When new videos are made and jammed into preexisting playlists with preexisting videos, they usually don't make sense. The new instructor cannot pick up properly from older videos. She is also pretty bad at presenting information in this video, particularly.
She never explains buffer capacity in its formal and easily understandable form. BC is just the number of moles needed to cause a change of 1.00 in the pH of a system.
It is a minor mistake, but the value 5,90 is pH and should not have a unit. As for the confusion about where the number 1,82 came from, it is simply the assigned value of the acid/base fraction.
i love that you can learn the key point of an entire topic by watching a 10min. video. I love it. Amazing. Thank you lady from Khan academy. thank you khan academy.
I have a question: Why do we assume that ALL of the added hydroxide ions react with the conjugate acid in the buffer solution? Is it just an approximation? Because to my knowledge, especially according to Le Chatelier’s principle, the system would reverse a change made to it; in other words, some hydroxide ion will be reproduced because the initial HA present in the buffer solution would dissociate into H+ to replace the conjugate acid consumed by the added OH-. Therefore, the pH change would technically be lesser than what was derived from the Henderson-Hasselbach equation. I am so confused - pls help :(
Do keep in mind that much of ionic equilibrium assumes approximations. Maybe the equilibrium constant of the reaction is very high so we can assume that all of hydroxide ions react
After finishing the video I even wondered whether the calculations in this video are correct. For buffer 1 the [A-]/[HA+]] would go from 1.82 to 2.08. This means that to get an new (acid) equilibrium with the given Ka the H+ concentration needs to go down which would give a reverse reaction (from right to left). This would mean the pH would go up (less H+ so less acidic). This is why in the previous video it was said that you can't just add the amounts of the base concentration to the ones in the acid equilibrium as a new equilibrium needs to be reached first. So also this is confusing. But maybe I got it wrong as this video is still on UA-cam after 8 years ???
Thank you for this video, but I have a question regarding a point of confusion: If a weak acid reaction has a dissolution constant of 1.8x10^-5, how is it possible for the ratio [A-]/[HA] to equal 1.8? I put your given values for [A-]=0.9M and [HA]=0.49M into the equilibrium equation (assuming [H]=[A-]) and the result is much larger than 1.8x10^-5. What am I missing?
Luminaut I guess because this video is about buffer capacity. Although in this video she talked only about how the concentration of the buffer affects how much the pH will change if you add strong acid/ strong base, also the fraction of HA and A- affect how much the pH will change if you add strong acid/base. CH3COOH is a weak acid, so its a very bad buffer if its in equilibrium, since there is only tiny bit of A- compared to HA. However you can alter the ration of A- and HA by adding strong base for instance, that increases the amount of A-. The buffer resists pH changes the most when [A-]/[HA]=1. I presume that in the next video this will be explained, but I have no idea why this video was overly complicated in the sense that its not explained. Thats the only reason I came up with.
This video seems to contradict the previous video as is in this video the OH- reacts with the HA in the buffer (left side of the buffer equilibrium) where in the previous video OH- reacts with the H+ in the buffer (right side of the equilibrium). The latter of course also creates a new buffer equilibrium as H+ disappears whereupon more HA is turned into H+ en A- which is similar to what happens in this video where also more HA is converted but I can imagine the calculation will be different. Anyhow this is confusinng.
Does this mean that we cannot change the pH by adding more acetic acid to the buffer? Proportions of A- and HA will not change if we add more acetic acid?
0.1M is equal to 1mol when in 1L of solution. M=mol/L. It is often useful to convert to moles, depending on what you need to solve for (volume, grams, etc.)
i love the instructor voice.
it's the only thing to love about her methods.
I luv the instructor! lol😋
😂😂 .Bade harami ho beta
it would be helpful if you gave the source of your values like where did 1.82 came from? Gizz created confusion
Thanks! BTW @9:30 you said that the pH is 5.90"M", but pH has no unit.
it was st a mstk
I usually find Khan academy videos very useful, but for some reason this video did not help me understand the topic at all.
I think it's the speaker.
When new videos are made and jammed into preexisting playlists with preexisting videos, they usually don't make sense. The new instructor cannot pick up properly from older videos. She is also pretty bad at presenting information in this video, particularly.
She never explains buffer capacity in its formal and easily understandable form.
BC is just the number of moles needed to cause a change of 1.00 in the pH of a system.
Thats how i feel about this entire channel. organic chemistry tutor makes better videos for sure.
Now im questioning why i clicked on this.
I will cry i cant believe i found this video THANK YOU ..
what a great video!!! Really helped me. Thank you very much!! xxx
I love this video.
Thank you ❤
It is a minor mistake, but the value 5,90 is pH and should not have a unit.
