How to derive the Henderson Hasselbach Equation for weak bases.

Added: 2026-05-15

[00:00:00]Hi everyone and welcome back to another chemistry video.

[00:00:04]Today we are going to talk about how to derive the Henderson-Hasselbalch equation for weak bases.

[00:00:11]The question is, what is this formula used for?

[00:00:15]This equation for weak bases is used for calculating the pOH of a buffer solution.

[00:00:23]When you are given the concentration of the weak base and the concentration of the conjugate acid.

[00:00:30]Now, how do we derive it?

[00:00:32]Well, since we're deriving the formula for weak bases or for calculating the pOH of a weak base, we are going to use a weak base to also do that. Follow me closely and attentively as we solve this question together.

[00:00:47]So, I'm going to say solution.

[00:00:50]Then, we are going to use a weak base.

[00:00:54]An example of a weak base is ammonia.

[00:00:58]And then, we are going to dissolve this ammonia in water. Water in this case is an acid. Remember, when water reacts with a base, it acts as an acid. And when water reacts with an acid, it acts as a base. Now, [snorts] this is it.

[00:01:15]This is an acid.

[00:01:18]And here is the weak base. Remember that ammonia is a weak base.

[00:01:23]So, by definition, an [snorts] acid is a proton donor.

[00:01:29]And what is it going to donate? This hydrogen.

[00:01:32]So, this H2O here will donate one of its hydrogens to ammonia.

[00:01:40]And when it does that, this now becomes NH4. Remember that this is a neutral compound, right? So, when it donates one hydrogen, which is a proton, protons are positively charged.

[00:01:54]Okay? They're positively charged. So, when it donates one proton to a neutral compound, it becomes more positive.

[00:02:03]And then, when it loses one positive, it becomes a negative. So, we now have OH- remaining. Remember, if you remove one hydrogen from two hydrogens, you only have one hydrogen left, right?

[00:02:18]Okay. So, this now becomes H and O remaining.

[00:02:22]And uh since they're going to lose a positive, they're going to become a negative, and that is where this negative is coming from. So, having written this dissociation reaction, the next thing I'm going to do is to write the KB, which is the base dissociation constant.

[00:02:41]Okay?

[00:02:42]For bases. Then, this is going to be the concentration of the products over the concentration of reactants.

[00:02:53]That's how to write your KB expression.

[00:02:55]And then, we have OH-.

[00:02:58]OH- rather.

[00:03:03]Over And then, we have the concentration of the reactant, which is NH3.

[00:03:10]Now, you might ask me, why did I exclude H2O? So, H2O was excluded because when writing this type of expressions, pure liquids and pure solids are not included. So, right here, H2O is a pure liquid because it appears as a reactant, so it is pure. Their concentration Its concentration will not change in the course of the reaction because it's a reactant. So, we have to exclude it.

[00:03:36]Remember, when writing this type of reaction, whether it's KA, whether it's KB, water as a reactant is a pure solid and must not appear in your KA or KB expression.

[00:03:52]Okay. So, next thing I'm going to do is to rearrange this. Remember that these two are multiplying, so their position does not matter.

[00:03:59]This becomes KB is equal to the concentration of OH minus multiplied by the concentration of NH4 [clears throat] plus over the concentration of NH3.

[00:04:17]Okay, so you might ask me why did I interchange them? Yes, they are multiplying, so their concentrations are being multiplied, and when you're multiplying things together, their position is irrelevant.

[00:04:28]Next thing, what am I going to do? I'm going to write this nicely to be KB is equal to So, I'm going to open a bracket. We have OH minus, this is a concentration.

[00:04:39]Then times we have the concentration of NH 4 plus over the concentration of NH 3.

[00:04:52]Okay, I haven't still changed anything yet at this very moment because they are still multiplying. So, the concentration of hydroxide ions, which is right here, multiplied by this one will still give you the original one back, okay? I'm only trying to manipulate the expression so that I can take the logarithm on both sides.

[00:05:12]So, next we take log on both sides.

