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Course: Chemistry library > Unit 11
Lesson 2: Acid-base equilibriaWeak base equilibrium
Quick overview of Kb and pKb. Examples of calculating the pH of a weak base solution.
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By finding the ph of ammonia, shouldn't we technically be solving for the pOH since ammonia is a base. It makes no sense to me that the pH is a larger number than the pOH. I would expect the pH to be 2.52 and pOH to be 11.48(0 votes)
Keep in mind that a pH >7 is classified as a base and a pH < 7 is considered acidic .
We know that ammonia is a base so we expect the pH to be above 7.
Remember that pH + pOH = 14
This can be rearranged to pH = 14 - pOH
If the pOH was larger than the pH then the solution would have to be acidic.(14 votes)
- I don't understand why we assume that .003 is for the OH. Couldn't it be for the NH4+ also since they are both set to x?!(4 votes)
You are correct,xrepresents both the [OH¯] and [NH₄⁺].
Why do you think that is a problem?
Remember the goal was to work out the pH, so [OH¯] matters while [NH₄⁺] does not ...(6 votes)
Why do we say that the concentration of pure liquids and solids is one? Is it molarity? If it is, are we saying that there is always one mole of the solid/liquid in every litre of the solid/liquid?(3 votes)- It is the activity of the liquids and solids that equals 1.
The concentration of water is 55.5 mol/L. The typical concentration of a solute in a reaction
is 0.1 mol/L to 1 mol/L.
Even if the water is a reactant in the reaction, its concentration stays almost constant.
We can consider that its constant is incorporated into the equilibrium constants K_a, K_b. and K_w.(5 votes)
how come we can assume that x<< 0.500 or x<< 1 in the two examples discussed above?(4 votes)
let us consider the case of NH3. x in the above example is the concentration of NH4+ and OH- formed ( also called degree of ionization), and we know that NH3 is a weak base. Since Weak bases ( or acids) only disassociate by a very small amount, the conc. of NH4+ and OH- ( which is x) is very small. Hence we ignore it to simplify our calculations, and assume that 0.5 - x is almost the same as 0.5. Hope this helped :)(1 vote)
I've also seen a version of finding the pH of a base by dividing the number you get from the ICE table calculation by 1.01x106-14. It got me a pH of 11.47 so I was wondering if it was the same thing or if it was just lucky for this calculation?(1 vote)
Assuming you mean x = [OH-] when you say what we get from the ICE table.
1.0 x 10^(-14) is Kw, or the autoionization constant of water at 25°C. We can relate the hydroxide concentration to the hydronium concentration by the autoionization of water reaction.
H2O(l) + H2O(l) ⇌ H3O^(+)(aq) + OH^(-)(aq) so Kw = [H3O^(+)][ OH^(-)] = 1.0 x 10^(-14). So if we know [OH-] then we can find [H3O^(+)] and therefore get pH because pH = -log([H3O^(+)]).
This basically the same method used in the video. The Kw equation from before can be written as pKw = pH + pOH by taking the –log of both sides. And pKw is 14 at 25°C. So if we know [OH-] then we can find pOH since pOH = -log([OH-]). And then we can use the pKw equation to find pH.
So in both methods you use the autoionization of water to relate hydroxide and hydronium concentrations.
