Markovnikov rule: In addition reactions to unsymmetrical alkenes, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component adds to the carbon atom with more hydrogen atoms bonded to it. Therefore you got the right answer but your logic was reverse. You should have been looking at just the two carbons making up the double bond (1 vs 2 Hs) instead of the two sides of the whole molecule (6 vs 2 Hs)
@@jesseosbourne it gets more complicated, there are such things as anti-markovnikov addition as well, based on if there is a more stable intermediates - 1,2 hydride shifts are an example of this I think - I'm weak in this area
9:45 for anyone watching this, it's actually the other way around. Markovnikov's rule states that the hydrogen joins the carbon that has the most hydrogens
Organic chem reactions are a nightmare, I'm also not strong in this area, good to see you scored 100 with incomplete knowledge of organic chem reactions, I think your advice of looking for patterns is prob best, I 'may' go back and revise some reactions though it's prob not the best use of time - I even tried to watch the crash course organic chem videos and they were too complicated
I'd say reviewing the different reaction types is still worth it if you're not comfortable with them, but use it as a chance to continue to identify visual patterns and summaries
Hello Jesse, thank you very much for the videos, very informative !! Could I please ask regarding to video time at 36:39, on the ICE table, why is it 1mol times half a litre, when the question says 1 litre? I have been watching this many times and still could not understand why. Thank you very much for the explanation.
Hello Jesse, great work with your crash course chemistry videos! I find them even as a science background student good ways to refresh old concepts. Regarding the Henderson hasselbach question on asking for the conditions that would give pH of over 16, shouldn’t the Pka be around 13 rather then 14, as Pka and Pka values add up together to 14?
We wouldn't want to assume that pKa is around 13. We can only say that pKa < 14 so it could still theoretically be pKa = 13.99999 which is effectively 14 I avoided using a pure inequality in the video to avoid confusion but will do it here so we can see how the maths plays out pKa < 14, pH > 16 pH = pKa + log(A-/HA) so pKa + log(A-/HA) > 16 ( 16 log(A-/HA) > 16 - 2 A-/HA > 100 Technically, pKa and pKb are mathematical indices representing orders of magnitude of relative dissociation so they can actually be negative in special circumstances and this would make a mess of the maths for this question, but this would be beyond GAMSAT knowledge Hopefully that clarifies things :)
Hi Jesse I Just had a question regarding the ICE table. I realised in the C you have Negative x on the L side but then 2xs on the R side. I'm not sure if I misunderstood but as far as I know you cant lose 1 x and add 2 xs on the other side. If you can please maybe clarify? Thank you so much for these videos, it has helped me remember so much from over 7 years.
Hey Gina, yep so this is because of the difference between the amount of a substance in mole and the mass of a substance in grams. Although there is a conservation of mass ie. If we lose or consume 1g of reactants we should produce a total of 1g of products (if there are more than one product then their total would be 1g), there is no conservation of moles between different species. With moles it's a little different because a mole is a measure of particle numbers. I'll exclude the complexity of avagadro's number for this explanation just to simplify the explanation. So if we have a reaction of A2B + 2C --> D + 3E then this means that 1 mole of A2B will react with 2 mole of C to produce 1 mole of D and 3 mole of E. Overall this is a net loss of 3 mole of stuff from the LHS and a gain of 4 mole of stuff on the RHS, however the mass lost on the LHS would be the same as the mass gained on the RHS to conserve mass. This is because mole is dependent on the molar mass of each compound and so this also means we wouldn't normally compare totals of moles of multiple compounds because they would not be equivalent in mass. eg. 1 mole of H weighs 1g but 1 mole of C weighs 12g. So in the scenario in the example in the video, the differences in the amount lost and gained on each side is due to the coefficients of each species and although it looks like we're making more stuff than we reacted, the mass would be conserved. What constitutes 1 mole on the RHS will be 'lighter' than 1 mole of the LHS because of the difference in mass number. Hopefully that clears things up for you and anyone else who might be reading this and wondering the same thing :)
hi laya, I'd say the depth that I go into in the crash course series is more than enough for GAMSAT level. The trap for NSB sitters is that you 'don't know what you don't know' and therefor assume that if a topic is challenging you must need to study it more deeply and broadly but I'd say it would largely be wasted effort. If you're still not confident with acid base chemistry after watching these videos, I'd suggest checking out some other resources that explain it but at the same level as my method might not have been he right one for you. Khan academy is a good option. The world of core sciences is vast so if you're studying without setting a boundary on how far or how deep you need to venture, it's easy to get lost in the weeds so use these videos as a way of understanding the boundaries of what is relevant when you study with other resources that are not GAMSAT specific. best of luck!
