Buffers and Henderson-Hasselbalch | Chemistry | Khan Academy
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Buffers and Henderson-Hasselbalch | Chemistry | Khan Academy

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–

About James Carlton

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100 thoughts on “Buffers and Henderson-Hasselbalch | Chemistry | Khan Academy

  1. 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

  2. @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!

  3. 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.

  4. 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.

  5. lol u just decided to do all of that on an envelope randomly while cleaning clothing? xD
    asian much? lolololol

  6. 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.

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

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

  9. 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])?

  10. we can get ph= -log H+ .

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

    answer your question?

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

  12. 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.

  13. 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

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

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

  16. 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. 

  17. 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?

  18. 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.

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

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

  21. 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

  22. 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😄😄😄

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

  24. 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…

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

  26. 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.

  27. 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.

  28. 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?!

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