As for the confusion about where the number 1,82 came from, it is simply the assigned value of the acid/base fraction.
study hard
Thank you!!! God bless you xx
i love that you can learn the key point of an entire topic by watching a 10min. video. I love it. Amazing. Thank you lady from Khan academy. thank you khan academy.
I have a question:
Why do we assume that ALL of the added hydroxide ions react with the conjugate acid in the buffer solution? Is it just an approximation? Because to my knowledge, especially according to Le Chatelier’s principle, the system would reverse a change made to it; in other words, some hydroxide ion will be reproduced because the initial HA present in the buffer solution would dissociate into H+ to replace the conjugate acid consumed by the added OH-. Therefore, the pH change would technically be lesser than what was derived from the Henderson-Hasselbach equation.
I am so confused - pls help :(
Do keep in mind that much of ionic equilibrium assumes approximations. Maybe the equilibrium constant of the reaction is very high so we can assume that all of hydroxide ions react
Great explanation
where do you get the 1.82?
her voice made it impossible to concentrate on the video...cute voice
+j4w7 haha was just about to comment this
thank you
where did 1.82 come from?
She just took it as an example
You save my live thanks
The equations are tidy!! Much easier to see!!
After finishing the video I even wondered whether the calculations in this video are correct. For buffer 1 the [A-]/[HA+]] would go from 1.82 to 2.08. This means that to get an new (acid) equilibrium with the given Ka the H+ concentration needs to go down which would give a reverse reaction (from right to left). This would mean the pH would go up (less H+ so less acidic). This is why in the previous video it was said that you can't just add the amounts of the base concentration to the ones in the acid equilibrium as a new equilibrium needs to be reached first. So also this is confusing. But maybe I got it wrong as this video is still on UA-cam after 8 years ???
You facilitated it alot , Grateful 🤍🤍
Thank you again Dr Harper
how did u get 1.82. how to solve buffering capacity and its maximum buffering capacity?
Thank you for this video, but I have a question regarding a point of confusion:
If a weak acid reaction has a dissolution constant of 1.8x10^-5, how is it possible for the ratio [A-]/[HA] to equal 1.8? I put your given values for [A-]=0.9M and [HA]=0.49M into the equilibrium equation (assuming [H]=[A-]) and the result is much larger than 1.8x10^-5. What am I missing?
Luminaut I guess because this video is about buffer capacity. Although in this video she talked only about how the concentration of the buffer affects how much the pH will change if you add strong acid/ strong base, also the fraction of HA and A- affect how much the pH will change if you add strong acid/base.
CH3COOH is a weak acid, so its a very bad buffer if its in equilibrium, since there is only tiny bit of A- compared to HA. However you can alter the ration of A- and HA by adding strong base for instance, that increases the amount of A-. The buffer resists pH changes the most when [A-]/[HA]=1.
I presume that in the next video this will be explained, but I have no idea why this video was overly complicated in the sense that its not explained.
Thats the only reason I came up with.
This video seems to contradict the previous video as is in this video the OH- reacts with the HA in the buffer (left side of the buffer equilibrium) where in the previous video OH- reacts with the H+ in the buffer (right side of the equilibrium). The latter of course also creates a new buffer equilibrium as H+ disappears whereupon more HA is turned into H+ en A- which is similar to what happens in this video where also more HA is converted but I can imagine the calculation will be different. Anyhow this is confusinng.
How to get the 1.82? how to calculate ? Thank you
Use a log table or a calculator.
Geekminer
the log of what
please answer
Just do 10^-pka (which on her case, pka is 4.74).
its agiven range
It will be given
Good video
Does this mean that we cannot change the pH by adding more acetic acid to the buffer? Proportions of A- and HA will not change if we add more acetic acid?
How to do this for a base do we need to convert to the respective acid
Ur voice is so sweet...
I m going have diabetes
also wait so what happens if the [HA] is greater than 1M?
Where is Coming from [A-]=0.90M and [HA]=0.49M & why we add and subtract 0.04M in buffer1&2 ???...
Should it not be the -log of acid/base?
not for henderson-hasselbach equation. watch the video where sal derived it if you're confused
Hi, may I know if 0.1M of acid/base added are actually same as 0.1 mol? Help me please.
Sorry for my bad english.
0.1M is equal to 1mol when in 1L of solution. M=mol/L. It is often useful to convert to moles, depending on what you need to solve for (volume, grams, etc.)
I understand the topic now but can someone tell me where is she getting the numbers from?
i got more confused after watching this tbh
She's great. I'm so sick of the other dude.
I love this info Tysm, but the voice not so much 😅😂
You added M to 5.90 I noticed.
very cute voice mam...
She sounded nervous
ahahahaha i love the video but...she sounds really insecure or unconfident for some reason 😂😂....
what she started and what she is doing completely unrelated; made it complicated unfortunately.