want to be able to just work it out. So, the manganate uh they often have manganate with iron as well in tablets.
They do the old tablet and then work out the percentage yield and that. Is there going to be a redux titration? Listen, I asked AI what's coming up. Um, and I don't I don't believe it.
If you're f trying to find the percentage between two molecules with different MR, do you have to use moles?
Um, so in this situation, we used moles.
We could have converted it to mass. We could have converted it to mass and then done it out of mass, we probably get the same thing, right? As long as we convert them both to their mass and used both their different MRS. Yeah, it would be exactly the same answer. So if I times 0.036 036 by the chromate potassium dromates MR cuz that's potassium dromate not CR 207 or whatever and that then yeah if I did that by that then yeah we could do the grams as well so no you don't have to go to moles but there was just no point in going there's like a further two steps to go back so there was no point um now I asked AI what did oh jeez all right well okay this one's not really a tie trade.
I don't know if people want to do this.
We don't have to do this one.
This one is I just pulled out literally like I only pulled out a bunch of things and then I asked AI what's coming up and it was none of the things I'm going over. I I pulled up. I was just looking through old past questions. I don't know whether we should do this one because this one looks um absolutely filth. So I don't know whether we should bother doing it because you don't want to just do loads and loads of blooming calculations. That's quite disgusting.
Uh, we can do this if you're if if you if you want to. We can do this.
It's a back titration.
I've done this quest. Yeah, I've done this in a video. Let's not do this one then. I have done this in a video. Do you want to do it? All right, we can do it. We can do it. We can do it. I have done it in a video. There's one catch here, but I have done it in a video.
There's only one tricky bit.
Do you know what? And in the video, I'll explain it a lot better than I will now.
Basically, so all those questions for videos I've done at different points, they're not all of these, but I have them obviously hidden on my computer in various files and stuff. And like an idiot, I don't label anything properly.
So, I'm just like, I'm going to do a chemistry live tonight. I put in things like Liv and then like I'm looking and just long. If I only was organized, I've got thousands of questions and they're all over the shop so I can't find like stuff. Anyway, what's going on with this one? I'll go through this one. You got hydrated aluminium sulfate Al2SO4. For those of you that can't see, it's Al2SO4 with a three outside the brackets. Dox H2O is soluble in water. The relative formula mass and the value of X can be found from a titration experiment. This is a back titration.
So aquous Al H2O63 plus ions react to form a stable complex when treated with an excess of EDTA4 minus ions. The excess EDTA4 minus ions is determined by titration with ZNSO4 solution. We don't really need to do half equations here because guess what they've done? They've done all of that business for us down there. They've given us the balanced equation. So lovely dissolve. Okay, let's get down to it. Dissolve that.
So, guess what? They're giving me mass.
But before you think, can I work out moles? We got a dirty X there. So, we can't work out any moles of that. So, we got mass, though. Um, and then make it up to 250. So, we've been given volume, but that's not necessarily going to help us because of this mass situation. Add 25 cm.
Look, they love the 25 cm one. So, look, we've already just put a note there.
Whatever you want to do you sometimes I write divide by 10 or oh I've got to times by 10. So it depends like you could write they've divided by 10 they've taken a tenth or you could say oh write a note to myself I need to times that I need to times something by 10 at the end. So of this solution to 50 cm cubed of a solution containing EDTA ions of concentration. Guess what?
Here's our first thing that is like solid we can work out. We've been given the volume and the conchk of EDTA.
Hallelujah.
Volume conchk. So straight away, those of you that are like in a panic when you see the size of this question, you can do N over CV and work out the moles of EDTA. Determine the excess EDTA by titrating with zinc sulfate in the presence of an indicator. Excess EDTA 4 minus ions requires 18 cm of 0.0105 or 105 zinc sulfate to react completely. Guess what's beautiful here? They've given us volume and conch. So the minimum that someone should get is two marks for working out moles of the zinc sulfate and moles of the original EDTA4.
So we got volume and conchk. Now here's the thing. This is in a back titration.
Some people start from here. Some people start at the bottom and work their way back up.
All you're doing is you're finding out how much zinc moles have been used. It's one to one look with the excess. That means we'll now know the excess moles of EDTA used. And up here is the original.
I'm going to start here with the first bit of information because it's the original EDTA. It's the thing that we added in excess. And then you titrate the excess against something. So I'm going to work out the original first. So original original EDTA is going to be 50 over a,000 times 0.01.
So what do we get? We get the original moles of EDTA. I don't know what that is. That's supposed to be a N. Come on, son. Sort it out.
So 50 divided by a,000 is 0.05.
I'll probably flop at some point guys.
So I got 5 * 10 -4 moles originally [snorts] [gasps] and this is original eda original. Okay, cool. Now, what I want to do, let's work out our zinc sulfate.
So, we're going to go 18 over a th00and times 0.0105.
If you've worked it out, give yourself a boom.
Uh uh. Okay. Right. Yeah. So, what is this?
18 / 1,000 = 0.0.18 * 0.0105. Watch me flop on the calculator. Right, that gives me 1.89* 10us4. Here's where you're checking it out. It's a one one. We've just worked out the um what what did they say with a zinc sulfate? We just worked out S Zn SO4, but it's one one. So I know excess you call it excess or lef over EDTA is actually 5 * 10 -4 - 1.89 * 10 to the minus4. So what do we call the EDTA that isn't the leftover and wasn't the original? I'm going to call it the reacted. It reacted in this equation with the aluminium. So I'm going to call it reacted ED TA uh whatever 4 minus. So the reacted EDTA 4 minus. So I need to do 5 * 10 -4 minus answer equals glasses on 3.11 * 10us 4. Yeah, I think 18 is a pretty good answer. There's one sticking point where people get flopshotted on. Uh 3.11 * 10us 4. Is that the same as 0.0 0.311?
I'm already falling apart. 3.11 * 10us 4. Let me go. 3.11 * 10us 4. Just check that. Yeah, it is. It is. It is.
You know, sometimes your brain just goes moldy. So I've probably got three or four out of seven marks. I don't know if it was six or seven marks. Probably got about three or four. Now we've worked out all the way along here. I now have the moles of the reacted. But remember this is in 25. They want me to calculate that. I need to times this by 10 at some point. So um I now know I've got this.
