Diprotic Buffers - Video Tutorials & Practice Problems

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Diprotic buffers involve the 3 dominant forms of a diprotic acid and its 2 K_{aÂ }Values.Â

Diprotic Buffers

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Diprotic Buffers

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So in our discussions of buffers up to this point we've discussed mono product buffers where the acid only has one acidic hydrogen. Now we're going to extend that further by taking a look at dia product buffer systems. We're gonna say here that a dia product buffer can be approached in a way similar to mono product buffers. The key difference is that a di protic acid has two acidic hydrogen. So we're gonna say because it has two acidic hydrogen that's gonna possess to P. K. A. Or to K. Values. So just a quick refresher if we're taking a look at a mono product buffer. So here H represents our acid form where it possesses the H plus ion K. A. Can be thought of measuring the strength of this acid. The higher the K the stronger it is, the more likely it is to give away an H plus giving away H plus creates its base form or conjugate base form. Now a good example here we have hipaa Cloris acid which is H. C. L. O. It's a week oxy acid here. The conjugate base, we've removed the H plus from it usually by losing an H. Plus from our acid form. We replaced it with a group one metal. And this way we keep it as a neutral compound. So here is the conjugate base because again it has one less hydrogen. We know that the Henderson Hasselbach equation is just simply P. H. Equals P. K. A. Plus log of conjugate base over weak acid here in a mono product system, we only have one K value associated with the weak acid, so we know that this P k A. Is just one possible value. When we deal with dia product systems, though, we're gonna have to acidic hydrogen, so we have to choose wisely. Do we use P k. One or P. K. To, within our calculations, click onto the next video and see how we approach die product buffer systems.

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Diprotic Buffers

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So now when we take a look at di protic buffers, we have to pay attention to the fact that we have more than one K. Value associated with our acid forms here we started with H. Two A. Which is the generic form of any dia protic acid here it's fully propagated. So it has it's too acidic hydrogen. We're gonna remove the first acidic hydrogen to give us a church a minus. If we're talking about removing the first acidic hydrogen then we're dealing with K. A. one and the H A minus represents the middle or intermediate form of our initial die protic acid. Now when we're in the H a minus form we can talk about removing that last and second H plus ion. Talking about removing the second H plus ions means that we're gonna deal with K. Two. We're moving it gives us a two minus which represents the base form will form that has no H. Plus whatsoever attached to it. So because we have the presence of two acidic hydrogen will be dealing with two K values. Within our calculations we just have to make sure um which form are we going to use? So if we take a look here because of this presence of two K. S. We have two different forms of the Henderson Hasselbach equation. If we're dealing with the acid form and the intermediate form then K one connects them together. So the Henderson Hasselbach equation becomes P. H equals P. K. One plus log of the the intermediate divided by the acid form. If we're dealing with the intermediate form in the base form then we're dealing with K two. So the Henderson Hasselbach equation becomes P. H. Equals P. K. Two plus log of the base form divided by the intermediate form. If we were to take a look here at these two examples that we have. So if we take a look at this first one here, I'm telling us that we're dealing with sulfurous acid. So sulfuric acid is H. Two S. 03. This is the acid form of our acid. If we're gonna remove its first acidic hydrogen, that means we're gonna deal with K. A. One. And that will give us a church S. 03 minus which is hydrogen sulfide or by cell fight. So this is my intermediate form and then finally removing the second and which represents the last acidic hydrogen. We're dealing with K. Two. That would give me your soul fight ion. So this is the base form in this first question. We're dealing with the acid form here and we can see that one of the hydrogen has disappeared and we've replaced it with a sodium. So this is my intermediate form. So in this example here we'd say P. H. Equals P. K. One Which is the negative log of K one plus log of the concentration of the intermediate form divided by the acid form. So here this will give us 1.89 as our possible answer. Now in this one we're given volume of malaria. T remember when we talked about simple buffer systems. When we're dealing with mono product acids in earlier videos, we said that moles equals leaders times more clarity. So if we don't change these mls into leaders by dividing them by 1000 and then multiplying by the polarity, we get the moles of both of these compounds. So it be that many leaders times point 10 moller Will be .003 moles. And then we have .020 L times 0.20 moller they give me .004 moles. Now in this example, what do we have? We have sodium sulfite. So here both acidic hydrogen are totally gone now. So this must represent the base form. And here we still have one acidic hydrogen remaining. So this is the intermediate form. So here because we're dealing with the base form in the intermediate form, we know that Henderson Hasselbach equation becomes now P. H equals P. K two which is just the negative log of K two plus log of the moles of my base form divided by the moles of my my intermediate form. And again remember also from earlier videos we said that when it comes to the Henderson Hasselbach equation, we could use either moles or polarity as the units um within these formulas. So when we work this out, that gives me 1.07. Actually no, sorry, not 1.07. So we were supposed to take the negative log here of 6.48 times 10 to the negative eight. And then we're gonna add a log of .003 divided by .004. And that's supposed to give us here three supposed to give us seven point 07 as our final ph in terms of determining what the ph of my total solution would be. So we can see that there is a difference when we do this calculation whether we're dealing with the acid forming its intermediate or the intermediate form with its base form. So we have to be wary when it comes to die. Product buffer systems, know what form you're dealing with, which will therefore determine which PK value to use, which at the end will determine what your final ph will be. As we continue, our discussion of buffers will continue to do calculations dealing with dia product um acids and later on we'll move on to polyp Roddick acids.

