Calculating the pH of a buffer made from a weak acid and strong base
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- Опубликовано: 9 сен 2024
- This calculation is as big as it's going to get. Thankfully there is a method with an explanation to help you understand what you are doing. Get your calculator out and get practicing!
This guy is seriously underrated. He's reduced my panic attacks about A level Chem honestly.
no ones cares mate but same
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This channel is the reason i bothered to turn up to my chemistry exams, it helped me more in a few days than school did in an entire two years,
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I never got these buffers earlier x)
Hah yes buffers are really tough. A funny one to get your head round. So pleased you now understand them. I hated them when I was doing my a levels!
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Thanks for the video!
How come you don't use the Henderson-Hasselbach equation?
Chris, as I no longer have a chemistry teacher for my A-level exams in June 2019, I am really grateful for your videos and presentations. Quick question though when looking at 2 similar videos with different calculations, here an excess is calculate, in the other video ("calculating the pH of buffer solutions), you have actually an excess of base. So if using the same technique as in this video, you would actually have an excess weak acid of zero and of course that is wrong. Could you please explain why the other video is different? Thanks
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wow, just wow , I have nothing else to say sir !! Can you please teach these things to my teacher so he could stop wasting my time in the class :p lol hahaha, thank you so much sir .
Awesome video man! Can we use the same sort of calculation steps to work out the pH of a solution of a weak acid and strong base, when the strong base is in excess? Would you have to work in terms of [A-] being excess?
+Taras Mogetich Thanks! You would use the same equation but have OH- ions in excess. You would have to use Kw=[H+][OH-] to work out [H+] then just use as normal. Kw would be 1*10-14 at standard room temp and pressure. Hope this helps.
Thanks
@@AlleryChemistry Sir, if the acid is limiting how can be create the acidic buffer? without an acid. I am a little confused.
Another great video - thank you so much for making them!! 😄💙
Quick question: why can we assume the moles of NaOH are equal to the moles of CH3COO- when it is actually their concentrations that are the same? (because there are square brackets around them which should mean concentrations, not moles, are equal right?)
Can you make a buffer with exess strong base and a weak acid?
Hi Sir, im really confused about why the number of moles in step 4 is exactly the same as the number ofmoles in step 2; please explain.
Yes. The amount of moles of strong base is the same as the number of moles of salt produced. This is because the only chemical that makes the salt is the base. It makes sense that the amount of salt depends on the amount of base added. Hope this helps!
Thank you again
How do you calculate the ph of a buffer made from a strong acid and weak base instead?
Hi. Good day is this the same way i should approach this problem:
You are required to conduct an experiment that should be buffered at pH = 5.2. The only
chemicals available are butanoic acid and sodium butanoate. Plan and design an experiment
to demonstrate how you would prepare 100 ml of this buffer using 10 ml of 0.05 M
butanoic acid.
Why do we take the unreacted acid into the calcualtion? Why not the reacted amount?
Do we need to always use excess acid when making a buffer consisting of a weak acid and strong base?
The moles of acid should be 5x10-4
i understand this well, but i find the exam questions harder than these type of questions... do you have any suggestions on how i could tackle the exam questions?
+Jay P Yes, you need to focus on exam technique. I say that there are 2 halves to revision 1. Content 2. Exam technique. It's like learning to drive you have the content (the highway code) and you have the technique (driving). Just reading the highway code doesn't mean you can get in a car and drive it. Just like reading a chemistry book doesn't mean you can then sit an exam in it. I do have videos that look into exam technique by going through past papers and more will be added. Have a look at my playlists to find them. Hope this helps!
Hi, your videos are really helpful. Would it be possible to do the A2 videos for OCR B Salters Chemistry? Like you did for the AS modules?
Yes I would like to. Just when I get the time to do so.
Thank u !!!
THNAK YPU SM!!
why is step 6 using the moles of acid in excess, rather than the moles of weak acid in step 1 ?
+S AC This is because some of the acid calculated in step 1 has reacted with your sodium hydroxide. You have to use the amount of acid that is left behind (excess) when you are using Ka. Remember Ka is just a type of equilibria equation, it is used to calculate amounts after a reaction has occurred. Hope this helps!
Thanks man I love you no homo
can we not use the formula ka = [H+]^2/[HA] for this question??
No. It cannot be used for buffer solutions because the number of hydrogen ions [H+] will not always be equal to the number of its corresponding conjugate base ions [A-] so the following assumption cannot be made. There is a difference between them.
Hi there in step 5 i was taught as the same number of moles of the NaOH is used then we dont have to calculate conc of HA & A- just use the number of moles of HA & A- . Could you please tell me why ?
+Omar Alias OK. This would only be true if the total volume adds up to 1000cm3 (1dm3) In the video example the volumes were different so will have varying concentrations.
I don't follow could you please explain ?
+Omar Alias You are referring to step 5 which concerns calculating the total volume of the solution after mixing the alkali and weak acid. In order to calculate the correct concentration of all the components of Ka expression you need the total volume. If the total volume comes to 1000cm3 (1dm3) then you can just use the mole values as mole/1dm3 is just concentration.
+A Level Chemistry by Allery Tutors like use the moles of HA and A- place in the weak acid equation and don't go further with calculating conc of HA and A- but just by using moles
Hi, I was wondering shouldn't the conc. of CH3COO- be higher because of the dissociation of the remaining CH3COOH which did not react with NaOH ?
No. The assumption is that salts dissociate fully and weak acids don't. In other words the amount of CH3COO- from acid is so small we ignore it.
Thank you :)
For step 8 I did the same thing as you but got 3.73x10^-5
+S AC Hmmm... it looks like the number you used from step 8 is incorrect. I checked the video and all the numbers are correct, please go back and check again and let me know if you need any more help. Thanks!
Please adjust your camera
New vids have new camera. All higher quality now.