Hi Jesse, I had a question with regards to the second last stoichiometry example question- why did we take 0 moles of water? We need water for the dissociation/reaction to occur and it is only when the acid is dissolved in water in dissolutes in A- and H3O right? Is it because we normally don't need to consider it? I am a little confused. Thanks in advance!!
Yep there is water present in the reaction but it is acting as a solvent (the thing we are dissolving the other compounds in) so we don't track the amount of water in our calculation. The reason for this is that concentration is a ratio of amount per unit of volume of solvent therefore we can't theoretically measure the concentration of water dissolved in itself! So we leave H2O blank in the ICE table but it technically is present, just not a part of the calculations.
Hey there! Unfortunately I'm not able to for a couple of reasons. I actually don't have access to these materials and copyright restrictions prevent me from showcasing privately owned material on screen. I'd also say that the Des stuff is probably out of date by this point, given the evolution of the test. I remember sitting back in 2012 when Des was all the hype and even then it was questionable as to whether it was truly a reflection of the ACER material. The test is certainly different now. Sorry to be giving a "no" as an answer, it's not something I usually do, but I'd say you'd be better off recycling ACER material than completing Des questions for Section 3 :)
@@jesseosbourne Thank you for doing this video, Are you in plan to talk about cope and claisen rearrangement topic. Also ACER material is very limited. What else would you recommend or just repeat ACER material?
Hey Kamal, I'd probably say that this particular reaction pathway is unlikely to come up in the GAMSAT and if it did, the basics would be explained in the stem. The pathways I've included in this video are the ones that I think people should have some familiarity with in order to reason with the GAMSAT style questions. You're absolutely right, the ACER material is pretty limited, unfortunately, which is part of what has motivated me to make the resources page in the first place. I'd say that recycling ACER material is a better use of time than non-ACER material (I'll be going into a little depth about this in a video next weekend). Outside of ACER material, I worked through UK BMAT tests but these are definitely very different to GAMSAT (more than 4 options, multiple possible correct options, a little more theory based than reasoning based, shorter stems etc). I used these to practice my pacing www.admissionstesting.org/for-test-takers/bmat/preparing-for-bmat/practice-papers/
In the ICE table, why would you get an increase in H3O+ if the solution is not suspended in H2O? To me it seems in the reaction, that there aren't any free oxygen atoms to make H3O+ and all you would get is an increase in H+? I understand that H30+ and H+ are basically interchangeable as a marker for decreased pH so am i technically correct but its just a semantical error, or am i missing something more serious? haha thanks! bloody love the videos. I appreciate them immensely.
Hey Keelan, I assume you're referring to the yellow ICE table around 35 mins? Yep so H+ and H3O+ are generally used interchangeably in acid-base chemistry although the reason why H3O+ is produced is because the mixture is an aqueous solution, suggesting the acetic acid/acetate mixture is dissolved in water. The H2O is acting as a base in the presence of acetic acid and is accepting protons to form H3O+
Hi Jesse, making my way through your crash course videos in preparation for September as a trial run and most probably March next year. I noticed in this video you mention you will go through organic and inorganic redox reactions in another video but I can't find that content anywhere... Would you recommend revising this topic again outside of what's covered in this video i.e. is it a prolific concept covered in the GAMSAT that we should have well understood? Much appreciated in advance.
