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## Acid/base equilibria

Current time:0:00Total duration:11:22

# Weak base equilibrium

## Video 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.