SAL: Let’s say I have

some weak acid. I’ll call it HA. A is a place holder for really a

whole set of elements that I could put there. It could be fluorine, it could

be an ammonia molecule. If you add H it becomes

ammonium. So this isn’t any particular

element I’m talking about. This is just kind of a general

way of writing an acid. And let’s say it’s in

equilibrium with, of course, and you’ve seen this multiple

times, a proton. And all of this is in

an aqueous solution. Between this proton jumping

off of this and its conjugate base. A minus. And we also could have written

a base equilibrium, where we say the conjugate base could

disassociate, or it could essentially grab a hydrogen from

the water and create OH. And we’ve done that

multiple times. But that’s not the point

of this video. So let’s just think a little bit

about what would happen to this equilibrium if we were

to stress it in some way. And you can already imagine that

I’m about to touch on Le Chatelier’s Principle, which

essentially just says, look, if you stress an equilibrium

in any way, the equilibrium moves in such a way to

relieve that stress. So let’s say that the stress

that I apply to the system– Let me do a different color. I’m going to add some

strong base. That’s too dark. I’m going to add some NaOH. And we know this is a strong

base when you put it in a aqueous solultion, the sodium

part just kind of disassociates, but the more

important thing, you have all this OH in the solution, which

wants to grab hydrogens away. So when you add this OH to the

solution, what’s going to happen for every mole that you

add, not even just mole, for every molecule you add of this

into the solution, it’s going to eat up a molecule

of hydrogen. Right? So for example, if you had 1

mole oh hydrogen molecules in your solution and you added 1

mole of sodium hydroxide to your solution, right when you do

that, all this is going to react with all of that. And the OHs are going to react

with the Hs and form water, and they’ll both just kind of

disappear into the solution. They didn’t disappear, they

all turned into water. And so all of this hydrogen

will go away. Or at least the hydrogen that

was initially there. That 1 mole of hydrogens

will disappear. So what should happen

to this reaction? Well, know this is an

equilibrium reaction. So as these hydrogen disappear,

because this is an equilibrium reaction or because

this is a weak base, more of this is going to be

converted into these two products to kind of make up

for that loss of hydrogen. And you can even play

with it on the math. So this hydrogen goes down

initially, and then it starts getting to equilibrium

very fast. But this is going to go down. This is going to go up. And then this is going

to go down less. Because sure, when you put the

sodium hydroxide there, it just ate up all of

the hydrogens. But then you have this– you can

kind of view as the spare hydrogen capacity here

to produce hydrogens. And when these disappear,

this weak base will disassociate more. The equilibrium we’ll move

more in this direction. So immediately, this will

eat all of that. But then when the equilibrium

moves in that direction, a lot of the hydrogen will

be replaced. So if you think about what’s

happening, if I just threw this sodium hydroxide

in water. So if I just did NaOH in an

aqueous solution– so that’s just throwing it in water– that

disassociates completely into the sodium cation

and hydroxide anion. So you all of a sudden

immediately increase the quantity of OHs by essentially

the number of moles of sodium hydroxide you’re adding,

and you’d immediately increase the pH, right? Remember. When you increase the amount of

