Trust in management as a buffer of the relationships between overload and strain . Journal of Kim, P. H., Ferrin, D. L., Cooper, C. D., & Dirks, K. T. (). This relationship is defined as the soil buffer capacity. (Bache, ). The determination The general factors responsible for soil BC and pH buffer- ing in soils, include the cultural Trust, Rothamsted Experimental Station. Claredon Press. pH = -log [H+]. • pOH = -log [OH-] ([H+] and [OH-] in M). • [H+] x [OH-] = 1 x M2 / pH + pOH = • An acid is defined as a proton donor. • AH = A- + H+.
But that's not the point of this video. So let's just think a little bit about what would happen to this equilibrium if we were to stress it in some way. And you can already imagine that I'm about to touch on Le Chatelier's Principle, which essentially just says, look, if you stress an equilibrium in any way, the equilibrium moves in such a way to relieve that stress.
So let's say that the stress that I apply to the system-- Let me do a different color. I'm going to add some strong base. I'm going to add some NaOH. And we know this is a strong base when you put it in a aqueous solultion, the sodium part just kind of disassociates, but the more important thing, you have all this OH in the solution, which wants to grab hydrogens away. So when you add this OH to the solution, what's going to happen for every mole that you add, not even just mole, for every molecule you add of this into the solution, it's going to eat up a molecule of hydrogen.
So for example, if you had 1 mole oh hydrogen molecules in your solution and you added 1 mole of sodium hydroxide to your solution, right when you do that, all this is going to react with all of that.
And the OHs are going to react with the Hs and form water, and they'll both just kind of disappear into the solution. They didn't disappear, they all turned into water. And so all of this hydrogen will go away. Or at least the hydrogen that was initially there. That 1 mole of hydrogens will disappear. So what should happen to this reaction? Well, know this is an equilibrium reaction. So as these hydrogen disappear, because this is an equilibrium reaction or because this is a weak base, more of this is going to be converted into these two products to kind of make up for that loss of hydrogen.
And you can even play with it on the math. So this hydrogen goes down initially, and then it starts getting to equilibrium very fast.
pH and pKa relationship for buffers
But this is going to go down. This is going to go up. And then this is going to go down less. Because sure, when you put the sodium hydroxide there, it just ate up all of the hydrogens. But then you have this-- you can kind of view as the spare hydrogen capacity here to produce hydrogens.
And when these disappear, this weak base will disassociate more. The equilibrium we'll move more in this direction. So immediately, this will eat all of that. But then when the equilibrium moves in that direction, a lot of the hydrogen will be replaced. So if you think about what's happening, if I just threw this sodium hydroxide in water. So if I just did NaOH in an aqueous solution-- so that's just throwing it in water-- that disassociates completely into the sodium cation and hydroxide anion.
So you all of a sudden immediately increase the quantity of OHs by essentially the number of moles of sodium hydroxide you're adding, and you'd immediately increase the pH, right? When you increase the amount of OH, you would decrease the pOH, right? And that's just because it's the negative log.
And just think OH-- you're making it more basic. And a high pH is also very basic.
- Acids, bases, pH, and buffers
If you have a mole of this, you end up with a pH of And if you had a strong acid, not a strong base, you would end up with a pH of 0. Hopefully you're getting a little bit familiar with that concept right now, but if it confuses you, just play around with the logs a little bit and you'll eventually get it.
But just to get back to the point, if you just did this in water, you immediately get a super high pH because the OH concentration goes through the roof. But if you do it here-- if you apply the sodium hydroxide to this solution, the solution that contains a weak acid and it's conjugate base, the weak acid and its conjugate base, what happens?
pH Scale: Acids, bases, pH and buffers (article) | Khan Academy
Sure, it immediately reacts with all of this hydrogen and eats it all up. And then you have this extras supply here that just keeps providing more and more hydrogens.
And it'll make up a lot of the loss. So essentially, the stress won't be as bad. And over here, you dramatically increase the pH when you just throw it on water. Here, you're going to increase the pH by a lot less. And in future videos, will actually do the math of how much less it's increasing the pH.
But the way you could think about it is, this is kind of a shock absorber for pH. Even though you threw this strong based into this solution, it didn't increase the pH as much as you would have expected. And you can make it the other way. If I just wrote this exact same reaction as a basic reaction-- and remember, this is the same thing.
So if I just wrote this as, A minus-- so I just wrote its conjugate base-- is in equilibrium with the conjugate base grabbing some water from the surrounding aqueous solution. Everything we're dealing with right now is an aqueous solution. And of course that water that it grabbed from is not going to be an OH. Remember, are just equivalent reactions.
Here, I'm writing it as an acidic reaction. Here, I'm writing it is a basic reaction. If you were to add a strong acid to the solution, what would happen? So if I were to throw hydrogen chloride into this.
Well hydrogen chloride, if you just throw it into straight up water without the solution, it would completely disassociate into a bunch of hydrogens and a bunch of chlorine anions. And it would immediately make it very acidic. You would get to a very low pH. If you had a mole of this-- if your concentration was 1 molar, then this will go to a pH of 0. But what happens if you at hydrochloric acid to this solution right here? This one that has this weak base and its conjugate weak acid?
Buffers and Henderson-Hasselbalch (video) | Khan Academy
Well, all of these hydrogen protons that disassociate from the hydrochloric acid are all going to react with these OHs you have here. And they're just going to cancel each other out. They're just going to merge with these and turn into water and become part of the aqueous solution.
So this, the OHs are going to go down initially, but then you have this reserve of weak base here. And Le Chatelier's Principle tells us. Look, if we have a stressor that is decreasing our overall concentration of OH, then the reaction is going to move in the direction that relieves that stress. So the reaction is going to go in that direction. So you're going to have more of our weak base turning into a weak acid and producing more OH.
So the pH won't go down as much as you would expect if you just threw this in water. This is going to lower the pH, but then you have more OH that could be produced as this guy grabs more and more hydrogens from the water.
So what we're gonna do, is we're gonna rearrange this equation to solve for this ratio that we might be interested in. And I don't know about you, but I actually find, well, laughs I find logs not super-intuitive sometimes.
So I'm actually going to get rid of the log by raising both sides to the 10th power. So what does this tell us? It may not look like it tells us a whole lot more, but actually, it tells us a lot. It tells us about the relative relationship and size between A minus and HA concentration.
So if we look at this, we can derive a couple relationships. So let's go ahead and look at all the possible scenarios for these three things. So anything to the zeroth power is equal to one. Which tells us that this ratio is equal to one. And if A minus concentration over HA concentration is equal to one, that means that they have the same concentration.
I forgot a minus sign there. This is a really helpful thing to remember.
Buffers and Henderson-Hasselbalch
And this comes up a lot not just when you're talking about buffers by themselves, but also when you're doing titrations. And the point in your titration where the HA is equal to A minus is called the half-equivalence point. And if you haven't learnt about buffers, that's okay. Oh, sorry, if you haven't learn about titrations yet, that's fully fine. Just ignore what I just said laughsbut if you have, the moral is just that, this is a really, really important relationship that is really helpful to remember.