Here comes the hard bit. I know now I know knew and he know that that is how many moles reacted with the aluminium.
Guess what? It's a one to one. But here's the effed up bit. We are looking for this. We are looking for Al2 S4.
But the sneaky little bumheads, which is what AQA are, dirty downright low down beat rat yellowbellied sneaks have given us AL H2O6 which only has one aluminium. Now if you look at it for every AL H2O6 I don't really want to draw this out because this is actually becoming painful watching me do this for every one of them if I want to make that if I want to go back to that what is the ratio between that and that well I need two of these bad man I need two of them to one of them.
No, we got that major wrong.
We got that major flop shot. We need to divide by two is what we need to do. So, um, yeah, ratio is 2:1. So, I need to divide by two. Does everyone get that? This is the hard bit. I'd say this the hardest bit in this question because you've got that and alum, you're actually trying to find out how much aluminium sulfate you've had. Well, you need two lots of them to get one lot of aluminium sulfate. So, we're only going to have half the amount of aluminium sulfate.
And I think that was the hardest bit in this whole question. So we have to go 3.11 * 10 to theus4 and we have to divide that by two. I think that's the hardest bit in the whole question. I'll be honest with you.
Yeah.
So that's the hardest bit in the whole question. And a lot of students, you know, I noticed that were saying they saw the Alh206 and they saw that was Al2 and they said, "Sir, why don't I times 3.11 10us 4 when I time the moles of that to buy two to make that?" But you need two of them for every one of them you make.
You see? So it's 2:1. So we've got to It's actually two lots of that to one lot of that. The ratio is 2:1.
the co if you put the big coefficients in front. Anyw who, so we're going to divide by two and that is going to disgustingly get us 1.55 * 10us 4. I'm running out of room.
Where the hell am I going to write this?
So, we've actually got 1 Oh my gosh, I'm sweating. 55 * 10. It's gone hot again.
10 the minus4 1.55* 10us 4 moles of the aluminium up here. the old sulfaticus.
So now what we can do? We can What's my next bloody step, people? Oh, we know the mass of that.
[sighs] I have the moles of that. That was 250. I need to times by 10. So why am I tsing by 10? Because I've got the mass of 250. I need to times by 10 so I get back to the amount of moles that would have been of aluminium sulfate that would have been in the original. So if we times by 10 we would have had in the original 1.55 * 10us3 moles of Al 2SO4.
Now I want to work out how many grams that is.
So I'm going to times by the MR. These kind people gave us the MR down here.
342.3.
So we're going to do moles time MR 342.3.
And that is going to give me the mass of aluminium sulfate. Not the mass of the hydrator, but the mass of aluminium sulfate. This taken me too long. 342.
What was it? Three.
And it is Oh, sugar. Hold on.
I think I've uh flopped this. Why am I writing over my face? Writing over my face and I can't see. It's all gone pear-shaped people. It's gone pear-shaped.
Yeah, I think uh Amos John, I think you got it right. Yeah, I think I forgot to times by 10. I think you're right. I didn't times by 10, maybe. Hold on.
Where am I?
Yeah, I've got a times by 10. Sorry, I didn't put the I put the times by 10 on the on the board, but I didn't put it in the calculator. I still have my 1.55 * 10us 3. So if you times by 10, and by the way, I don't know if I can't see cuz I'm blind as a bat. Whether uh times 34, what was it? 2.3.
So if you do that, you'll actually get 0.53 g. And then you go, right, I know that's the aluminium sulfate. How much is the water? Work out the moles of water. So we're going to go, oh my gosh, he's run out of room. 0.036, 036 which is the total mass take away the aluminium sulfate gives me the grams of water. So we're going to go 1.036 036 takeaway answer equals bish bash bosch 0503 g divide that by the mr of water which is 18 right gives you I'm going to lose the will to live this gives me what a disgusting question the only good thing about a question like this is um that is worth quite a lot of marks. So, um, yeah, that gives me the moles of water. I now have moles of water. Times that by [sighs] moles of water, do I? What am I doing?
I've got the mass. I've got the moles of water. What am I doing, people? Oh, I want to work out the ratio between the XH2O and the aluminium sulfate. So, what's the ratio? Well, um, the water is obviously bigger because it's 0.027 027 and the aluminium sulfate moles were not originally we had them right at the start. Where's the bloody L? 5 * 10 the minus no what were the aluminium sulfate moles originally guys?
I've lost a plot 1.55 * 10us 3. So let's just smash those in our moles of aluminium sulfate divide by that and then we'll get x. So x equ= 18 people are saying. So, let me divide that by 1.55 * 10us 3. They were the original moles. And I got 18.05, but you're going to keep x as 18. It's an integer. It does say calculate x as an integer. So, keep it as 18. Don't put 18.05. And then all you're going to do if you want the total MR, um, it's 342.3 + 18 * 18, I guess. And that will give you the total MR which is I don't know what that is 660 something like that.
Anyway people let's not do any more calculations. I might just break down seven marks. This was foul and disgusting but all right.
That was absolutely horrific.
What is that?
Oh, that's easy half equations. These are easy half equations. We don't want to do those ones, do we? Uh, you do need to know your reducing ability of the haly ions. So, um, I thought let's go let's do some equations. [sighs] I just randomly clicked on this. Uh, give a half equation to show the oxidation of iodide. So, really dead easy. Oxidation of iodide ions. Let's go. Who can do the oxidation of iodide ions?
Reducing ability increases down group seven absolute more and we can see that here. So, give a half equation to show the oxidation of iodide ions. oxidation oil rig I it's loss. So the oxidation of iodide ions is going to be I minus goes to I2 plus E minus and then it's just a case of balancing it. So that is easy peas right and then we want to do the give a half equation to show the reduction of concentrated sulfuric acid to hydrogen sulfide. I ain't got a bubble. But we would imagine that we're just going to go sulfuric acid. And then we're going to go like this arrow. And we're going to do the same thing we did in that big dirty iron thingy dromate thing. So we're just going to go hydrogen sulfide is H2S.
And we're going to go I got the the four oxygens. I must be making four waters.
And uh got balance this off. All right. What we got to do? We got that. We got that.
That that that that that. Um how many hydrogen ions do I need to add?
I've got two there. There's four hydro.
No, there's eight. What?
What?
