So with our discussion of mono product buffers. Di product buffers were now left with poly product buffers. So here we're gonna say polly product buffer is usually synonymous with a try product buffer where we have three acidic hydrogen. So we're gonna stay here try product buffer can be approached in a way similar to die product buffers. The key difference is that a try protic acid has now three P K values or three K values. Um Again when we say polly product we're usually referring to try product acids but there are acids with more than three acidic hydrogen, those are fewer. So we tend to just stick to three acidic hydrogen. Um Now we're gonna say for polyp product buffers. Remember we have different forms that are polyp protic acid can take we have the fully acidified version or it's fully predominated version which is H three A. Here it has all three of its acidic hydrogen, we can talk about removing the first acidic hydrogen. And if we're talking about removing the first acidic hydrogen, that means we're talking about K one, removing that first acidic hydrogen gives us our first intermediate form which is H two a minus. Then we remove the second acidic hydrogen, giving us K two involved. So now we get the second intermediate form which is H. A two minus and then finally removing the last acidic hydrogen which will give us our basic form a three minus. Remember in earlier videos where we talked about polyp Roddick acids in general if we could talk about K. A. Going one way in removing acidic hydrogen, we could talk about KB when adding back acidic hydrogen. So remember here this would be KB one. This would be K. B two and this would be KB three. If you don't quite remember that, make sure you go back and take a look at our polyp protic acid videos. We talked about the relationships between your acid dissociation constants and your based association constants when looking at protic dia protic and polyp Roddick assets. Now with these different forms of our polyp product buffers, we have different forms of the Henderson Hasselbach equation. In this first one here we're talking about relationship between the acid form in the first intermediate form. Then we know we're dealing with K. A. one. And so the equation is P. H equals P K. One plus log of the intermediate one over the acid form. Remember the bottom portion of the Henderson Hasselbach equation is the portion that contains more acidic H plus ions. And the top version which represents the more basic form has one less acidic H plus ion. Then if we're talking about our second form of the Henderson Hasselbach equation, we're talking about P. K. Two, which means we're talking about the relationship between our first intermediate and the second intermediate. So we've removed the second acidic hydrogen. So that's K two. Which is why we're dealing with P. K. Two. So here would be plus log of intermediate two divided by intermediate one. Again, the more acidic form, the one with more hydrogen goes on the bottom, the more that the one that's more basic, the one that has less acidic H plus ions goes on top. Then finally, in this form we're talking about intermediate two going to the basic form. So we're dealing with K3 just as before we put the more acidic form on the bottom and the more basic form on top. Now here we could take a common example, Citric acid, citric acid represents a try protic acid. So it has three acidic hydrogen. So in these three different scenarios in this first one, we have 10.10 moller of citric acid and then we have 0.15 molar of the first intermediate form here, it's lost one of its acidic hydrogen and it's been replaced by a sodium ion here. Since we're talking about the asset form in the first intermediate form, we're dealing with K- one. So here this would be P H equals P K A. Plus log of intermediate one divided by the asset form. So that would be negative log of 7.4 times 10 to the negative four plus log of the conjugate base form, which is 40.15 moller Divided by 0.10 Moeller. So that would give us 3.31 as our ph at the end for the next form here we're dealing with moles. Remember in the Henderson Hasselbach equation, we can plug in polarity or moles into these areas. And the first one? We have more clarity for both. So we plugged in polarity now we have moles so we'll plug in the molds now. So here this would be P H equals negative log of 1.7 times 10 to the negative five plus log of the more basic form goes on top the one with less acidic hydrogen, which would be this one on top. So 10.25 moles divided by the more acidic form, the one with more H plus ions on the bottom. So here that would give me 4.94 at the end. Then finally here we have volume of polarity. Remember in this last situation, moles equals leaders times polarity. So if you divide these uh mls by 1000 you'll get their leaders then you just multiply by their polarities. Right? So you'd get this many leaders times the polarity and then that would give us point 016 moles. And then we have .060 L times .025.25 Moeller And then give us .015 moles. So P H R equals negative log of K three times 10 to the negative seven plus log of more basic moles on top divided by the more acidic moles on the bottom. So at the end that will give us 6.43. So just remember when we're dealing with a polyp product buffer. It is important that, you know which forms are we dealing with in the question, This will dictate if you're going to use K one, K two, or K three, in terms of the Henderson Hasselbach equation. So as we delve deeper and deeper into these different types of buffers, always keep this in mind.