[00:05:20]It could be both sides of of both sides or on both sides, anyone is okay. So, we introduce logarithms, okay? So, once you're introducing logarithm on both sides, please make sure you do it on both sides, not just one side. So, this becomes log KB is equal to log open bracket concentration of OH minus times the concentration of NH4 loss over the concentration of NH3.

[00:05:58]Okay. Now, what next are we going to do here? The next thing that we're going to do is to remember that we have a logarithm property that says, so I'm going to say recall log A times B is equal to log A plus log B. So, this is the logarithm property that says when two logarithms are multiplying, you can separate them.

[00:06:26]Okay? But then they have to be added when you separate them.

[00:06:30]So, these two here, we have this and this ratio, they're multiplying each other right? Correct. So, we can also use the same logarithm property to separate them.

[00:06:40]So, this is now going to be log KB equals So, separating these two, we now have log concentration of hydroxide ions plus >> [clears throat] >> log We have this ratio here.

[00:07:01]So, I'm going to open a bracket.

[00:07:03]We have NH4 loss over NH3.

[00:07:12]Okay. So, they've been separated using this logarithm property.

[00:07:16]Now, what next am I going to do here?

[00:07:18]I'm going to interchange their positions. So, this log hydroxide ions will move to the left while the log KB goes to the right.

[00:07:28]So, this [snorts] is now going to be negative log concentration of hydroxide ions is equal to negative log KB.

[00:07:39]Okay, plus log open brackets concentration of NH4 plus over the concentration of NH3.

[00:07:53]Then you can close the brackets.

[00:07:55]Now, look at this very closely here.

[00:07:57]This logarithms they became negative because I interchange the position. This one was a plus, when it goes over the equality sign, it became a negative.

[00:08:06]This one as well was a positive logarithm, when it crossed the equality sign, it became a negative. So, we need to also remember that we have I'm going to say recall that pOH or we can say recall that whenever you have negative logarithm of hydroxide ions this equal to pOH.

[00:08:32]So, this is the well-known formula. It's part of the big six equations in acids [clears throat] and bases. And then whenever you have negative log KB, this is equal to pKB.

[00:08:46]Okay? So, this is how you're going to write it in your exam. Whenever you have negative logarithm the negative logarithm of hydroxide ions, it's equal to pOH.

[00:08:56]And whenever you have the negative logarithms of KB, it's equal to pKB. So, we do the replacement. Whatever we see negative log OH minus, we put its value.

[00:09:08]So, this now becomes pOH is equal to whatever we see negative log KB, we put pKB.

[00:09:18]Then plus log And here we have the concentration of NH4 plus over the concentration of NH3.

[00:09:35]Okay. Now, this is the formula that we're looking for, but we need to understand and know which one is the conjugate acid and which one is the weak base. Let me take you back to the original um chemical reaction we started with.

[00:09:49]So, look at this.

[00:09:51]This is an acid, okay?

[00:09:53]When an acid donates a proton, whatever remains becomes a conjugate base.

[00:10:04]Let me repeat this again. When an acid donates a proton, whatever remains of that acid becomes a conjugate base.

[00:10:16]So, when the H2O donated a proton, what remained of it was OH-. So, this OH- is our conjugate base.

[00:10:26]When a base, okay, accepts a proton, so this base accepts a proton from H2O, it's going to change into something else. That thing it's going to become will be a conjugate acid.

[00:10:49]So, when the base accepts a proton, it becomes a conjugate acid. Don't forget.

[00:10:57]And when the acid donates a proton, it becomes a conjugate base.

[00:11:02]So, in this case right now, we can see that NH4+ which is right here, is our conjugate acid. And NH3 is our weak base. So, we now say, since NH4 is our conjugate acid, and NH3 is a weak base.

[00:11:38]The formula will now be written as Therefore, pOH is equal to pKb plus log We have concentration of conjugate acid all over the concentration of the weak base.

[00:12:08]Okay. So, this is how to derive the formula for calculating the pOH of a buffer solution given the conjugate acid and the weak base.

[00:12:18]Thank you so much for watching this video. If you have any questions, comments, or concerns, you can reach out to me and I'll be right in the comment section waiting for you. Thank you and see you in the next one.