OH, you would decrease the pOH, right? And that’s just because

it’s the negative log. So if you increase OH, you’re

decreasing pOH, and you’re increasing pH. And just think OH– you’re

making it more basic. And a high pH is also

very basic. If you have a mole of this, you

end up with a pH of 14. And if you had a strong acid,

not a strong base, you would end up with a pH of 0. Hopefully you’re getting a

little bit familiar with that concept right now, but if it

confuses you, just play around with the logs a little bit and

you’ll eventually get it. But just to get back to the

point, if you just did this in water, you immediately get a

super high pH because the OH concentration goes

through the roof. But if you do it here– if you

apply the sodium hydroxide to this solution, the solution that

contains a weak acid and it’s conjugate base, the weak

acid and its conjugate base, what happens? Sure, it immediately reacts with

all of this hydrogen and eats it all up. And then you have this extras

supply here that just keeps providing more and

more hydrogens. And it’ll make up a

lot of the loss. So essentially, the stress

won’t be as bad. And over here, you dramatically

increase the pH when you just throw

it on water. Here, you’re going to increase

the pH by a lot less. And in future videos, will

actually do the math of how much less it’s increasing

the pH. But the way you could think

about it is, this is kind of a shock absorber for pH. Even though you threw this

strong based into this solution, it didn’t increase

the pH as much as you would have expected. And you can make it

the other way. If I just wrote this exact

same reaction as a basic reaction– and remember,

this is the same thing. So if I just wrote this as, A

minus– so I just wrote its conjugate base– is in

equilibrium with the conjugate base grabbing some

water from the surrounding aqueous solution. Everything we’re dealing

with right now is an aqueous solution. And of course that water that

it grabbed from is not going to be an OH. Remember, are just equivalent

reactions. Here, I’m writing it as

an acidic reaction. Here, I’m writing it is

a basic reaction. But they’re equivalent. Now. If you were to add a strong

acid to the solution, what would happen? So if I were to throw hydrogen

chloride into this. Well hydrogen chloride, if you

just throw it into straight up water without the solution, it

would completely disassociate into a bunch of hydrogens and

a bunch of chlorine anions. And it would immediately

make it very acidic. You would get to

a very low pH. If you had a mole of this– if

your concentration was 1 molar, then this will

go to a pH of 0. But what happens if you at

hydrochloric acid to this solution right here? This one that has this

weak base and its conjugate weak acid? Well, all of these hydrogen

protons that disassociate from the hydrochloric acid are all

going to react with these OHs you have here. And they’re just going to

cancel each other out. They’re just going to merge with

these and turn into water and become part of the

aqueous solution. So this, the OHs are going to

go down initially, but then you have this reserve

of weak base here. And Le Chatelier’s Principle

tells us. Look, if we have a stressor that

is decreasing our overall concentration of OH, then the

reaction is going to move in the direction that relieves

that stress. So the reaction is going to

go in that direction. So you’re going to have more of

our weak base turning into a weak acid and producing

more OH. So the pH won’t go down as much

as you would expect if you just threw this in water. This is going to lower the pH,

but then you have more OH that could be produced as this guy

grabs more and more hydrogens from the water. So the way to think about it is

it’s kind of like a cushion or a spring in terms of what a

strong acid or base could do to the solution. And that’s why it’s

called a buffer. Because it provides a

cushion on acidity. If you add a strong base to

water, you immediately increase its pH. Or you decrease its acidity

dramatically. But if you add a strong base

to a buffer, because of Le Chatelier’s Principle,

essentially, you’re not going to affect the pH as much. Same thing. If you add and acid to that same

buffer, it’s not going to affect the pH as much as you

would have expected if you had thrown that acid in water

because the equilibrium reaction can always kind of

refill the amount of OH that you lost if you’re adding acid,

or it can refill the amount of hydrogen you lost

if you’re adding a base. And that’s why it’s

called buffer. It provides a cushion. So it give some stability

to the solution’s pH. The definition of a buffer is

just a solution of a weak acid in equilibrium with its

conjugate weak base. That’s what a buffer is, and

it’s called a buffer because it provides you this kind

of cushion of pH. It’s kind of a stress absorber,

or a shock absorber for the acidity of a solution. Now, with that said, let’s

explore a little bit the math of a buffer, which is really

just the math of a weak acid. So if we rewrite the equation

again, so HA is in equilibrium. Everything’s in an

aqueous solution. With hydrogen and its

conjugate base. We know that there’s an

equilibrium constant for this. We’ve done many videos

on that. The equilibrium constant here is

equal to the concentration of our hydrogen proton times

the concentration of our conjugate base. When I say concentration,

I’m talking molarity. Moles per liter divided by the concentration of our weak acid. Now. Let’s solve for hydrogen

concentration. Because what I want to do is I

want to figure out a formula, and we’ll call it the

Hendersen-Hasselbalch Formula, which a lot of books want you

to memorize, which I don’t think you should. I think you should always just

be able to go from this kind of basic assumption

and get to it. But let’s solve for the hydrogen

so we can figure out a relationship between pH and

all the other stuff that’s in this formula. So, if we want to solve for

hydrogen, we can multiply both sides by the reciprocal

of this right here. And you get hydrogen

concentration. Ka times– I’m multiplying

both sides times a reciprocal of that. So times the concentration of

our weak acid divided by the concentration of our weak base

is equal to our concentration of our hydrogen. Fair enough. Now. Let’s take the negative

log of both sides. So the negative log of all of

that stuff, of your acidic equilibrium constant, times HA,

our weak acid divided by our weak base, is equal to the

negative log of our hydrogen concentration. Which is just our pH, right? Negative log of hydrogen

concentration is– that’s the definition of pH. I’ll write the p and the

H in different colors. You know a p just means

negative log. Minus log. That’s all. Base 10. Let’s see if we can simplify

this any more. So our logarithmic properties. We know that when you take the

log of something and you multiply it, that’s the same

thing as taking the log of this plus the log of that. So this can to be simplified to