Hey Mitchell, hmm that's a good point, I may have actually missed doing a video on redox! I wouldn't say it's overly useful as it's much more limited in its ability to be abstracted into reasoning problems. So that you can keep moving on with your study, I'd say the most important aspects to understand are: 1. The concept of an oxidation number and how to calculate it based on a compound formula and charge 2. Knowing that oxidation is an increase in ox number (or a gain of O or loss of H or loss of electrons) and reduction is the reverse. 3. The overall 'potential' of a redox reaction or cell is the Eº(reduction) - Eº(oxidation) assuming the Eº values are from an electrochemical series with reactions written with the reduction reaction going forward Anything beyond this would likely be introduced in the stem for you. I've seen some des questions and some from prep companies that show anodes/cathodes/salt bridges etc and while yes they are also parts of redox, this is all very technical and is more aligned with memorised knowledge in a Y12/uni chem subject so is really just lazy question writing that doesn't involve reasoning. I'll see if I can get a short redox crash course video done ahead of September but in the meantime you could also check out some of my sample question sets as I think a few of them involve redox chem and the solutions videos will help in explaining the relevant concepts :)
Hi Jesse, just wondering the importance of knowing to use ICE calculations for the actual gamsat testing. I know this is more of a content video, but would gamsat expect us to know how to use ICE? Thank you
I'd say it's still pretty unlikely that you'd need to actually draw up the table and work through it so exactly as this would be largely repetitive and would also advantage those that have studied chemistry formally. That being said, anything I cover in these crash course videos are things that I would consider either directly required or useful for problem solving. ICE tables are a good methodical way of dealing with equilibria, especially for people less familiar with chemistry concepts and it also helps demonstrate that chemical equilibrium is not the same as a reaction in which all reactant is consumed. These would be the key things that are important to consider in equilibria questions
Hey Jesse, awesome videos man... I'm just stuck at 31:00. If you're adding a base to the acid buffer solution, wouldn't it be shifting to the right since the HA is absorbing OH- to counteract the change in pH by converting it into a weaker base (its conjugate base)? Is it the case that you need to add an extra step by saying it will then shift back to the left since HA has dropped? Especially because if it favours the back reaction like you said, then H3O+ would get consumed when a base is added thus increasing the pH even further which I can't make sense of (I could be wrong by Chem is rusty since 1st year uni)
Hey Zaid, I think I see what you mean but there might be a couple of confusions surrounding the phrasing "shifting to the left/right" with regard to the equilibrium position and the relative strengths of conjugates, that I'll clarify. 1. We describe the relative position of the equilibrium as "shifting left/right". So if a system favours the forward reaction (rightward arrow) to reachieve equilibrium after an adjustment, then we say that equilibrium has shifted to the right. 2. With conjugate pairs acids have conjugate bases and vice versa, and weak compounds have strong conjugates and vice versa. So in this instance, a weak acid buffer is used (which would have a strong conjugate base). The reason for using a weak acid as a buffer is that 'weak' means it will not readily dissociate and donate its proton (ie. it maintains a high amount of HA). 3. The achievement of equilibrium is through favouring the forward or back reaction until the ratio of the concentrations of LHS and RHS are such that their reaction quotient can equal the equilibrium constant, K again. This means that equilibrium is a mathematically determined position and does not mean that LHS and RHS are equal in concentration. So, when an equilibrium system such as a buffer system is 'adjusted' or a change is introduced, it will really only shift in one direction in order to re-establish equilibrium and bring its reaction quotient back to the value of K. This also means that when a system reachieves equilibrium, it does not reachieve the same concentrations as before but rather it reachieves the same ratio of concentrations. So in this case, what is happening is We have an existing buffer equilibrium system HA + H2O --> A- + H3O+ We then add some base (OH-) which increases the OH- concentration and threatens to increase the pH (you're also right that an alternative way to consider this is that the introduced OH- will react with the H3O+ and lower the H3O+ and therefore threaten to increase pH) However, the acid buffer (HA) will work to resist this potential change in pH. Because the H3O+ is being consumed, it throws the system out of equilibrium and the system favours the forward reaction (shifts right) with HA being consumed to produce more H3O+ until the ratio (Q) of RHS and LHS reaches K again. As it does this, H3O+ being produced helps offset the pH increase so that it's only a very small change. Now, yes, HA did ultimately get consumed in the process, but it being a weak acid means that very little of it got consumed (this is the ideal characteristic of a buffer and is the reason we always choose a weak acid/base for the purpose). This might make you think that the system should now try to push back to replenish the lost HA but there are two reasons why it won't. 1. Where would the 'ingredients' for this HA's production come from? It would have to come from the RHS A- and OH- but you can't take from here or else you throw the system out of balance which brings up the second reason; 2. It's already in a new state of equilibrium so it is happy as it is. Hopefully this clears things up
@@jesseosbourne Thanks for the quick reply! This explanation makes perfect sense now... Thinking in terms of Q and K clears things up too. So if I'm understanding correctly, adding a base (to weak acid buffer) would ultimately favour the forward reaction until the reaction quotient Q reaches K again (close to the 'original' ratio). I think I misunderstood your wording in the video since you had the arrow pointing to the backward reaction when Base is added but I understand the original meaning now. Thanks again!!
