Main content

## Chemistry library

### Unit 13: Lesson 2

Acid-base equilibria# Relationship between Ka and Kb

The relationship between Ka and Kb, and pKa and pKb. Examples of finding Ka of a weak acid given Kb of the conjugate base.

## Want to join the conversation?

- What is the relationship between( PKa+PKb=14)and (PH +POH=14)(34 votes)
- There a slight relationship (see bottom), due to the fact that their measurements are very similar. For PH, the concentration of H3O+ is measured, for PKa, products over reactants is used.

If you think of it mathematically:

Ka x Kb = ( [H3O+]{Base] / [Acid] ) x ( [OH-][Acid] / [Base] ) = [H3O+][OH-] = 1.0x10^(-14)

and:

pKa + pKb = -log( [H3O+]{Base] / [Acid] ) + -log( [OH-][Acid] / [Base] )

= -log( [H3O+]{Base] / [Acid] ) x ( [OH-][Acid] / [Base] )

= -log( [H3O+][OH-] ) = 14

So: pH x pOH = pKa x pKb

This is why the conversions between pH and pOH are so similar to conversions between pKa and pKb.(41 votes)

- Can we compare pKa to pKb? So for example, if pKa for NH4+ is 10, and pKb for NH3 is 4, in this situation, can we say that NH3 is a stronger/better base, then NH4+ is an acid?(5 votes)
- Thar's exactly what you can say.

NH₃ is a stronger base than NH₄⁺ is an acid.(11 votes)

- What is the relationship between pKa/pKb and pH/pOH?(4 votes)
- pH and pOH add up to equal 14

pKa and pKb are the -logs of Ka and Kb; Ka and Kb, when multiplied, are equal to Kw (1.0x10^-14). Kw is also equal to the hydronium and hydroxide concentrations when multiplied.(6 votes)

- What is the physical sense of adding the two equilibrium reactions ...is it just mathematical?(3 votes)
- It's justified because NH4+ and NH3 are a conjugate acid-base pair. The transfer of a proton converts one to the other.

Remember, the two reactions that are being added together are each reversible. Each reaction is interconverting NH4+ and NH3. If you have NH3 in water, both reactions will take place (although the equilibrium constant if different for each).(6 votes)

- I entered 1x10^-14 / 3.7x10^-4 to my graphing calculator and it gives me the answer 2.7027x10^-19. Does anyone know why I am getting the wrong answer?(2 votes)
- Yes. You are probably using the "x" sign to make 1x10^-14.

Don't do that. Your calculator should have a key that says EE or maybe EEX. You use that to create scientific notation.

When you do it the way you are currently doing it, your calculator thinks you are doing multiplication and division of 4 numbers:

1

10^-14

3.7

and

10^-4

It uses standard order of operations, doing multiplication and division from left to right, so you get

1 x 10^-14 = 1x10^-14

1x10^-14/3.7 = 2.7 x 10^-15

2.7 x 10^-15 x 10^-4 = 2.7x10^-19

If you want to insist on doing it this way, you need to use parentheses

(1x10^-14) / (3.7x10^-4)

Then you will get the right answer

But you really need to learn how to use your calculator properly. You need to be able to enter 3.7E-4 as a single number, not as the product of 3.7 and 10^-4. It's much faster, and much more accurate.(6 votes)

- This is more of a math than conceptual question, but what is a quick way to convert the pKa or pKb back Ka or Kb? Like for example if a question asks what is the Kb of a base with a pKb of 5.4?(3 votes)
- this is also a conceptual question because it looks like you might not be fully comfortable with the idea of Ka/Kb yet. you can work with these numbers the same way you would convert an acid/base concentration to pH/pOH. so if the pKb is 5.4, you can take 10^-5.4 to get the Kb.(3 votes)

- Is the equation Ka x Kb = Kw (similarly, pKa + pKb = 14) only applicable for reactions where the H2O is the solvent?(3 votes)
- Yes but all acids and bases are dissolved in H2O so this is always the case.(3 votes)

- What does the numerical value of Ka and Kb signify? I get its the equilibrium constant, but what does it mean for something to have a constant of 5.6 X 10^-10? Is it the rate that the forward/backward reactions occur?(1 vote)
- The K values give us an idea of the relative amounts of products and reactants at equilibrium.