Hope that helps.(5 votes)
At 0.05, why does the water donate proton and ammonia don't?Why ammonia can't act as acid? How to figure out which one donates proton and which one not?(2 votes)
That is because NH2 (it's apparently written as H2N- and called azanide) is a better base than water. So for that reason NH3 will have to be the Bronsted-Lowry Acid here. You'd usually know which way it goes by practicing and seeing questions after all the range of compounds you will be given will be limited and you'd normally instantly be knowing the answer.(1 vote)
for example at0:45(as with previous videos), why is the B-L definition of a base used when the Lewis definition is more inclusive?(3 votes)
The B-L definition touches more on bases having one less proton (H+) that acids, relating more on the Arrhenius definition.(1 vote)
why did you mention concentration as I and not C?(1 vote)
I stands for initial concentration
C stands for change in concentration
E stands for equilibrium concentration(4 votes)
- Does Pka for bases exists and can we predict ionization strenth of the same(2 votes)
- We can only make these mathematical assumptions with weak acids or bases, correct?(1 vote)
- You only need to do so with weak acids and bases yes(2 votes)
0:45Video transcript
- Ammonia is a weak base, and if ammonia reacts with water water is gonna function
as a Brønsted–Lowry acid. Water is going to donate
a proton to ammonia, and ammonia is going to accept the proton. It's going to be a Brønsted–Lowry base. Lone pair of electrons in the nitrogen pick up this proton which
leaves these electrons behind on the oxygen. If you protonate ammonia
you form ammonium, the ammonium ion which is NH4 plus. Let's show those electrons. These electrons, these
lone pair of electrons here on the nitrogen pick up this proton and form this bond right here. We form NH4 plus, and these electrons in blue
come off on to the oxygen. The oxygen had two lone
pairs of electrons around it. And the electrons in
blue are now around it which give it a negative
1 formal charge here. We form the ammonium ion, NH4 plus, and the hydroxide ion, OH minus. If ammonia functioned
as a Brønsted–Lowry base over here would be the conjugate acid. Ammonium is the conjugate acid to NH3. Water functioned as a Brønsted–Lowry acid. Over here would be the conjugate base, the hydroxide anion. Instead of using ammonia
let's just do a generic base. Here I have B written. Some generic base reacts with water. It takes a proton from water to form BH plus. If you take a proton away from H2O you would form OH minus. Once this comes to equilibrium we could write an equilibrium expression. We're gonna write Kb here. Instead of Ka we're now writing Kb because we're talking about a base. Kb can be called, you can call this the
base ionization constant. Let me write base ionization constant here. Or you could call this the
base dissociation constant, so base dissociation constant. When you're writing an
equilibrium expression remember it's the
concentration of your products over your reactants. Over here we have the
concentration of BH plus times the concentration of OH minus. That's all going to be over the concentration of your reactants, so we leave out water so we
have only the concentration of our generic base, B. You could think about Kb the
same way we thought about Ka. The higher the value for Kb the stronger the base because the more of your products you are going to make here. Let me go ahead and write that. The stronger the base the larger the value for large for Kb. Let's talk about two weak bases here. We've already mentioned ammonia, and we're going to compare
ammonia to aniline. NH3 is ammonia, and the Kb for ammonia is 1.8 times 10 to the negative 5. Aniline is C6H5NH2. Notice it has a lower value for the Kb, 4.3 times 10 to the negative 10 is a much smaller number,
a much smaller value than 1.8 times 10 to the negative 5. Ammonia has a higher value for the Kb, and therefore it's a
stronger base than aniline. Once again, both of them are weak bases but ammonia is the stronger of the two. Remember when we talked about Ka we also talked about pKa. We're gonna do the same thing here. We've talked about Kb, and so now let's talk
about calculating the pKb. The pKb is equal to the negative log of the Kb. Let's say we wanted to
find the pKb for ammonia. All we'd have to do is
plug in the Kb value. The pKb would be equal to the negative log of 1.8 times 10 to the negative 5. Let's get out the calculator
here and do the problem. Negative log of 1.8 times 10 to the negative 5. We get 4.74 here. The pKb for ammonia is 4.74. You could do the same
