pKa and pKb relationship | Acids and bases | Chemistry | Khan Academy


SAL: In the last video we
learned that if I had some– let’s say I have
some weak acid. So it’s hydrogen plus some–
the rest of whatever the molecule was called. We’ll just call it A. And I think this tends to be
the standard convention for the rest of the acid. They can disassociate, or
it’s in equilibrium because it’s a weak acid. So it can be in equilibrium
with– since it’s a weak acid, it’s going to produce some
hydrogen proton. And then the rest of
the molecule is going to keep the electrons. So it’s going to be
plus– oh, this is in an aqueous solution. Let me do that. Aqueous solution. And then you’re going to have
the rest of the acid, whatever it might be. A minus, and that’s also going
to be in an aqueous solution. And that’s the general
pattern. We’ve seen the case where A
could be an NH3, right? If A is an NH3, then when you
have this, you have an NH4 plus, and this would
be ammonia. And this is just NH3. A could be a fluorine molecule
right there, because then this would be hydrogen fluoride
or hydrofluoric acid. And this would just be the
negative ion of fluorine. Or a fluorine with an
extra electron. So it could be a
bunch of stuff. You can just throw in anything
there, and it’ll work. Especially for the weak acids. So we learned a last video that
if this is the acid, then this is the conjugate base. And we could write the same
reactions, essentially, as kind of more of a
basic reaction. So we could say, if I start
with A minus– it’s in an aqueous solution– that’s
in equilibrium with– This thing could grab a hydrogen
from the surrounding water and become neutral then. It’s still in an aqueous
solution. And then one of those water
molecules that it plucked that hydrogen off of is now going
to be a hydroxide molecule. Right? Because it’s hydrogen. Remember, whenever I say pluck
the hydrogen, just the proton, not the electron for
the hydrogen. So the electron stays on that
water molecule, so it has a negative charge. It’s in an aqueous solution. So we could write the same
reaction both ways. And we can write equilibrium
constants for both of these reactions. So let’s do that. Let me erase this, just because
I can erase this stuff right there, and then
use that space. So an equilibrium reaction
for this first one. I could call this the K sub
a, because the equilibrium reaction for an acid. And so this is going to be
equal to its products. So the concentration of
my hydrogen times my concentration of whatever my
conjugate base was, divided by my concentration of
my original acid. My weak acid. So this would be the
concentration of HA. Fair enough. I could also write an
equilibrium constant for this basic reaction. Let me do it right down here. So I’ll call that my K sub b. This is a base equilibrium. And so this is equal to the
concentration of the products. It’s becoming tedious to
keep switching colors. Actually, I’ll do it. Because it makes it easier
look at, at least for me. HA times the concentration of
my hydroxide ions divided by my concentration of
my weak base. A minus. Remember, this can only
be true of a weak base or a weak acid. If we were dealing with a strong
acid or is or a strong base, this would not be an
equilibrium reaction. It would only go in
one direction. And when it only goes in one
direction, writing this type of equilibrium reaction makes
no sense– or equilibrium constant– because it’s
not in equilibrium. It only goes in one direction. If A was chlorine, if this was
hydrochloric acid, you couldn’t do this. You would just say look, if
you have a mole of this, you’re just dumping a mole of
hydrogen protons in that solution and then a bunch of
chlorine anions who are not going to do anything. Even though they are the
conjugate base, they wouldn’t do anything. So you can only do this,
remember, for weak acids and bases. So with that said, let’s see if
we can find a relationship between Ka and Kb. What do we have here? We have an A minus on
both sides of this. We have H over– OH
over A minus. Let’s solve for A minus. Right? If we multiply both sides of
this equation by HA over H plus, on the left-hand
side we get Ka times the inverse of this. So you have your HA over
H plus is equal to your concentration of your
conjugate base. A minus. And let’s do the same
thing here. Solve for A minus. So to solve for A minus here,
we might have to do 2 steps. So if we take the inverse of
both sides, you get 1 over Kb is equal to A minus over H,
the concentration of my conjugate acid times the
concentration of hydroxide. Multiply both sides by this. And I get A minus is equal
to my concentration of my conjugate acid times
concentration of hydroxide. All of that over my base
equilibrium constant. Now, these are the
same reactions. Right? In either reaction for given
concentrations, I’m going to end up with the same
concentration. This is going to equal that. Right? These are two different
ways of writing the exact same reaction. So let’s set them equal
to each other. So let me copy and paste
it, actually. So I’m saying that this thing,
copy, is equal to this thing right here. So this is equal to– let me
copy and paste this– that. That’s equal to that. So let’s see if we can
find a relationship between Ka and Kb. Well, one thing we can
do is we can divide both sides by HA. Right? So if we divide both