[sighs] Good.
Thank you for the eight. Thanks, peeps.
Well, we got eight, too. Yeah. So, there's 10 over here and only two here.
So, you're absolutely right. Where can I fit that in?
8 H+. Now, here's the coup.
Here's the bit.
We need to balance the charges.
How are we going to balance the charges, peeps? Look over here. If I was balancing this, you go 2 minus 2 minus.
Yay. It's balanced. But if we come down here, when you do your half equations, electrically, they have to balance. I've now got 8 plus and I've got zero.
How are we going to balance it?
Add electrons. So, we're going to add uh we can do an electrode one next.
Actually, add eight electrons. We'll do an electrode potentials one next if you want. Someone said, "I hate electrode potentials." I hate them, too. Um especially it's uh Yeah. Right. And now you got to combine. Hopefully, we get this right. I haven't flopshotted this.
Um now we're going to combine. Look, what's the multiplier?
How do I How do I balance so that I have eight electrons up here? I need to multiply everything up here by time 4.
So, I'm going to come and do that and write them down here. I don't want to change this half equation. So, I'm going to say 8 I minus plus.
Now, we can get rid of the electrons.
They're going to cancel out. So, we're going to go plus H2 SO4 plus HH+.
And afterwards, the electric charges should all cancel out on either side.
And we know we've flopped if they don't.
And then we're going to go with the we're going to have that and we're going to have four waters and we're going to have four Whoops. + 4 I2.
And does that work?
Yeah, we got eight pluses, eight minuses. Look, we don't need the electrons anymore because we've got these eight iodide ions, so they cancel that out. And that should work. There we go. Bosch. Right. Um, super.
Some people like redux, some people hate redux.
Um, let's have a few warm up multiplechoice questions here.
[sighs] Yeah, bit of multiple choice, bit of warm up.
What are these about? Well, stuff that could come up.
[music] HERE WE GO. WHAT?
[music] [music] YES.
INTERESTING.
That's how I feel like the lady at the end.
It's hot. It's too hot for work. Too hot.
Which occasion? My god, I really can't speak. Which equation represents the process that occurs when the standard enthalpy of atomization of iodine is measured? So, this is testing you on your definitions for the atomization of iodine.
Which one is correct? Some people might say, "Oh, it's I2 solid, isn't it? To two I gaseous because you're going to gaseous." Atomization is you going to gaseas from a solid. But what are the standardized things? Is it one mole of the reactant or is it one mole of gaseous atoms? What what's the what's the answer here, peeps? Who's got the first one? Yeah, you guys have got it.
It's a. And the reason it's a is because in the definition of the atomization definition, it is to form one mole of gaseous atoms. That's the only one that's forming one mole of gaseous atoms. Half a mole of iodine will form half will form one mole of gaseous atoms. Great uber duba suba beluba. Now um latis enthalpy values can be obtained from born harbor cycles and by calculations based on the perfect ionic model. which compound shows the greatest percentage difference between these two values. So what we're looking at is you have something called the perfect ionic model. I hope you guys all know what that is. Perfect ionic model is like that where you have what do they call them? Point spheres or something like that. Uh where they're perfect spheres and the only thing existing between them is the electrostatic force of attraction between oppositeely charged ions. What we do know is that when you have a small seriously powerfully positive ion and you have a large negative ion you get polarization and what we mean by that is this gets distorted and it gains some covealent character generally makes the lattice a bit stronger actually a bit more exothermic. So um and the more polarization you have the Bourne Harbor calculation predicts this it assumes that everything is a point sphere when you do the calculation it doesn't take into account polarization the Borne Harbor cycle when you do the experiment and you find the experimental evidence you find the polarization exothermic amount and then you see who's got basically peeps I'm waffling who's got the biggest difference and what you're looking at is which is the smallest positively highly charged ion and which is the biggest annion what are we going with let's have a look are we going all right so the trend is small positive ion big fat annion so we're going to go all of these are plus one ions so when we're looking at this cesium is + one lithium is + one but who's smaller lithium or cesium ium and what we find is lithium is smaller. So lithium is basically like that and seesium is bigger. So lithium will have more polarization. So we can get rid of those two. Right? Lithium is more polarizing.
Now we look at the annion. Now we've decided which cation is the most um highest charge density. We look at the annion and fluoride ions and iodide ions. Again, iodide ions are big and much more easily polarized because their electron cloud is bigger. So, it's going to be D.
Anyone go for D?
Definitely Li. Close Newton. But it is going to be L I with uh not LF Lif because fluoride ions although more electro negative and that might does that distort it no the electrons are going to be really held tightly to floor the florine nucleus right so the iodide has that extra shielding and you know it's a big fat annion and the the electron cloud is easy to get pulled away towards the really strongly positive high dense ion Right, let's go and check out this one down here. Which one of the following has the most covealent character? What are they asking us? Again, they're asking us very similar. Which one has been distorted the most? They're asking us the same question just in a different way. Which one has been distorted the most? So, let's do the same thing. Let's compare the cations ions. Work out which cation has the highest charge density and then work out which annion uh is the biggest. So with the ion annion, you're just looking for the most negative and big.
It's easy to polarize a big annion. You want your annion to be big and your cation to be insywincy small and have lots of charge. So this one is slightly hot, maybe slightly harder. This one, are we going with D again for this one?
Yeah. So I think you guys might be right if you're doing that. If you look at magnesium and you right, it's got a atomic number breaking down here. It's got an atomic number of plus two. Uh what the fudge am I on about atomic number? It's got an atomic number of 12. It's you know aluminium's got a slightly smaller atomic radius, hasn't it? Why? Why?
We're going across the period, aren't we? They're in the same period, these guys. And um this guy has got higher nuclear charge. Not only that, he's a plus three ion. He forms a plus three ion. So this one is definitely more dense than that. So let's get rid of the magnesiums. They're outdone by the high electron density of the aluminium ion.
Not only does it have a smaller radius, but it's also highly more highly positively charged. You know the idea.
You know the coup. 13 protons in that nucleus compared to 12 there. And this has got a higher charge density. So boom, bam bish. And then we look at again fluoride and broomemide. And broomemide is bigger, easier to distort for that positive pulling power.
Something I lacked, right? Um, yeah. So essentially your aluminium 3+ highly chargely high charge density. It distorts the big fat bromide ion and you get coalent character. The answer is D again. D.