I'm assuming you mean the yellow question at around 35 minutes? Yep, absolutely but obviously I'm looking to demonstrate an alternative way for people to problem solve in this area as I follow up with another example that then uses the Henderson Hasselbalch equation so that people can choose what works best for them
Hi Jesse, for the example of figuring out the pH of the buffer, I am a little confused by the ICE table you used. In the table are you doing number of moles and not concentration? I was looking at a chem textbook and they used concentrations in the example rather than moles as you seem to. I did the question before watching and used concentration and ignored/didn't see the bit which says you produce 1L from equal parts, and got the same answer. Why did you use the moles - are there other times where it would give diff answers? I hope this made sense and thanks for the vid!
Great question, Priya! So when doing ICE tables, you are effectively working out the 'amount' of each compound reacting or being produced, and the amounts at equilibrium. We always need to use the number of mole of each when adding and subtracting because concentrations are actually a ratio of moles to volume whereas the number of moles are absolute amounts. Another way to think about this is that when a reaction occurs, it reacts certain amounts of mole of compound rather than certain amounts of concentration. A non-chemistry way to think about this is to consider how we add 20% of 100 and 30% of 200. The % are ratios (like concentration) and the figures 100 and 200 are like volumes. We could not simply ignore the volumes and just add 20% + 30% = 50% because the volumes are different. We would instead have to first determine the absolute value that they represent 20% x 100 + 30% x 200 20 + 60 = 80 The reason why you may have seen other videos using concentration is that they may have been using a system with a total volume of 1 litre and so since n = cV and V = 1, then we have the special case that n = c and so we can kind of shortcut the maths and use the concentration. I use this same volume in this question, but still calculate the number of mole at the start anyway so that people are aware that outside of this special case, you should always use the number of mole in ICE tables.
Hey Emily, that's my bad! I actually miscounted when I filmed this one haha. This is actually crash course #3. I've numbered the video titles now to avoid any confusion for people. It should be 1. chemical equilibria and stoichiometry 2. intro to organic chemistry 3. organic chemical pathways and buffers 4. Stereochemistry
Markovnikov rule: In addition reactions to unsymmetrical alkenes, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component adds to the carbon atom with more hydrogen atoms bonded to it.
Therefore you got the right answer but your logic was reverse. You should have been looking at just the two carbons making up the double bond (1 vs 2 Hs) instead of the two sides of the whole molecule (6 vs 2 Hs)
Thanks for the correction! I must've somehow gotten through all of uni chem without this misunderstanding ever catching me out haha.
@@jesseosbourne it gets more complicated, there are such things as anti-markovnikov addition as well, based on if there is a more stable intermediates - 1,2 hydride shifts are an example of this I think - I'm weak in this area
9:45 for anyone watching this, it's actually the other way around. Markovnikov's rule states that the hydrogen joins the carbon that has the most hydrogens
Good catch! Someone else called me out on this too actually haha
@@jesseosbourne Does that mean minor product will be the major one?