If K is large, there is a large amount of H⁺ or OH⁻ at equilibrium and very little undissociated acid or base. We have a**strong**acid or base.

If K is very small, there is very little H+ or OH- at equilibrium, and mostly undissociated acid or base. We have a**weak**acid or base.(5 votes)

- Is there a maximum and a minimum value of pKa and pKb? If there are, are they 14 and 0 respectively, where 14 would be a very weak acid/base and 0 a very strong acid/base?(1 vote)
- There aren't really any mathematical reasons that pKa (or pKb, since they are measured the same way, just measuring different properties) would be limited to any particular range, but there are some chemical reasons. To get to greater extremes of acid or base, you have to use more extreme materials, and you would be working against what the atoms generally want to do. So, there are some practical limits.

Lots of familiar acids have negative pKa: hydrochloric acid, HCl, has a pKa of about -6. Sulphuric acid, H2SO4, is around -3 (for the first proton, the second is harder to remove).

Some of the strongest acids have much lower pKa: Fluoroantimonic acid is mind-bogglingly acidic, with a pKa of -25!

On the other hand, pKa values can go very high for basic compounds, or for other molecules that don't really act as acid or base. For example, methane, CH4, is pretty happy where it is, and doesn't especially want to give up or to take a proton, and has one of the highest pKa values I've seen, at around 45.

Mind, these are extreme extremes, and most of what you see will be in that middle range of around -7 to 15 or so. So, keep this range in mind for normal stuff, and the extremes in mind as a "sanity check" if you get an answer that seems way out there.(3 votes)

- What are the ka and kb values of water??(2 votes)
- Ka is 10^-7 because once you calculate -log(Ka) you get pH=7, which is neutral water.

The same goes for Kb (also 10^-7). You can do the same log calculation but solve for pOH=7(1 vote)