Okay.
Oh my good gosh.
Um so basically your rules are you have to look at the positive iron and the negative ion and if you're given multiple choice you decide which negative ion is the largest.
If they've all got the same negative charge, it makes it really easy because we just go for the bigger one. The bigger negative ion will be more easily polarized. Electrons are further from its nucleus. They're more easy to be pulled away and like distorted out of shape. If they start adding extra negative charge, um then it starts to get more complicated, but they don't often do that. If we look at the positive charge because then you're looking at size and the number of electron and the negative charge. If you're looking at positive ions, it's not big positive ions that are very polarizing. It's small, highly charged, dense, positive pulling power that disrupts big annions. Anywh who, so you're looking for the opposite. You're looking for small and a high charge. If they've all got the same charge, just go with the smallest when you're looking at the cat ions, right? Uh, oh my gosh. All right, let's do a Gibbs free energy as it's here and then we'll bust on to something else. What the fudge is this? Calculate the entropy for the reaction between ammonia and oxygen.
Okay, so just a Gibbs free energy.
Yeah, I aed a they they I AIed it. By the way, I'll show you the predictions on AI in a second. I asked AI, but I wouldn't believe it. So, I'm not I don't believe it, but you know, AI said time of flights coming up. It said low. It said about 13 marks on buffers. I did a buffer question in the last live. You can check that out. It was right at the start of the last live. It all went a bit pear-shaped after that, but the first question was a buffer calculation and that will help you revise buffers.
So I did that in a couple of days ago and uh that's predicted to come up for about 12 or 13 marks. Uh and a weak acids and buffers question according to artificial intelligence say.
Yeah. Time of flight PV equals NRT.
That's another one that carries over from AS. So we'll go through one of them. Uh right. Um, what else did they say? I'll have a look. I'll have a look. Right.
Calculate the entropy change for the reaction between ammonia and oxygen. So, what do you want to do when you're doing entropy? The entropy change is equal to products minus reactants. So, all we want to do here is go four * n, which is 4 * 21, and we want to add that to anyone already calculated the entropy change. Good luck if you have. and 6 * H2O. So we want to add that to 6 * H2O.
And that is going to give me the products entropy. And then I want to minus from that the reactants entropy.
We got four ammonia and we got five oxygen. So I'll smash that in the calculator. Peoples. How much is that? What what what uh oxygen is 205. Okay. So 5* 205. Okay.
Okay.
What'd you get? 181 naturra 7126.
Come on, Tom. Get your calculator working. You you reit right. So, I got and then plus 6 times what? Why did I put 6* H2O? What a joker. 6 * 189 equals Okay, so we got one. That was supposed to be 6 * 189, but you get the C. Uh, there's 1 978.
And then we're going to minus um - 4 * 1 93. Oh my days. Oh my god. I put 1,993.
What the blow? Um Jes um I'm glad you guys have already done it.
I'm I'm struggling with the old calculator blues. 772 plus 1,025 1 978.
Gosh, do you know how long that took me?
That took me 482 years. We're in the year 200 something. Okay. Yeah, I finally got there, guys. I hope other people struggle with basic maths.
Calculate a value for gives free energy in kilogjles per mole blah blah blah.
Right? You need to know this off by heart. So if you don't know, get to know. You've got to do delta G is delta H minus delta no T delta S. There we go. By the way, people might flop on this sometimes.
Remember minus minus. So put your brackets in for this one. This um is exothermic. So it's - 9005. So we want to work out delta G. Um do they want us to calculate delta G? Yep. Gibbs free energy. That's delta G. We don't have to rearrange. We've not been given Gibbs.
So delta G is going to be equal to -905 KJ. Check out that. That's in kilogjle per mole. But check out over here entropy is in jewels per kelvin.
Sometimes uh people think of that that that's kilogjles. That's not. That's per kelvin per mole. So that is not kilogjles that's in jewels. Sometimes they put it in kilogjles and sometimes they put the enthalpy in jewels. So really really really really read the question carefully and check it out peeps. Now that's 600 degrees. We need to add 273 don't we? So it's 8 73. I think my face is all over my working there. Let me move my fat face for those of you that are um There we go. Okay, we're back.
>> [sighs] >> and minus. Come on, Tom. Get your pen working in the right place. Right. Times 181. Um, and then someone mentioned in the comments, yeah, we probably got to divide that by a th00and because we don't want that in kilogjles. We want that in jewels. So, what does that equal, people? Has anyone worked it out, by the way?
Let me just see if I'm Oh, that don't look right.
Hold on.
I don't even know if I've typed that in right. Um, doesn't look right to me. Don't look right to me.
- 95. What do people get, by the way, if you don't mind me asking?
Hey, let me check.
Well, I'd love to know if I had the answer somewhere here. Uh, hold on.
You get minus uh 1063.01.
Hopefully you guys got that. Minus 1063 and I'm not going bonkers. Everyone confirm that I'm not completely insane.
Now, if we look here, the answer, you could have done this all in jewels, by the way, and then converted back to kilogjles if you wanted to. So, you could have done this. You could have converted that to jewels and done 95,000 take away 873 * 181. But you're going to have to come back to kilogjles because gibbs, they wanted Gibbs in KJ per mole to the minus one. Okay, so they wanted that in kilogjles. These were all to the mole, but they were in different units.
So there we go. The reaction between ammonia and oxygen was carried out.
Moles affect KP. Um, we can look at a KP question. I'm not sure. Moles do affect KP. What do you mean by moles? um KP you're looking at equilibria at the moles and you do the mole fraction times the pressure mole fraction times the pressure to get the individual pressures.
Um I think you mean I think you might mean does why does pressure not affect KP?
Temperature is the only thing that affects the K's. So whether it's KC, KP, K A, KW, it's the temperature that affects it, not the pressure. And it's because it's a ratio at equilibrium and it doesn't change the ratio.
Yeah.
The ratio will remain the same.