minus log of our Ka minus the log our weak acid

concentration divided by its conjugate base concentration. Is equal to the pH. Now, this is just the pKa of our

weak acid, which is just the negative log of its

equilibrium constant. So this is just the pKa. And the minus log

of HA over A. What we can do is we could make

this a plus, and just take this to the

minus 1 power. Right? That’s just another logarithm

property, and you can review the logarithm videos if

that confused you. And this to the minus 1 power

just means invert this. So we could say, plus the

logarithm of our conjugate base concentration divided by

the weak acid concentration is equal to the pH. And this right here,

this is called the Hendersen-Hasselbalch

Equation. And I really encourage you

not to memorize it. Because if you do attempt to

memorize it, within a few hours, you’re going to

forget whether this was a plus over here. You’re going to forget this,

and you’re going to forget whether you put the A minus or

the HA on the numerator or the demoninator, and if you forget

that, it’s fatal. The better thing is to just

start from your base assumptions. And trust me. It took me a couple minutes to

do it, but if you just do it really fast on paper– you don’t

have to talk it through the way I did– it’ll take

in no time at all to come to this equation. It’s much better than memorizing

it, and you won’t forget it when you’re

30 years old. But what’s useful about this? Well, it immediately relates pH

to our pKa, and this is a constant, right, for

an equilibrium? Plus the log of the ratios

between the acid and the conjugate base. So the more conjugate base I

have, and the less acid I have, the more my pH is

going to increase. Right? If this goes up and this

is going down, my pH is going to increase. Which makes sense

because I have more base in the solution. And if I have the inverse of

that, might be just going–

Thanks for the great videos

Thank you Sal 🙂

Thank You!

Remake? No, it's the same video.

Nooooooooooooooo~~ It's too late.

I took the exam on this on the previous Monday…And I've done it wrong T_T.

But anyway, Thanks Sal

Did it get cut off at the end…?

@norwayte He cut off the end since he said that volume doesn't matter.

I love you Sal

@savvy22dew Nearest I can tell, you have mistake the "p" in this equation as a regular variable. It is actually the short hand for -log(). Thus p(Ka) = -log(Ka) and p(HA/A) = – log(HA/A). Log functions have a few unique properties, one of which is that log(X*Y) = Log(X) + Log(Y), which explains why

p(K * [HA]/[A-]) = pK – log [HA]/[A].

Sal has a proof of this concept on his website in the algebra section if you are interested in why this is so. Hoped that helped!

ur awesome!!!!!!!!!

I have done HH a dozen times never even though it was derrived from the original equation

tks

thanks for the conceptual idea of hendersen hasselbalch but can you teach us ICE tables???

Shouldn't the weak acid be AH^+?

@hkk313 Well, yeah, that makes sense. I just meant for the basic formula. You're right, though.

They Give us a the equations , so fuck it I'm memorizing it .

I remember deriving the Hendersen-Hasselbalch on the back of an envelope when I while doing my laundry. I was an undergrad at the time.

Ka = [H+][A-]/[HA]

-logKa = -log([H+][A-]/[HA])

-logKa = -log[H+] – log ([A-]/[HA])

pKa = pH – log ([A-]/[HA])

pKa + log ([A-]/[HA]) = pH

I aways found equations easier to remember when I undertand how they come about.

Also I realize that the H.H. Equation shows that pKa is the pH at which half a quantity of acid is disassociated. This showed that pKa relates to an acid's strength.

lol u just decided to do all of that on an envelope randomly while cleaning clothing? xD

asian much? lolololol

I'm not sure I understand what you mean. I happen to ethinically irish and italian

racism much?

BASIC assumption; hahahahaha i'm retarded….

wow… u know nothing. lol

Buffer…. I don't even know her!!!

I won't forget it when I'm 30yrs old… oh, wait, I am 30 years old and I'm here because I forgot it. As well as a bunch of other equations. Touché.

no wonder he's so smart. he has the voice of kahn academy explaining stuff to him all day long.

its too old vids when ipad wasnt even invented

it's liquid so you don 't use it homie

WTF where are these videos in his playlist??!?!?!?!?!?!?

It's applicable to dilute aqeous solutions, which means [H2O] changes only negligably so you needn't include it.

Accompanied by the destruction of an equal amount of hydronium or hydroxide which take up the same amount of volume.

