I explained that the volumes cancel out (I’ve written it in blue). You can work out the concentrations here but you don’t need to with buffers. Either way, you get the same answer. Try it!
Your videos are very helpful!:) Could I just ask; why with some do you write out the equation and work out the amount of product produced via the limiting reactant (eg 16:30) and others you don't? I understand how to use both methods, where you go straight into kaacidoversalt/rearranging kaacidoversalt and then when you instead use the equation to calculate moles produced but not when to use them?
This depends on how the buffer is formed. If I'm simply adding a salt to a weak acid, there is no reaction. We immediately arrive at a point on the titration curve. If, however, I add a strong alkali to an excess of weak acid, then there is a reaction (partial neutralisation) and I have to work out how much weak acid is left afterwards and how much salt is formed (I didn't have any before I added the weak acid and strong alkali together). This is why I did a number of examples. You always end up using the expression Ka = but how you determine the values to plug into this may vary depending on wht exactly it is you've added together to form your buffer.
Hello. Really helpful video. I just had one question as to why in the very last problem, when you added HCL, you took 'before' moles and not the 'after' ones to add to the moles of HCL
Hi sir. For the last one, when calculating the concentrations after adding an extra 5cm3 of hcl. Would you include that volume in the total volume when calculating the new conc. I know it cancels but if you did wanna work it out would you add the 5cm3 to the total vol or is it negligible? Thanks.
I understand the principle for why we added and subtracted mol of hcl due to equilibrium at 36:52 but how come you didn’t use the ICE table format from before?
There are no ICE tables in this video. ‘Before’ in the table you’re referring to is about the number of moles BEFORE you add the reactants together. This isn’t the same as an ‘Initial’ concentration in an equilibria. However, I didn’t need to do the ‘Before’ table again because I only needed to work out which of the weak acid and salt were being added to / decreased and just do the sum. Say if that doesn’t make sense.
29:16 Isn't the concentration of ch2clch2c00h = 0.019767.... as that is how much acid is remaining. I think you plugged the moles of the acid remaining into the ka eq but not its concentration?
I don't need to plug in the concentration of the acis remaining providing I also only plug in the moles of the salt. Concentration is moles / volume. As the volumes are the same for the acid and salt, they cancel out and I can just use moles. Say if I've misunderstood your query. Thanks.
In the first question why did you use the concentrations of each substance in the final calculation but in the other questions you used moles in the final calculation. How are we meant to know when to use moles and when to use concentrations?
I used concentrations in the first two questions because this required either the same or fewer calculation. In the third, I explain why I don’t use concentrations - because the volumes cancel out. Providing your rearranged expression has the same number of [ ] above and below the fraction line (so the volumes cancel out), you can use moles or concentration. I’ve just don’t whichever is easier. I hope this helps.
unless i'm doing something very wrong at 19:45 i think the mols of CH3COOH should be 9.25x10^-3 incase anyone doing calculations along with the video
You're right - I'm sorry about that.
Brilliant video, this really helped me understand buffer calculations, I'll be sharing it with my classmates
nice, any chance you can share me the link to the video too?
@@samiultr huh?
@20:42 shouldnt the moles at equilibrium of ethanoic acid be 9.25x10^-3?
You’re right - a bad calculator mistake on my part. Hopefully the method still helps.
thanks for pointing that out mate i was so confused at what i had done wrong
i have an acids , bases and buffers test tomorrow . its 2 am and u have saved my life honestly . : ))))
I hope it went well.
Smashing video, thank you for your hard work
For the last example I keep getting 1.312x10-4 as H+, is there something wrong 😅
You’re quite right - I mistyped a value into my calculator. I’m really sorry about that. The final pH is 3.88.
@@simonflynn4515 ah ok thank you 😊 the video was really helpful
For the question at 23:41
you used the moles of the acid and the salt, why didn't you divide them by the total volume to get their concentrations
I explained that the volumes cancel out (I’ve written it in blue). You can work out the concentrations here but you don’t need to with buffers. Either way, you get the same answer. Try it!