Yeah. So um temperature will affect all the K's KP K A KC they're all going to be affected by temperature so they're all those constants are always at you get given a constant 1.7 * 10us 4 they're always for a specific temperature changing the temperature will affect the position of the equilibrium but um changing the other ones they eventually if you change the concentrations s in a closed system it kind of like eventually the ratio stays the same so it doesn't affect the overall uh position or it doesn't affect KC the constant so the reaction between ammonia and oxygen was carried out at a higher temperature explain how this change affects the value of delta G yeah so oh yeah so just talking of that that's probably why you said it yeah the reaction of meroxy is carried out at a higher temperature how does well it's definitely going to change so if it's carried out a higher temp. Oh, look. We can do it. We can do it. We can do it right here. We got minus 1063. So, what if we let's say what did we get for that over there? 158 we got for so del Okay, it makes delta G more negative. So, it makes delta G more negative. And I would say because t * delta s is bigger.
So let's let's stick it in. We've got the chance to stick it in. We got 600 + 273. Let's make it up and say, right, temperature was 2,00 Kelvin. So I did 2,00 Kelvin* 181. And then minus that from 9005. You're going to get a more negative delta G because you ended up with a bigger T delta S. then you're always going to be taking T delta S from your enthalpy. So the bigger T delta S S is the smaller or the more negative delta G is going to be. So the explanation is it makes delta G more negative because T delta S is bigger and my face is all over it. That's the problem when um when I when I write like that. So yeah.
So yeah, whatever. Okay, let's get off of this. Let's try something else. Let's try something else. What should we try?
Oh no, I'm not doing one of these disgusting foul things. We did one of these the other day and get myself in a twist.
Yeah, these are the answers for the entropy question. -81 - 9005.
Oh yeah, minus 1063.
They accept minus 1060. Someone said something about what's the accuracy needed.
Well, there you go. Any anywhere between minus 1063 and minus 106 entropy, what was this? Delta G becomes more negative. Yeah. Yeah. Yeah. Yeah.
Yeah. Yeah. We said that the entropy change delta G is positive. Okay. Yeah.
Oh, I said that. TS gets bigger. Boom.
Lovely. Right. Let's have a go at a um Got to check these things cuz I don't know when I'm make a calculation error.
Right, let's have a go at this one.
Um inorganic reaction. It's always a niche equation. Uh for me, titration, it's the titration one markers. What?
You don't like the titration one markers?
Okay, we can do that. We can do that. We can do that. We can do that. Yeah, you're right there. The weakest oxidizing agent in the table is going to be Z N2+.
Give the conditions under which the electrode potential of ZN2 plus ZN electrode is minus0.76.
This is just asking you for the standard conditions. So when we use the hydrogen fuel cell, what are the standard conditions? I can't even remember them.
Is it 298K? What is it again?
298 100 kpa, one mole per day dm. Boom.
Nice one. I'm not going to write them down. Um, well done. Well done. Well done. Right. Two half cells involving a species in the table are connected together to give a cell with a EMF of plus0.48 volts. Use the data in the table to deduce the convention all representation of this cell. Write the half equation for the reaction that occurs at the negative electrode. I'm going to move my fat head. Right. Um down here. Here we go. I'm down here for this one.
All right, peeps. Let's see if you can do this one. I'll I'll do a tit some titration questions. Somebody said they don't like the titration questions. I'll lick up some titration questions if you want to go through some of those. The one markers. I I'm happy to go through some titration one markers be a relief from this stuff. Use the data from the table to deduce the conventional representation of this cell. Write the half equation for the reaction that occurs at the negative electrode. So look, two things here give plus 0.48.
Okay, so we need two half cells that are going to give plus 0.48. Use the data convention for this cell, right? So 0.48. I mean, you can already see it, right? Let's take the most negative one to start.
That's going to be the your left hand side.
I hate that thing. I don't know what that's on my electro that's on a couple of my powerpoints. Right, the left hand side is always the more negative one. So if I take the minus0.76 and then oh I can see what it is, right?
Let's do the EMF equation. That's easy peasy, isn't it? It's right hand side minus left hand side. So the way I do this, let me just show you. EMF electrootive motive force equals right hand side of your cell minus the left hand side. Now the left hand side is always the more negative one. So straight away have a look peeps. If I put in on the left hand side, so if I was setting up my fuel cell and I went like this, I put my I put my uh that one there, zinc 2+, and I put Oh god, I can't write. It's all gone.
It's all gone terrible. Right. So, you're going to do leftand side minus the right hand side.
And you can see that that means I'm no right hand side minus the left hand side. So I'm going to go minus 0.28 minus minus 0.76.
And I've put a big fat line over it. So I don't even know what that is. Right?
And if you do that, you're going to get plus I mean you can put that in brackets if you like. You're going to get plus 0.48. So now you need to write the half cell for these two. What is the half cell going to look like for those two?
What's the conventional cell diagram going to do?
So what you're going to do for the conventional cell diagram here now that you've worked out what the half equations are that give you and I mean you can look there another way to do this by the way peeps look you can look what's the difference between those two numbers the potential difference. So you can see the difference is 0.48. If you were climbing up a ladder from 0.76 you've got to go up 0.48 points to get to minus0.28. You can do this. Uh I like you know right hand side minus left hand side business right. So all we got to do is write the half cell for that. Now the left hand side is always put on the left because your half cell has to show the order of how it's set up. So and you put the oxidized forms on the inside and the reduced forms on the outside. So reduce means when that thing has electrons. I don't really want to put the state symbols but you know what you going to do? I don't want to put state symbols. Right. two lines for the um salt bridge and then that is is that cobalt? I think that's cobalt. My eyes cannot see.
What is cobalt? Cobalt 2+ is going to be aquous. It's an ion. And then cobalt is going to be solid. There we go. That's going to be your uh half conventional representation half equation. What's the half equation for at the negative electrode, right?
the negative electrode is always the left hand side and the more negative ion uh the more negative the electrode potential is the one that's going to be getting oxidized and losing electrons.
So we're going to be going Zn to Zn 2+ plus 2 E minus and then we come over here use data from the table to identify a cobalt species that can react with water. write an equation for the redux reaction that occurs and and identify the oxidation product in the reaction.
So, anyone got an idea for the cobalt one Right. Well, we're going to have a little butcher's hook here.
Oh, man. What's this? What's wrong with this pen? This blasted pen. Come on, pen. Do it. All right. Um, what you're looking at here is who can Let's get the granny glasses on because I can't see. These are the the clown show glasses.