## Video transcript

- We've already seen that NH4 plus and NH3 are a conjugate acid-base pair. Let's look at NH4 plus. The ammonium ion would function as an acid and donate a proton to water to form H3O plus. If NH4 plus donates a proton you're left with NH3. The Ka for this reaction is 5.6 times 10 to the negative 10. Now let's look at NH3 which
we know is a weak base, and it's going to take
a proton from water, therefore forming NH4 plus. If we take a proton from water we're left with OH minus. Since we talked about a base here we're gonna use Kb, and Kb for this reaction is
1.8 times 10 to the negative 5. What would happen if we add these two reactions together? We have two water molecules
for our reactants, so let me go ahead and write H2O plus H2O here. What about ammonium? We have ammonium on the
left side for reactant, we also have ammonium
over here for our product. That cancels out. Same thing happens with ammonia, NH3. We have NH3 on the left. We have NH3 on the right. We have NH3 as a reactant,
NH3 as a product. We can cancel those out too. Our only reactants would be two H2O. For our products we would
get H3O plus and OH minus, so H3O plus, hydronium, and hydroxide. This reaction should
sound familiar to you. This net reaction is the
auto-ionization of water where one water molecule acts as an acid, one water molecule acts as a base. We get H3O plus and OH minus. The equilibrium constant for
the auto-ionization of water you've already seen
that Kw is equal to 1.0 times 10 to the negative 14. We added these two reactions together and we got this for our net reaction. What would we do with Ka and Kb to get Kw? It turns out that you multiply them, Ka times Kb for a conjugate acid-base pair is equal to Kw. Let's do that math. Ka is 5.6 times 10 to the negative 10. So 5.6 times 10 to the negative 10. Kb is equal to 1.8 times 10 to the negative 5, 1.8 times 10 to the negative 5, and let's get out the
calculator and let's go ahead and do that math. We have 5.6 times 10 to the negative 10. We're going to multiply that by 1.8 times 10 to the negative 5. We get 1.0 times 10 to the negative 14. This is equal to 1.0 times
10 to the negative 14 which is our value for Kw. When you add reactions
together to get a net reaction you multiply the equilibrium constants to get the equilibrium
constant for the net reaction which in this case is Kw for
the auto-ionization of water. Let's go ahead and go in
even more detail here. Ka, that's your products
over your reactants. That'd be H3O plus, the concentration of H3O plus times the concentration of ammonia. Let's go ahead and do that. The concentration of H3O plus times the concentration of ammonia, and that's all over the
concentration of ammonium. This is Ka. This is all over the
concentration of ammonium. This represents, let me go
ahead and highlight this here. This represents Ka. Next let's think about Kb. Over here is Kb, that'd be the concentration
of your products, so NH4 plus times OH minus. Let's go ahead and do that. Let's put this in parenthesis here. We have the concentration
of ammonium, NH4 plus, times the concentration of OH minus. That's over the concentration of NH3. That's over the concentration of NH3 here. What do we get? The ammonium would cancel out. That cancels here. Then the NH3 cancels out. We're left with H3O plus times OH minus which we know is equal to 1.0
times 10 to the negative 14. Just another way to think about this. This can be important, relating Ka and Kb to Kw. If you know one you can
figure out the other. Let's think about a strong
acid for a second here. Let's think about HCl. The conjugate base to
HCl would be Cl minus, the chloride anion here. Let's think about what
this equation means. HCl is a strong acid which means a very high value for Ka. An extremely, extremely high value for Ka. What does that say about
Kb for the conjugate base? The conjugate base here
is the chloride anion. If Ka is very large then
Kb must be very small for this to be equal to Kw. Kb is extremely small here, so a very small value for Kb. This mathematically describes
what we talked about earlier the stronger the acid the
weaker the conjugate base. HCl is a very strong acid,
so it has a very, very high value for Ka. And the conjugate base
is the chloride anion, and it must have a very,
very, very low value for Kb which means it's an extremely weak base. This is mathematically how to
think about that relationship. Next let's look at a problem where we're calculating
one of those values. Methylamine is a weak base, and the Kb for methylamine is 3.7 times 10 to the negative 4. Our problem asks us to
calculate the Ka value for the methylammonium
ion which is CH3NH3 plus. We're talking about a
conjugate acid-base pair. There's one proton
difference between those. Therefore we can use our equation, Ka times Kb is equal to Kw. We can plug in Kb here. Now we have Ka times 3.7 times 10 to the negative 4 is equal to Kw which is 1.0
times 10 to the negative 14. Let's do the math and solve for Ka. 1 times 10 to the negative 14. We need to divide that by 3.7 times 10 to the negative 4. Ka is equal to 2.7 times
10 to the negative 11. Ka is equal to 2.7 times 10 to the negative 11. We're done. That's our answer, 2.7 times 10 to the
negative 11 is the Ka value for the methylammonium ion. Let's go a little further. Let's take our equation here, Ka times Kb is equal to Kw. Let's take the log of both sides. That would be the log of Ka times Kb is equal to the log of Kw. The log of Ka times Kb is the same thing as the log of Ka plus the log of Kb equal to the log of Kw. If we take the negative of everything, let's go ahead and do that, the negative of everything, negative log of Ka. I'll put that in parenthesis, plus the negative log of Kb is equal to the negative log of Kw. The negative log of Ka, we know that this is equal to the pKa. The negative log of Ka was
our definition for our pKa, and the negative log of Kb was our definition for pKb. pKa plus pKb is equal to finally the negative log of Kw. That would give you 14. 14.00. The negative log of 1.0
times 10 to the negative 14 is 14.00. Now we have something else
that we can work with, so let me go ahead and
box this right here. Let's take the Ka value
that we just found. Let's find the pKa. The pKa would be equal to the negative log of 2.7 times 10 to the negative 11. Let's do that on our calculator here. Let's get some room. The negative log of 2.7 times 10 to the negative 11 gives us 10.57. We have to round that. 10.57. These are our two significant figures because we have two
significant figures here. Let's go back up to our problem here, so that's the pKa for
the methylammonium ion. Let's say you're given the
pKa for the methylammonium ion and asked for the pKb for methylamine. What is the pKb for methylamine? All we have to do is
plug in to our equation. 10.57 was our pKa value. Let's go ahead and write that in here. We have 10.57 plus pkb is equal to 14. When we solve for the
pKb that would give us 3.43. So the pKb is equal to 3.43. We could double-check that. Let's go back up here and
we could double-check that, because if we took the
negative log of this number that's what we should get. Let's go ahead and do that. Let's take the negative log of 3.7 times 10 to the negative 4. We get 3.43. That's what we just calculated down here. The pKb is equal to 3.43. That's the relationship between Ka and Kb, and you can also talk about
the relationship between pKa and pKb.