Monoprotic & Diprotic Buffer Calculations

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example

Monoprotic & Diprotic Buffer Calculations 1

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Here, it states what is the ph of a solution consisting of 2.5 molar potassium di hydrogen phosphate and 2.75 moller a phosphorous acid. Now here we're given two K. S. K one and K two. And if we take a look at phosphorous acid, it's H three P. 03. And you might think that it's a try protic acid because it has three hydrogen in front of it. But in reality because of its structure, phosphorous acid actually only has two acidic hydrogen, it looks something like this. So the only acidic hydrogen are these two here. So it's actually a die protic acid, not a try protic acid as like it would seem. Now here we're dealing with it's fully acidic version when none of its hydrogen have come off yet. So this is it's acid form. And then here we've lost one of the hydrogen. So not only has two and the hydrogen that we've lost has been replaced by a potassium ion. So this potassium here, so this represents our intermediate form. Now, remember if we're dealing with the asset form in the intermediate form, that means we're talking about K one. So we're gonna have to use K one in this question. So here we're gonna say ph Equals p. K one plus log of the intermediate form over the acid form. So here that would be negative log of 3.0 times 10 to the -2 plus log of 2.5 Moller The concentration of my intermediate form divided by 2.75 Molar, which is the polarity of the asset form when we plug these in, that gives me a value of 1.48 as my ph So remember for questions like this we have to ascertain for sure what forms that we're dealing with? Are we dealing with the asset form with the intermediate form? In that case we're dealing with K. One? Are we dealing with the intermediate form and the base form then we would be dealing with K. Two. Now take the same approach as we move on to example to try to work out the question yourself to see how to answer it. But if you get stuck don't worry. Just come back and take a look at the next video. Look to see what exactly are we dealing with here, what forms are involved and therefore what K. A. To use.