Uh, you identify a cobalt species that can react with water. So, how do you work out what's going to react with water? Uh, it's going to be one of these. It's either got to be this one or this one. And where's the water equation? There's the water equation. All right. It's going to react with water. So, water needs to go this way to react. This half equation needs to shift to the left. This has to be the one that is getting oxidized and going in that direction. It's the only thing with water in here. So, that has to be going there as a reactant. That means it can only react with something more positive than it. So, um, you kind of like, and you might be like, "Sir, how do I know that will react with the one down there?" So, I'll show you, right? You can write out the two equations and you write them like this.
Like that. And then um the the more negative one you write on top and the uh less negative one we write at the bottom. I think that's CO3 plus and that's plus E minus. And you write that down there. This is just one way you can do it. And basically the product this one is shifting in that direction. The more positive uh the less positive one, the more negative one is always going to be getting oxidized. And then this one is shifting in that direction. So it proves that water can re can react with cobalt 3+. It can react with cobalt 3+. If I was to if you look at these, this is the more negative up here. I don't know if you can see this, but um if you try to react water with cobalt 2+, it can't react. The reason is cobalt can react with these hydrogen ions and the oxygen to make water. But um yeah, that is going this one is going to shift in that direction and this one is going to shift in this direction this time to make water. So no, water can't react with CO2 plus. So the cobalt species it can react with is CO3 plus. Oh, look. I put my fat head right down where I was writing those equations. Never mind.
Oh my good god.
Boy, it's hard, man. It's so hot. It is difficult. I ain't built for this.
Right. Write an equation for the redux reaction that occurs and identify the oxidation product in the reaction.
Right, guys? You got to be good at combining equations. So you just got to combine this equation essentially.
Combine the equation mish bash bosch and you'll be all right.
The oxidation product well the products are over here and here. What's happened to that? That was + three. That went to + two. So that got reduced.
The water was minus2 and it went to zero. You can't really see that. It's all a mess. Essentially, I can't be bothered to combine the equation. You got to combine those two equations together. Oxygen is the oxidation product. The reason is because it is minus2 when it's in water and then when it becomes elemental oxygen, it goes to an oxidation state of zero. So, it has lost electrons. And you can tell that the cobalt 3+ has been reduced because it's gained electrons. So, that's the reduction product. And the equation we would just need to combine H. We just need to write like um H2O plus CO3 plus E minus goes to CO2 plus plus half oxygen plus 2 H+ plus 2 E minus. And then we could cancel out and blahy blahy blah. Right. Let's zing. And somebody said they want to do titrations um enthalpy of solutions. God too many calculations. Let me just see what else.
What else? What else? I've got about eight billion questions. Hey, talking of that, actually, it's getting quite late, isn't it? I might have to like chip out actually. So, what what do I want to do?
Um, yeah, let me show you the AI prediction.
>> What's your strategy?
>> Don't need any.
>> What's your prediction?
>> Prediction?
>> Yes. Prediction.
>> Pain.
>> Yep. Unfortunately, it could be a painful one. Uh here are my AI predictions. They're not my predictions.
This is AI. Look, I asked AI started telling me, oh, one chemistry tutor has said that um 13.2% is acids, base and buffers, blah blah blah. Weak acid calculations, titration curves, knowing that half the equival equivalence point is if you go to half the equivalence point, the equivalence point is the straight line on a acid base graph where you get the Scurve and you get that straight line.
That's the equivalence point. If you go halfway, that is where the pH is equal to the pKa.
And learning to choose the indicator that falls within the straight range of the curve. So, you must only use an indicator that falls within the vertical line on the titration curve. So, they're saying that that's coming up. They said that transition metals, aquous ions, etc. is coming up. This is what AI said coming up.
So uh yeah why why we give off colors the whole orbit 3d10 orbital splitting electrons moving to the higher then falling down in specific wavelengths of light all of that business being transmitted or whatever uh thermodynamics born harbor cycles actually for about four years born harour cycle was the opening question to the Alevel paper and hasn't been the opening question for like 3 years, two, three years. So, um maybe people are predicting the opening question will be the Born Harbor cycle.
Boom. Um and then we just did that Gibbs free energy. Obviously, if you're more negative, the more negative you are, the more spontaneous the reaction is likely to occur. This comes up all the time, time of flight. So, that's not really a big one to predict there. I mean, KC, KP, yeah, they're quite easy. uh periodicity of group three.
Okay. Um and then yeah, this is weird. They said the pract I I asked AI specifically based on the historical data what practicals are coming up said volutric analysis and and qualitative analysis of the group two cations ions and the negative halide ions. Pretty crazy, right?
One thing that might be worth you studying that you've got time for to study is um obviously you can just ask AI say you should use like Google AI but don't get me wrong they can make mistakes Google can make mistakes but if you say to AI like make I just said it two seconds before the live I just said make me a table of all the complex iron color changes in Alevel chemistry for AQA and it did this. So AI just come up with this. Obviously it cut off half the stuff because the thing's bloody dumb.
But um yeah, so it did these change. It just came up with a table like that. Now you can find that in any textbook obviously, but AI just did that. And what you notice is the only time you add sodium hydroxide in excess the precipitate. When you add the sodium hydroxide, you get the specific precipitate. So when you add sodium hydroxide to iron 2+, you get a green precipitate, you get a blue precipitate with copper, easy, and a brown precipitate with iron 3+ and a white precipitate with aluminium. The only time that precipitate disappears is with the uh is in excess sodium hydroxide is with the aluminium. It will go colorless and that's part of the amphoteric nature, the acid base nature of aluminium. Now um then you notice you alternatively can add the ammonia and you get the exact same substitution product with the lians um whether you add ammonia or sodium hydroxide. Look the product is exactly the same. So there's not a lot to revise there. The product is exactly the same and the colors are exactly the same. Green, blue, brown and white with ammonia. So that's nice to know. And the only time that adding excess ammonia makes any change is with the copper where it goes to deep blue and you get those NH3 lians. So um you get two H2O lians and four NH3 lians. And then they say add the sodium carbonate. And again, there is only two that give off carbon dioxide gas. And that's iron 3+ and aluminium 3+ that give off carbon dioxide gas. I don't know why there's a minus there. There shouldn't be a minus there, but you get the coup, right?