@@sonidhillon5244 Yep! It should be the other way around from what I mention in the video
The Markovnikov's rule helped a lot. I thought of it like a concentration gradient of Hydrogens.
Awesome! Just watch out because I actually stated the rule the wrong way round! I've pinned a comment from someone with the correction
Organic chem reactions are a nightmare, I'm also not strong in this area, good to see you scored 100 with incomplete knowledge of organic chem reactions, I think your advice of looking for patterns is prob best, I 'may' go back and revise some reactions though it's prob not the best use of time - I even tried to watch the crash course organic chem videos and they were too complicated
I'd say reviewing the different reaction types is still worth it if you're not comfortable with them, but use it as a chance to continue to identify visual patterns and summaries
Jesus I don't know how I'd get through this stuff without your videos
Just realised you have a resources page, so happy! This vid was great...again!
Oh yeah, there's plenty to work through there! Glad it was helpful :)
Hello Jesse, thank you very much for the videos, very informative !! Could I please ask regarding to video time at 36:39, on the ICE table, why is it 1mol times half a litre, when the question says 1 litre? I have been watching this many times and still could not understand why. Thank you very much for the explanation.
Hello Jesse, great work with your crash course chemistry videos! I find them even as a science background student good ways to refresh old concepts. Regarding the Henderson hasselbach question on asking for the conditions that would give pH of over 16, shouldn’t the Pka be around 13 rather then 14, as Pka and Pka values add up together to 14?
We wouldn't want to assume that pKa is around 13. We can only say that pKa < 14 so it could still theoretically be pKa = 13.99999 which is effectively 14
I avoided using a pure inequality in the video to avoid confusion but will do it here so we can see how the maths plays out
pKa < 14, pH > 16
pH = pKa + log(A-/HA) so pKa + log(A-/HA) > 16
( 16
log(A-/HA) > 16 - 2
A-/HA > 100
Technically, pKa and pKb are mathematical indices representing orders of magnitude of relative dissociation so they can actually be negative in special circumstances and this would make a mess of the maths for this question, but this would be beyond GAMSAT knowledge
Hopefully that clarifies things :)
Thank you, Jesse. great explanation!
No worries, glad it was helpful :)
at 39:00 you had x=[OH-] = 9*10-6. how does that equal [H+]? Thank you for the videos!
i though [H+][OH-] = 10^-14
Hi!! Thanks so much for making this video. I was just wondering why it is that we can't substitute hydrogens that are attached to a carbon? Thanks!!
how did you estimate log 10 2 to be around 0.3 without using a calculator at 41:52?
Hi Jesse
I Just had a question regarding the ICE table.
I realised in the C you have Negative x on the L side but then 2xs on the R side.
I'm not sure if I misunderstood but as far as I know you cant lose 1 x and add 2 xs on the other side. If you can please maybe clarify?
Thank you so much for these videos, it has helped me remember so much from over 7 years.
Hey Gina, yep so this is because of the difference between the amount of a substance in mole and the mass of a substance in grams.
Although there is a conservation of mass ie. If we lose or consume 1g of reactants we should produce a total of 1g of products (if there are more than one product then their total would be 1g), there is no conservation of moles between different species.
With moles it's a little different because a mole is a measure of particle numbers. I'll exclude the complexity of avagadro's number for this explanation just to simplify the explanation.
So if we have a reaction of A2B + 2C --> D + 3E then this means that 1 mole of A2B will react with 2 mole of C to produce 1 mole of D and 3 mole of E. Overall this is a net loss of 3 mole of stuff from the LHS and a gain of 4 mole of stuff on the RHS, however the mass lost on the LHS would be the same as the mass gained on the RHS to conserve mass. This is because mole is dependent on the molar mass of each compound and so this also means we wouldn't normally compare totals of moles of multiple compounds because they would not be equivalent in mass. eg. 1 mole of H weighs 1g but 1 mole of C weighs 12g.
So in the scenario in the example in the video, the differences in the amount lost and gained on each side is due to the coefficients of each species and although it looks like we're making more stuff than we reacted, the mass would be conserved. What constitutes 1 mole on the RHS will be 'lighter' than 1 mole of the LHS because of the difference in mass number.