calculation for aniline and plug in this Kb, and you would get a pKb of 9.37. Comparing two weak bases, ammonia and aniline, ammonia is the stronger of the two because it has a higher value for the Kb. And notice that the pKb is a lower value, so once again analogous to
what we talked about for pKa. Next we do a calculation for a solution of ammonia. Our problem asks us to calculate the pH of a 0.500 molar solution
of aqueous ammonia. We have ammonia in water. We have NH3 plus H2O. Ammonia is going to
accept a proton from water and turn it into NH4 plus, ammonium. If water loses a proton H2O turns into OH minus. Let's go ahead and write our
initial concentration here. For ammonia we have 0.500 molars. We write 0.500 molar here. And we're pretending like
nothing has happened, so the concentration of our products is 0. Nothing has happened yet so the concentration of ammonium is 0, and the concentration
of hydroxide is also 0. Next let's think about the change. Since ammonia, since
NH3 turns into NH4 plus, the concentration of ammonia that's lost is the same concentration
as ammonium that is gained. If we represent that
concentration using X, if we lose a certain
concentration of ammonia that's the same concentration
of ammonium that we gain. This would be plus X
over here for ammonium. It'd be the same thing for hydroxide. This would be a plus X here
for the hydroxide anion. Therefore at equilibrium, for ammonia, the concentration
of ammonia at equilibrium we would have 0.500 minus X. For ammonium we would have X, and for hydroxide we would also have X. Next we write our equilibrium expressions, so we write Kb here is
equal to the concentration of our products, so that's NH4 plus, so I write NH4 plus here, times the concentration of hydroxide. That's all over the
concentration of ammonia. We leave water out, so this all over the
concentration of ammonia here. Let's plug in what we know. At equilibrium the concentration of ammonium is X. Let's write in an X here for
the concentration of ammonium. And it's the same thing for the
concentration for hydroxide. For hydroxide, the
concentration at equlibrium is also X. We write an X right here. This is all over the
concentration of ammonia and that would be the
concentration of ammonia at equilibrium is 0.500 minus X. We put in 0.500 minus X here. This is all equal to the
base ionization constant for ammonia. Let's go back up to our table above to remind ourselves of what it is. Ammonia has a Kb of 1.8 times 10 to the negative 5. We're gonna plug that
value in for Kb here. This is equal to 1.8 times 10 to the negative 5. Once again we're going to assume that X is much, much smaller than 0.500 molar because
that makes our life easy for the math. We don't have to do the
quadratic formula if we do that. If we make this assumption
that 0.500 minus X, if X is extremely small, this is pretty much the
same thing as 0.500. We're going to leave that out. Let's rewrite what we have here. We have 1.8 times 10 to the negative 5 is equal to, this would be X squared over 0.500. All we have to do now is solve for X. We would have X squared is equal to, let's see, 1.8 times to 10 to the negative 5, we have to multiply that by 0.5 and we would get 9.0 times 10 to the negative 6. Next we just take the square root of 9.0 times 10 to the negative 6 to solve for X. Let's go ahead and do that. Let's get out the calculator here and let's take the square root of 9.0 times 10 to the negative 6. We get 0.003. Let's go ahead and write, X is equal to 0.0030. Remember what X refers to. Let's go back up here. Let's think about what X is talking about. X is talking about a concentration and notice we have an X
here for our concentration of hydroxide anions. That's what's going to allow
us to eventually get the pH which is what the question asked for. X is equal to the concentration
of hydroxide anions. This would be molar. This is our concentration
of hydroxide anions. Our goal is, once again, to
find the pH of our solution. First we could find the poH. That's one way to do it. The poH is equal to the negative log of the concentration of hydroxide ions. We could go ahead and
take the negative log of 0.0030 to calculate the poH of our solution. Let's do that on our calculator. Negative log of 0.0030 is equal to 2.52. The poH is equal to 2.52. Then to find the pH all we have to do is subtract from 14 because pH plus poH is equal to 14. I believe in a previous video I forgot to put these zeros here. We're concerned about significant figures. You plug in your poH into here, so your pH plus 2.52 is equal to 14.00. You just solve for pH. pH is equal to 14 minus 2.52 which is 11.48. Finally we've calculated
the pH of our solution of ammonia.