There should be a three there, but whatever. Yeah. So, you can ask AI to do some stuff for you and it can help you with your revision. That's just one part. Of course, you got to know all those dirty venadium ones and all the color changes for that. But this is just one part of the complex iron changes you need to know. Um, so what I would say is that is predicted to come up and it has come up in the past with like a six marker on that. So yeah, what I've learned with AI is um yeah, last year to didn't come up. Maybe it won't come up for a few years actually. To's been quite consistent. Um it might not come up for a few years, but they've AI predicts that it's coming up with AI.
You got to be really specific. So, I only use Google cuz I'm sitting here and I'll just go AI mode next to the Google search bar on Chrome and then I type in but you got to be specific. So, you got to say something like based on all the AQA A level chemistry paper ones from 2016 I know there wasn't one in 2016 but you get get what I mean from 2016 16 whatever to 2026 say what is the most likely uh heavily assessed questions that will come up. So you have to be really specific. Take your time over the thing that you put in. Take your time over it and say like what is the likely based on all the data from all the previous past papers which is the most likely practical to be heavily assessed in chemistry A level paper one. Now it might be talking out his backside because half the time AI is talking out his backside but it's worth a shot and then give you something to focus your revision on. give you a little bit of hope. They might turn out to be absolute cabals. Um, I'm going to shoot off in 10 minutes. Did anyone want to go through?
Someone was saying something about Oh, that's group two. Someone was saying something about titrations. I want to have a look at titrations. That's group two. That's unlikely to come up. Oh, I'm not doing that. We already did that. We already did that. We already did that.
Oh, s that. That's the old um enthalpy of solutions one where you got to draw the That's that one where you got to draw the practicals and the extrapolate and the line of intersection. Uh I'm not going over that now. I'm too tired, man.
Too tired.
Uh someone was saying titration one markers. I might just do a couple of titration one markers. you know that uh all you need to know uh is that at half the equivalence point. So I think I'm going to shoot off guys. I am knackered and then I'm going to I'll come back on tomorrow and do some more crap. But yeah, all you need to know is that at half the equivalence point. So, so as you're adding your sodium hydroxide to your hydrochloric acid or something like that, right, you get there and you get there. Whatever. I know it's a bit of a crap graph, but this is going to be your volume of like NaOH added that you're adding. So, you're increasing the amount of NaOH added and then you got your pH up here or whatever whatever's up the side there. I can't even remember what's going to be up the side there. Uh, might be the conchk. What's it concentration up the side? Whatever. But, um, what you want to do is at half the volume. So when you go down to say I've added 50 cm cubed to get to this point where we reach neutralization then it's telling me that at 25 cm cubed along here half that the pH which I can read off of here. So it might be 3.1 is equal to the pKa. So now you know the pKa is 3.1 right? If the pH is equal to the pKa, then we know that pKa equals 3.1. And that means I can do my 10 the minus 3.1 and I can give myself the Ka and then I can use that in an equation because they want me to calculate the Ka. So I can do something in that with that and calculate whatever. Okay. So that's just like a brief little skiff scuffle of what they're saying about that.
Okay. And how would you know the indicator is suitable? Yeah, it's not very good when I got this crappy graph on here. I should really go to a blank page, shouldn't I? So, the indicator is going to be suitable.
Now, certain graphs, there's no suitable indicators. So, this is like your pH and this is your volume added. If the graph is like that then nothing is going to be suitable because there's no definite vertical point. You need a definite vertical point otherwise there is no indicator suitable and that will happen with like weak acid and weak base. So think about it a weak base is going to start here and if you're neutralizing it sorry a weak acid is going to start here like pH3 or something and then if you're neutralizing it with a weak base it's going to happen so slowly that there's no indicator that works. Um, now you're looking at the vertical section.
So here, let's say this is pH10 and this is pH4. Then I need an indicator. Indicator X changes color between 6.9 and 8.0 whatever that is suitable. So if that changes color, if the color changes completely within that range, it is suitable. If the color changes anywhere outside of that straight line, the indicator is not suitable. So you have to check the range. This might change between 6.9 and 7.25 that changes color. X would be suitable, but something that changed color at like PH5 to PH2.5 would not be suitable because it means it would start changing color even when we haven't achieved neutralization and that's no good. So, it's got to fall within the vertical section of the graph.
Yeah, I could have just said that, right?
Yeah, I could have just said that. Would have been a lot quicker.
Oh my days. Let's have a look. One second.
I had What the fudge is this? This is titrations.
Oh my god. I'm not doing fudge that.
Um boy, it is so damn hot.
Um, oh my gosh, it is too hot.
This is just outrageous.
I think I might have to dash off in two seconds. Let me just somebody said earlier about doing some titration questions. So um I just thought let me do two seconds of these. So um so someone was mentioning earlier about these before adding the solution from the bureete in the rough titration there was an air bubble like that below the tap. At the end of this titration the air bubble was not there. Explain why this air bubble increases the final preoret reading. So remember that part below the tap is included in the measurement on the on the scale on the the bureete. So if I think I've added 25.5 cm cubed of hydrochloric acid in my titra, I've actually haven't cuz the air bubble has a volume and which was not acid. So I get an overread of my titra. So you're going to get an air bubbles trapped in the tap will give you an overestimation of your titra. During the titration, the student washed the inside of the conicle flask with some distilled water. You have to know why that is not a bad thing. You are allowed to wash the conicle flask with distilled water because your unknown is going in there. And it doesn't really matter actually what goes in there. You're not changing the moles.
Now you might say, "But sir, I'm changing the concentration. I'm diluting it."
But the reaction is between the moles of acid in the bureete and the moles of alkali in the conical flask. And so it's the ratio we're trying to find. Uh we want to know how many moles were in the conicle flask. And that's not going to change. So, it doesn't matter how dilute I make the conical flask, it's going to require the same amount of moles of hydrochloric acid to neutralize it. So, I'm going to have to add the same exact amount to neutralize it because you haven't changed it. But if you wash the bureete with water, why will this mess up your results? Because this will dilute your known concentration. So when you come to do your calculation, your again, if you've diluted it, you will have got an overestimation of your titra because you will have added extra um a bigger volume than you needed to because you don't really know the concentration anymore that that you decreased the concentration. So you'll have had to add too much and it's going to mess the whole thing up. So essentially, you need to remember that stuff. Uh what is this? The student identified the use of the bureete as the largest source of uncertainty and use the same operator to suggest how the student could improve to reduce the percentage error. Oh, I got the answers down there.