You are very good at explaining , keep up the good work !!
Thank you!
Your videos are very helpful!:) Could I just ask; why with some do you write out the equation and work out the amount of product produced via the limiting reactant (eg 16:30) and others you don't? I understand how to use both methods, where you go straight into kaacidoversalt/rearranging kaacidoversalt and then when you instead use the equation to calculate moles produced but not when to use them?
This depends on how the buffer is formed. If I'm simply adding a salt to a weak acid, there is no reaction. We immediately arrive at a point on the titration curve.
If, however, I add a strong alkali to an excess of weak acid, then there is a reaction (partial neutralisation) and I have to work out how much weak acid is left afterwards and how much salt is formed (I didn't have any before I added the weak acid and strong alkali together).
This is why I did a number of examples. You always end up using the expression Ka = but how you determine the values to plug into this may vary depending on wht exactly it is you've added together to form your buffer.
@@simonflynn4515 Thank you so much! You have just saved my chemistry exam tomorrow morning:)
@@simonflynn4515 that makes so much sense, thanks so much!
Hello. Really helpful video. I just had one question as to why in the very last problem, when you added HCL, you took 'before' moles and not the 'after' ones to add to the moles of HCL
I'm not sure I did - I use the 'after' moles of 4.25 x 10^-3 weak acid and 3.75x10^-3 salt. Do you mind double-checking? Thanks.
Yes, actually you are right. My bad. Thanks again for the helpful video
Hi sir. For the last one, when calculating the concentrations after adding an extra 5cm3 of hcl. Would you include that volume in the total volume when calculating the new conc. I know it cancels but if you did wanna work it out would you add the 5cm3 to the total vol or is it negligible? Thanks.
You would add it. But the point is that it does cancel
@@simonflynn4515 thanks sir.
Your video are useful. I will defo be sharing your channel with new year 12’s!
Why not just use the Henderson-Hasselbalch equation?
It doesn't feature in AQA's specification.
Will you do predictions for the papers based on the advance info? And will you be doing any last minute tips/ revision for the paper 1-3?
I'm afraid not. My aim is to do videos that have longevity - sorry.
I understand the principle for why we added and subtracted mol of hcl due to equilibrium at 36:52 but how come you didn’t use the ICE table format from before?
Also this video was very helpful ty
There are no ICE tables in this video. ‘Before’ in the table you’re referring to is about the number of moles BEFORE you add the reactants together. This isn’t the same as an ‘Initial’ concentration in an equilibria. However, I didn’t need to do the ‘Before’ table again because I only needed to work out which of the weak acid and salt were being added to / decreased and just do the sum. Say if that doesn’t make sense.
@@simonflynn4515 ah yes I went through more examples and I understand better now, and I did mean the before and after. Thank you!
@@simonflynn4515 your buffer videos saved me for the 5 marker in p1 2023, thank you!
29:16 Isn't the concentration of ch2clch2c00h = 0.019767.... as that is how much acid is remaining. I think you plugged the moles of the acid remaining into the ka eq but not its concentration?
I don't need to plug in the concentration of the acis remaining providing I also only plug in the moles of the salt. Concentration is moles / volume. As the volumes are the same for the acid and salt, they cancel out and I can just use moles. Say if I've misunderstood your query. Thanks.
In the first question why did you use the concentrations of each substance in the final calculation but in the other questions you used moles in the final calculation. How are we meant to know when to use moles and when to use concentrations?
I used concentrations in the first two questions because this required either the same or fewer calculation. In the third, I explain why I don’t use concentrations - because the volumes cancel out. Providing your rearranged expression has the same number of [ ] above and below the fraction line (so the volumes cancel out), you can use moles or concentration. I’ve just don’t whichever is easier. I hope this helps.
@@simonflynn4515 howcome ka is in cm3 but the concentrations are in dm3 shouldnt they all be in the same units?
Terrible
Can I ask why? I’m always up for constructive criticism as I’m sure you are.