Very informative 😀

Got a final tomorrow. Even Le Chatelier's principle can't relieve this stress.

Why do we separate the H+ from the rest of the equation when we apply -log to it?

-logKa = -log([H+][A-]/[HA]) to -logKa = -log[H+] – log ([A-]/[HA])?

Why not use a strong acid and base in order to nuertralize a solution ph= 7 right?

we can get ph= -log H+ .

OooooH!! I think I got it. log functions log ( ab) = Log a+ log b.

answer your question?

I think so, I am just confused by why we separate H+ from the rest of the fraction and take its log separately.

I understand you, having chemistry on Monday o_O

I definitely thought I was done with this when I took Gen Chem 2 in undergrad, and now at Vet school, they're all "ACID-BASE PHYSIOLOGY!" Damn.

LOL

"You won't forget it when you're 30 years old."

Khan, your buffer video from 2009 had better pixelation than this one that is from 2010. What are you from the future or something? :O

thank you for the refreshment 🙂 its actually the first time i truly understood buffers ! though i took many chemistry courses !

You lost me after the first 3 minutes

UC RIVERSIDE shout out!!

OHHH it was never explained to me (or i just never understood) that buffers were connected le chateliers principle ah i understand thankyouthankyou

This one was probably the first of yours that wasnt so clear to me…

I don't know why, but I prefer to just memorise equations rather than being able to derive them; the derivations confuse me a lot more.

QMUL shout out!!

Can someone full explain the log property of raising it to the -1 and reversing the sign from negative to positive?

Please help me with this question-

What is the pH of this solution – 10 ml 0.1N acetic acid + 10.1 ml 0.1N sodium hydroxide?

I think you explained it great. Now I understand what is actually going on and why the graph looks the way it does when we use an amino acid as a buffer in lab.

Yassss omg thank you!

omgggggg you've enlighten me!!! thank u so much!

the funny thing .. when he said you will not forget it till you r 30 years old… oops I am already 34!!!

your acadmy does not explain how buffer capacity related with henderson equation why cocentration of salt =acid or base having maximum capacity

i wait ur msg or video related wirth buffer capacity

does khan academy have anything on H-H/buffers but with amino acids??

This isn't khan's voice!

Dude you like make sense…its weird.

This MAKES sense. No more memorizing formulae.

dood u r the most amazing teacher ever i watch ur vidios like an addict thank u so much

Thank you so much I love your videos they make things very easy to understand and I don't know if I would have as much confidence in bio as I do right now if not for your videos

Awesome video thank you so much

Thank you Khan Academy 🙂

Thank you so much !!

I have been struggling wid d exact meaning of buffer for a long time but dis video relating wid d le chatelier principle is just awesome and it explains everything about it😄😄😄

I FINALLY UNDERSTAND BUFFERS HOLY CRAP

thanks super explained

awesome! thank ya! just helped me with my introduction to molecular and celular biology class for college lol

You said it's just a weak base with its conjugate weak acid or vice versa but for me that doesn't make a lot of sense since the conjugate base of a weak acid has to be strong…

I’m still so confused. need an some one to dumb this down more for me. lol

can someone tell me why buffer concentration affect ph of solution when adding acid or base into the buffer?

what is wrong with 30 years old, that is not old nor is it the future for some. I am almost 40 reviewing this so I guess above statement would not apply to me…hehe.

Ayy, what about us 30yo first time chem majors? HAHA!

but aren’t buffers made of acid s and their respective salts

Good explainable

AN acid

Ayy-kwee-uhs. AQUEOUS

This video seriously saved me!!! Thank you!

how does confirm a buffer solution???

Thanks sal, but it’s “le CHA tel ire’s”, not le sheTAL i er’s…

Why do you stammer in your videos

Best explanation so far

+10 y a favoritos

Amazing.

The video's not finished where can i get a complete version

i love you

Even khan academy couldn’t help me w this one….

Thanks that is amazing💞

Thank you Khan academy now I understood about buffers

I was so confused about this in the textbook, now I completely understand!!! thank you so much!!!

Got any pixels?

this is the best video ive ever seen

ohhhhhhh now i get it

I can't describe how happy I was when I clicked on the next video in the Khan Academy app to hear Sal's voice and not one of the other 2 guys that do chemistry videos there.

2019

12:17

"P just means -log"

That sentence alone made me understand the whole concept.

Why haven't any of my professors ever said that?!

It ends in the middle?

what if a buffer is made up of strong acid and strong base. what happens to the equilibrium

Thanks, finally understand.