Thank thankfully. Okay, so if you have to use the same bureete, if you're not allowed to change your equipment in chemistry, you can reduce the percentage uncertainty by using larger amounts.
How can we use more acid? If I've said I've got half a mole of sodium hydroxide in the conical flask, half a mole, I'm going to need the same amount of moles of it's a one to one. I'd need 0.5 moles of HCl to neutralize it. I want to increase the amount of HCl I have to add. So I use a weaker concentration will give you more accurate results.
Well, it will give you less uncertainty.
If I use a smaller concentration, I will add a greater volume of acid. So if it took me 20 cm of 0.5 moles HCl to dilute to neutralize the sodium hydroxide if I if I make the HCl instead of 0.5 if I reduce the concentration to 0.25 moles per decimeter then maybe I now need 40 cm double the amount. Now the percentage error will stay the same for my equipment because it's got that built into the measurements but now it's a smaller oh sorry the the un the the the measurement of uncertainty is the same in the equipment but the percentage error gets smaller because I'm now doing the uncertainty over 40 rather than over 20 if you see what I'm saying. So yeah, [sighs] so basically use a larger um volume in your bureete, which means decrease the concentration of the acid.
Boom. Right. Whatever. Um what else?
The student used a wash bottle containing de I've got the answers right here, which is nice. The student contained a wash bottle containing deionized water when approaching the end point to rinse the inside of the conical flask. Explain why this improved the accuracy. Obviously on the sides of the conical flask you will have moles of your reactants. This ensured that all of the reactants reacted. Okay. So it improves the accuracy. It ensures that I don't know what was in there. Acid or whatever. It ensure the deionized water pushes the reactants down the flask together to ensure that all the reactants react. Give the meaning of the term concordant titras. What is it within 0.1 on AQA? Let's have a look.
0.1. There you go. AQA. idea that ensures oh I didn't know what the reactants were but ensures all the reactants in the mixture react or come into contact anything that's on the side and 0.1 is that all right let's go let's go what else what else I'm not going to do a fat question I'm just going to finish with a couple more little oh god NMR let's get away from NMR I've just seen NMR down the page let's get away from some of that uh which is the correct Oh yeah, the pipet that's correctly filled. Where should the meniscus be?
The bottom of that line. It should be resting right there. So the answer would be C. If you didn't know, it should be resting there.
Huh? Oh, I don't understand that. Oh, it's in this context.
Sort that. It's too much to read. I can't be reading that right now.
Concordant results. you guys are cool with that. Um, I can come back and do some of these if you really are interested in me going through titrations. Uh, uh, what's this? Standard solution. So, when you're making a standard solution, weigh a clean, dry, empty container on a balance 2DP. Add 2.5 g of solid sodium carbonate to the container. Tip the solid into a beaker. Add approximately 100 cm cubed of water to the beaker and stir until all of it has dissolved. Pour in a 250 volutric flask. Add distilled water until the top of the miniscus is level with the graduation mark. What three ways could you improve this experiment?
I'm finishing after this guys after this question.
>> [sighs and gasps] >> Well done. Do the before and after weighing method to find the actual mass that's been added. You can call that the before and after weighing method.
Add approximately 100 cm cubed of distilled water to the beaker and stir until all the solid has dissolved.
Uh they didn't add their washings. Did you notice that they didn't add their washings? So they should have tipped their washings into the volutric flask.
They did make it up to the level, but what else didn't they do? Once you've made it up to the level, what you supposed to do?
Yeah, add the washings. And then what are you supposed to do right at the end that they didn't do? They made it up to the graduation mask. What else should they do?
And also, how can they make sure they're accurately? How can they make sure they get this accurately? Might be on the mark scheme. Guess what? I got the mark scheme for this one. So, it's all good.
How can they make sure the meniscus they don't override? Look, I've overread the meniscus there. How can they make sure that drop by drop? We're near the graduation mark. Nice. Oh, yeah. Invert it. So, put a stopper in and invert.
Nice. So, I reckon there's like four things there. I reckon the before and after weighing method. Add your washings, use a dropping pipette close to the graduation mark, stopper and invert. And you might be asked to explain those.
So using a dropper close to the graduation mark ensures that the meniscus sits on the line on the graduation line and doesn't go over because that would be pear-shaped. then you would have a more dilute concentration than you wouldn't have a known concentration if it goes above the graduation line. Um the other thing is the before and after weighing method allows you to know exactly how much solid you've added because some solid will be left behind on the walls of the container, the weighing boat or whatever. Um what else? You stopper and you invert.
Why? That's to spread evenly so you have the same concentration throughout the volutric flask. Because if I now take 25 cm sample to titrate, you know, if it's near the bottom, the solid may it may be more concentrated there. So, you want to shake, invert a few times, make sure it's evenly spread so the concentration is the same everywhere, basically. Yeah, those are you don't need to explain that for this, but yeah. Well, I'm going to finish there, guys. If you want, I'll go I don't mind doing NMR tomorrow if you guys want.
Oh, yeah. Pipet.
Oh yeah, they just said add approximately 100 cm. Do they need a pipette? Maybe. Maybe. Let's see the marks. I don't think a pipette, but you know, could slap myself in the face.
There's the answers, guys. Yeah, I don't think a pipette because with this one, you don't really need a pipette because you're just adding the water to like dissolve everything and then you're bunging it in a volutric flask and making it up. You're not transferring an exact amount. So, for this one, we don't need a pipet.
I don't mind doing NMR tomorrow if you if you want and period 3. All right, we'll do a little bit of NMR and period 3 tomorrow.
All right, nice one. We'll do some NMR and period 3.
Nuclear magnetic resonance spectroscopy and period 3 we can do a bit of. All right, cool.
Oh yeah, we could do could do that. I might do that on the day before actually.
I'm not sure yet what I'm doing. So, probably be late, like 7, 8 o'lock.
All right, guys.
I will see you later.
Yeah, enjoy it when it's cooler anyway because it's too hot in the daytime. So, you know.
Oh my gosh. A levels. What you going to do? What you going to do?
>> I can't handle the stress no more, dog.
These damn A levels.
You're still here.
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