sides by HA. Actually, I could probably have
that earlier on to the whole thing. If we ignore this part right
here, this is equal to that. Let me erase all of this. Oh. I’m using the wrong tool. So we could say that
they both equal the concentration of A minus. So that’s equal to that. We can divide both
sides by HA. This over here will cancel
with this over here. And we’re getting pretty close
to a neat relationship. And so we get Ka over our
hydrogen proton concentration is equal to our hydroxide
concentration divided by Kb. You can just cross-multiply
this. So we get Ka, our acidic
equilibrium concentration, times Kb is equal to our
hydrogen concentration times our hydroxide concentration. Remember, this is all in
an aqueous solution. What do we know about this? What do we know about our
hydrogen times our hydroxide concentration in an
aqueous solution? For example, let me review just
to make sure I’m jogging your memory properly. We could have H2O. It can autoionize into H plus. Plus OH minus. And this has an equilibrium. K sub w. You just put the products. So the concentration of the
hydrogen protons times the concentration of the
hydroxide ions. And you don’t divide by this
because it’s the solvent. And we already figured
out what this was. If we have just completely
neutral water, this is 10 to the minus 7. And this is 10 to the minus 7. So this is equal to 10
to the minus 14. Now, these two things
could change. I can add more hydrogen, I
could add more hydroxide. And everything we’ve talked
about so far, that’s what we’ve been doing. That’s what acids
and bases do. They either increase this
or they increase that. But the fact that this is an
equilibrium constant means that, look, I don’t care
what you do to this. At the end of the day, this
will adjust for your new reality of hydrogen protons. And this will always
be a constant. As long as we’re in an aqueous
solution, a solution of water where water is a solvent
at 25 degrees. I mean, in just pure water
it’s 10 to the minus 7. But no matter what we do to this
and this in an aqueous solution, the product is always
going to be 10 to the minus 14th power. So that’s the answer
to this question. This is always going to
be 10 to the minus 14. If you multiply hydrogen
concentration times OH concentration. Now they won’t each be 10 to the
minus 7 anymore, because we’re dealing with a weak
acid or a weak base. So they’re actually going
to change these things. But when you multiply them,
you’re still going to get 10 to the minus 14. And let’s just take the minus
log of both sides of that. Let me erase all this stuff
I did down here. I’ll need the space. Let’s say we take the
minus logs of both sides of this equation. So you get the– let me do a
different color– minus log, of course it’s base 10, of Ka. Let me do it in the colors. Ka times Kb is going to be equal
to the minus log of 10 to the minus 14. So what is this equal to? The log of 10 to the minus 14
is minus 14, because 10 to minus 14th power is equal
to 10 to the minus 14. You take the negative of that,
so this becomes 14. So the right-hand side of your
equation just becomes 14. And this one, we could
use log properties. This is same thing as
the minus log of Ka. We use the colors. Ka plus the minus log of Kb. Or, though you could think of
this– this is your pKa, this is your pKb. So you can say, this
is pKa plus pKb. Oh, I wanted to use blue. Plus pKb, and all of that’s
going to be equal to 14. Now why is this useful? Well, if you know the pKa for
a weak acid– For example, let’s say we have NH4 plus. This is a weak acid, right? It can donate an H, but it’s not
an irreversible reaction. That H can be gained back. So this is a weak acid. If you look it up on Wikipedia,
it’ll say, hey, the pKa of NH4 is equal to 9.25. Right? So this is 9.25 for NH4. For ammonium. So what is going to be
the pKb for ammonia? Right? Let me write that
reaction down. So this is NH4. Is in equilibrium. This is plus. It can get rid of one of its
hydrogen protons, and you’re just left with ammonia. So this is the acidic
reaction. So this is what the pKa
is associated with. So the equilibrium constant
for this reaction is– the negative log of the equilibrium constant is equal to 9.25. And if I had the reverse
reaction, the conjugate base reaction, so ammonia converts
to ammonium. Plus it grabbed that hydrogen
proton from a water molecule. If I wanted to figure out the
pKb, or the equilibrium constant, or the negative log
of the equilibrium constant for this reaction, what is it? Well, this one’s 9.25. And 9.25 plus this pKb have
to be equal to 14. So what’s 14 minus 9.25? It’s what, 4.75. So we immediately know the
equilibrium constant for the conjugate base reaction. So it’s a useful
thing to know. That the pKa plus the
pKb is equal to 14. And always remember, you see
these pKa and pKb, and you say, what is that? Well, if you see a p, it’s a
negative log of something. And in this case, it’s the
negative log of the equilibrium constant for
an acidic reaction. Plus the negative log of the
equilibrium basic reaction, where this is the conjugate
base of this acid. It’s always going to be equal to
14 if we’re dealing with an aqueous solution at 25 degrees
Celsius, which is essentially room temperature, which is
usually going to be the case.

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