Hopefully that clears things up for you and anyone else who might be reading this and wondering the same thing :)
hi jesse! amazing video btw
hi laya, I'd say the depth that I go into in the crash course series is more than enough for GAMSAT level. The trap for NSB sitters is that you 'don't know what you don't know' and therefor assume that if a topic is challenging you must need to study it more deeply and broadly but I'd say it would largely be wasted effort. If you're still not confident with acid base chemistry after watching these videos, I'd suggest checking out some other resources that explain it but at the same level as my method might not have been he right one for you. Khan academy is a good option.
The world of core sciences is vast so if you're studying without setting a boundary on how far or how deep you need to venture, it's easy to get lost in the weeds so use these videos as a way of understanding the boundaries of what is relevant when you study with other resources that are not GAMSAT specific.
best of luck!
Thankyou Jesse !
Hi Jesse,
I had a question with regards to the second last stoichiometry example question- why did we take 0 moles of water? We need water for the dissociation/reaction to occur and it is only when the acid is dissolved in water in dissolutes in A- and H3O right? Is it because we normally don't need to consider it? I am a little confused.
Thanks in advance!!
Yep there is water present in the reaction but it is acting as a solvent (the thing we are dissolving the other compounds in) so we don't track the amount of water in our calculation. The reason for this is that concentration is a ratio of amount per unit of volume of solvent therefore we can't theoretically measure the concentration of water dissolved in itself!
So we leave H2O blank in the ICE table but it technically is present, just not a part of the calculations.
Hey Jesse, do you think you could make some Des O'Neill's section 3 practice questions videos with explanations? Thanks.
Hey there! Unfortunately I'm not able to for a couple of reasons. I actually don't have access to these materials and copyright restrictions prevent me from showcasing privately owned material on screen. I'd also say that the Des stuff is probably out of date by this point, given the evolution of the test. I remember sitting back in 2012 when Des was all the hype and even then it was questionable as to whether it was truly a reflection of the ACER material. The test is certainly different now.
Sorry to be giving a "no" as an answer, it's not something I usually do, but I'd say you'd be better off recycling ACER material than completing Des questions for Section 3 :)
@@jesseosbourne Thank you for doing this video, Are you in plan to talk about cope and claisen rearrangement topic. Also ACER material is very limited. What else would you recommend or just repeat ACER material?
Hey Kamal, I'd probably say that this particular reaction pathway is unlikely to come up in the GAMSAT and if it did, the basics would be explained in the stem. The pathways I've included in this video are the ones that I think people should have some familiarity with in order to reason with the GAMSAT style questions.
You're absolutely right, the ACER material is pretty limited, unfortunately, which is part of what has motivated me to make the resources page in the first place. I'd say that recycling ACER material is a better use of time than non-ACER material (I'll be going into a little depth about this in a video next weekend). Outside of ACER material, I worked through UK BMAT tests but these are definitely very different to GAMSAT (more than 4 options, multiple possible correct options, a little more theory based than reasoning based, shorter stems etc). I used these to practice my pacing
www.admissionstesting.org/for-test-takers/bmat/preparing-for-bmat/practice-papers/
In the ICE table, why would you get an increase in H3O+ if the solution is not suspended in H2O?
To me it seems in the reaction, that there aren't any free oxygen atoms to make H3O+ and all you would get is an increase in H+?
I understand that H30+ and H+ are basically interchangeable as a marker for decreased pH so am i technically correct but its just a semantical error, or am i missing something more serious? haha thanks! bloody love the videos. I appreciate them immensely.
Hey Keelan, I assume you're referring to the yellow ICE table around 35 mins?
Yep so H+ and H3O+ are generally used interchangeably in acid-base chemistry although the reason why H3O+ is produced is because the mixture is an aqueous solution, suggesting the acetic acid/acetate mixture is dissolved in water. The H2O is acting as a base in the presence of acetic acid and is accepting protons to form H3O+
Hi Jesse, making my way through your crash course videos in preparation for September as a trial run and most probably March next year. I noticed in this video you mention you will go through organic and inorganic redox reactions in another video but I can't find that content anywhere... Would you recommend revising this topic again outside of what's covered in this video i.e. is it a prolific concept covered in the GAMSAT that we should have well understood? Much appreciated in advance.