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example

Monoprotic & Diprotic Buffer Calculations 1

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here it states that sulfurous acid which is H two S. 03 is a major component in the creation of commercial fertilizers here it asks, what is the buffer component concentration ratio of a buffer that has a ph of 1.15. Alright so we have a buffer here. So we know we're dealing with the Henderson Hasselbach equation. So P H. Equals P. K. Plus log of are conjugate base over our weak acid. When it's when it asks for the component concentration ratio, it's really asking what is the ratio of conjugate base too weak acid. So what is that value there? We run into the problem though, we're dealing with a dye protic acid here and that's illustrated by having two K values as well. We don't know are we dealing with P. K. One or P. K. Two within this question. How do we determine that? Well, we can take a look at all the forms that sulfurous acid takes when we begin to remove its H plus ions. So here we have sulfurous acid which is the acid form. We remove the first acidic hydrogen. So that means we're dealing with K. One. So that's H. S. 03 minus. Then we remove the last acidic hydrogen. So we're dealing with K. Two to give us S. 032 minus. Now we're gonna say here that the relationship between the acid form and the intermediate form. We're gonna say we have this imaginary line here we're gonna say this imaginary line here, we're gonna say this represents the line for P. K. One. So here if we took the negative log of K1 which is this number right here, take the negative log of it. We'll get PK1 equals 1.86. Then we're gonna say there's an imaginary line here which separates the intermediate form from the base form. So we take the negative log of this K. Two to give me P. K. Two. That comes out to 7.17. These P. K values are important because we're gonna say here, we can relate them in terms of P. H. So basically if our ph is equal To 1.86. So let's say P. H. equals P K. One, what would that mean? What would that would mean that my asset form would be equal to my intermediate form. And how do we know that? Well, if we're dealing with a buffer, if my asset form which would be this form is equal to my intermediate form, which is this form? All of this was just equal one log of one would equal zero. So that would just all drop out. And so ph would equal P. K. One. And we're gonna say here, if my P. H. Equals P. K. Two, that would mean that that my intermediate form equals my basic form for the same exact reason. Now if your ph is less than your P. K. One, what does that mean? That means you're dealing with a give and take between the acid form and the intermediate form. So our P K. A. R. P H. R is 1.15. Which means we fall somewhere in here. Which means we're talking about removing that first acidic hydrogen in order to create this intermediate form. If your ph happened to be a number that was greater than 1.86 and Less than 7.17, then you would fall somewhere within here. Remember this goes all the all the way back to our principal species when we're talking about the form that predominates depending on the ph and when we compared to RP K. Value and if we had a ph greater than this P. K. Two then this would be the dominant form. Now going back we said the ph is 1.15. So we know we're dealing with basically the acid form as the predominant form. And so we're talking about removing its first acidic hydrogen to make the intermediate form. So that would mean our equation now is P. H equals P K. One plus log of my intermediate form divided by my asset form. So PH one is 1.15. Then we're going to say here, let me take myself out you guys. P H equals 1.5. P. K. One equals 1.86 plus log of H. S. +03 minus divided by H two S. +03. Subtract both sides by 1.86. So when we do that we're gonna get initially negative .71 equals log of the intermediate form divided by the asset form. Then all we do now is we take the inverse log Which just means here 10 to the negative .71 equals the ratio of my intermediate form divided by my asset. For So when we punch that number into our calculators, that's gonna give us .195 over H two S. 03. So what is this number telling me while this number here is telling me since the ratio whatever number I get is always is always going to be over one. So what this ratio is telling me is that for every one Um weak acid form we have .195 of the intermediate form. That's the ratio this ratio is telling me that I have more of the acid, weak acid form than I do of the intermediate form. So our ratio here would be .195-1. To represent the ratio of my conjugate base or intermediate to my weak acid form. So just remember remember remember with uh polyp Roddick acids and die protic acid. We have to be aware of which K are we dealing with that will determine which PK we're gonna deal with within our Henderson Hasselbach equation to find either the ph the ratio of the conjugate base to the weak acid. Or possibly just the amount of the conjugate base by itself, or the weak acid by itself. So now that we've seen these two examples attempt to do on the practice question that's left on the bottom of the page, attempted on your own, come back and take a look and see how I approach that same question.

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Problem

Problem

Calculate the pH of a solution made by mixing 8.627 g of sodium butanoate in enough 0.452 M butanoic acid, HC_{4}H_{7}O_{2}, to make 250.0 mL of solution. K_{a} = 1.5 x 10^{-5}.