Hey Mitchell, hmm that's a good point, I may have actually missed doing a video on redox! I wouldn't say it's overly useful as it's much more limited in its ability to be abstracted into reasoning problems. So that you can keep moving on with your study, I'd say the most important aspects to understand are:
1. The concept of an oxidation number and how to calculate it based on a compound formula and charge
2. Knowing that oxidation is an increase in ox number (or a gain of O or loss of H or loss of electrons) and reduction is the reverse.
3. The overall 'potential' of a redox reaction or cell is the Eº(reduction) - Eº(oxidation) assuming the Eº values are from an electrochemical series with reactions written with the reduction reaction going forward
Anything beyond this would likely be introduced in the stem for you. I've seen some des questions and some from prep companies that show anodes/cathodes/salt bridges etc and while yes they are also parts of redox, this is all very technical and is more aligned with memorised knowledge in a Y12/uni chem subject so is really just lazy question writing that doesn't involve reasoning.
I'll see if I can get a short redox crash course video done ahead of September but in the meantime you could also check out some of my sample question sets as I think a few of them involve redox chem and the solutions videos will help in explaining the relevant concepts :)
Hi Jesse, just wondering the importance of knowing to use ICE calculations for the actual gamsat testing. I know this is more of a content video, but would gamsat expect us to know how to use ICE? Thank you
I'd say it's still pretty unlikely that you'd need to actually draw up the table and work through it so exactly as this would be largely repetitive and would also advantage those that have studied chemistry formally. That being said, anything I cover in these crash course videos are things that I would consider either directly required or useful for problem solving. ICE tables are a good methodical way of dealing with equilibria, especially for people less familiar with chemistry concepts and it also helps demonstrate that chemical equilibrium is not the same as a reaction in which all reactant is consumed. These would be the key things that are important to consider in equilibria questions
Hey Jesse, awesome videos man... I'm just stuck at 31:00. If you're adding a base to the acid buffer solution, wouldn't it be shifting to the right since the HA is absorbing OH- to counteract the change in pH by converting it into a weaker base (its conjugate base)? Is it the case that you need to add an extra step by saying it will then shift back to the left since HA has dropped?
Especially because if it favours the back reaction like you said, then H3O+ would get consumed when a base is added thus increasing the pH even further which I can't make sense of (I could be wrong by Chem is rusty since 1st year uni)
Hey Zaid, I think I see what you mean but there might be a couple of confusions surrounding the phrasing "shifting to the left/right" with regard to the equilibrium position and the relative strengths of conjugates, that I'll clarify.
1. We describe the relative position of the equilibrium as "shifting left/right". So if a system favours the forward reaction (rightward arrow) to reachieve equilibrium after an adjustment, then we say that equilibrium has shifted to the right.
2. With conjugate pairs acids have conjugate bases and vice versa, and weak compounds have strong conjugates and vice versa. So in this instance, a weak acid buffer is used (which would have a strong conjugate base). The reason for using a weak acid as a buffer is that 'weak' means it will not readily dissociate and donate its proton (ie. it maintains a high amount of HA).
3. The achievement of equilibrium is through favouring the forward or back reaction until the ratio of the concentrations of LHS and RHS are such that their reaction quotient can equal the equilibrium constant, K again. This means that equilibrium is a mathematically determined position and does not mean that LHS and RHS are equal in concentration. So, when an equilibrium system such as a buffer system is 'adjusted' or a change is introduced, it will really only shift in one direction in order to re-establish equilibrium and bring its reaction quotient back to the value of K. This also means that when a system reachieves equilibrium, it does not reachieve the same concentrations as before but rather it reachieves the same ratio of concentrations.
So in this case, what is happening is
We have an existing buffer equilibrium system
HA + H2O --> A- + H3O+
We then add some base (OH-) which increases the OH- concentration and threatens to increase the pH (you're also right that an alternative way to consider this is that the introduced OH- will react with the H3O+ and lower the H3O+ and therefore threaten to increase pH)
However, the acid buffer (HA) will work to resist this potential change in pH. Because the H3O+ is being consumed, it throws the system out of equilibrium and the system favours the forward reaction (shifts right) with HA being consumed to produce more H3O+ until the ratio (Q) of RHS and LHS reaches K again. As it does this, H3O+ being produced helps offset the pH increase so that it's only a very small change.
Now, yes, HA did ultimately get consumed in the process, but it being a weak acid means that very little of it got consumed (this is the ideal characteristic of a buffer and is the reason we always choose a weak acid/base for the purpose). This might make you think that the system should now try to push back to replenish the lost HA but there are two reasons why it won't.
1. Where would the 'ingredients' for this HA's production come from? It would have to come from the RHS A- and OH- but you can't take from here or else you throw the system out of balance which brings up the second reason;
2. It's already in a new state of equilibrium so it is happy as it is.
Hopefully this clears things up
@@jesseosbourne Thanks for the quick reply! This explanation makes perfect sense now... Thinking in terms of Q and K clears things up too. So if I'm understanding correctly, adding a base (to weak acid buffer) would ultimately favour the forward reaction until the reaction quotient Q reaches K again (close to the 'original' ratio). I think I misunderstood your wording in the video since you had the arrow pointing to the backward reaction when Base is added but I understand the original meaning now. Thanks again!!
I am confused in how u rounded (9* 10^-6 ) to (10 * 10^-5)
in your acid base question, you could have same henderson hasselback pH equation to solve the first part as well, provided we know that log2 is 0.3
I'm assuming you mean the yellow question at around 35 minutes? Yep, absolutely but obviously I'm looking to demonstrate an alternative way for people to problem solve in this area as I follow up with another example that then uses the Henderson Hasselbalch equation so that people can choose what works best for them
@@jesseosbourne fair enough. btw, really appreciate all your GAMSAT videos, very helpful indeed
Hi Jesse, for the example of figuring out the pH of the buffer, I am a little confused by the ICE table you used. In the table are you doing number of moles and not concentration? I was looking at a chem textbook and they used concentrations in the example rather than moles as you seem to. I did the question before watching and used concentration and ignored/didn't see the bit which says you produce 1L from equal parts, and got the same answer. Why did you use the moles - are there other times where it would give diff answers? I hope this made sense and thanks for the vid!
Great question, Priya!
So when doing ICE tables, you are effectively working out the 'amount' of each compound reacting or being produced, and the amounts at equilibrium. We always need to use the number of mole of each when adding and subtracting because concentrations are actually a ratio of moles to volume whereas the number of moles are absolute amounts. Another way to think about this is that when a reaction occurs, it reacts certain amounts of mole of compound rather than certain amounts of concentration.
A non-chemistry way to think about this is to consider how we add 20% of 100 and 30% of 200. The % are ratios (like concentration) and the figures 100 and 200 are like volumes.
We could not simply ignore the volumes and just add 20% + 30% = 50% because the volumes are different.
We would instead have to first determine the absolute value that they represent
20% x 100 + 30% x 200
20 + 60 = 80
The reason why you may have seen other videos using concentration is that they may have been using a system with a total volume of 1 litre and so since n = cV and V = 1, then we have the special case that n = c and so we can kind of shortcut the maths and use the concentration. I use this same volume in this question, but still calculate the number of mole at the start anyway so that people are aware that outside of this special case, you should always use the number of mole in ICE tables.
@@jesseosbourne thank you so much that really clears things up!
hi you said this is the 4th video, i cant find the 3rd one ? looking for the chemistry crash course #3?
Hey Emily, that's my bad! I actually miscounted when I filmed this one haha. This is actually crash course #3. I've numbered the video titles now to avoid any confusion for people. It should be
1. chemical equilibria and stoichiometry
2. intro to organic chemistry
3. organic chemical pathways and buffers
4. Stereochemistry
@@jesseosbourne thank you so much! was just worried i had missed something.