So fractional compositions are just a way of determining the amount of acid to base at a given or specific ph as our ph decreases, our solution becomes more acidic because the amount of acid present is increasing as we increase, ph it becomes more basic because we're increasing the amount of conjugate base. Now remember with weak acids we have the formation of an equilibrium because weak acids are weak electrolytes, so they don't completely ionized into their products. Here are weak acid, which is just a simple generic formula. H. A partially ionized is to form H plus ion and a minus. As our conjugate base setting up, the equilibrium expression gives us the acid dissociation constant K. Equals the ions as products on top divided by the initial concentration of my weak acid. Now the formal concentration of my solution equals the concentration of my weak acid plus the concentration of my conjugate base. Now, when we talk about the fraction of acid molecules present, we use the variable alpha to represent H. A. Here if we take a look here what this chart is showing me is at a ph the lowest possible ph The composition of my solution will be 100% in the acid form and 0% of the basic form. As the ph begins to increase. As we move from left to right, we're going to have an increase in the amount of conjugate base and a decrease in the amount of weak acid here they intersect at this point here, at that ph we have 50% of the weak acid form and 50% of the contact base form. This is true when the ph is equal to my P. K. A. So when those two equal each other, it's because the amount of conjugate base and weak acid are equal in amount. Remember this coincides with the whole Henderson household back equation, which is P. H equals P. K. A. Plus log of conjugate base over weak acid. When they're both equal to each other, This all becomes equal to one. The log of one is simply equal to zero and that's why P. H. Equals P. K. A. So when P H equals P K. A. The amount of weak acid is equal to the amount of its conjugate base. And if we continue past this ph point we're going to continue continually. Have a drop in my weak acid form And in a steady rise in the amount of my conjugate base form until we get to a ph where the weak acid form is basically near zero and the conjugate base is near 100%. Now here we say that the concentration of my weak acid is equal to the concentration of H plus times of formal concentration of my solution divided by H plus plus que. Here, if we want to find out the fraction that exists in the acid form that just simply gets broken down into the concentration of H plus divided by the concentration of H plus plus K. So that their represents the fraction in the acid form. And here the fraction in the conjugate base form or a minus is represented by alpha sub a minus here. That would just be equal to K. A. Divided by H plus plus K. A. So these are the two equations we can utilize in order to figure out the fraction of either my conjugate acid or my weak acid form. And remember together both of them would equal one. And if you multiply that by 100 would be 100%. Okay, so if you know one, you know what the other one is because together they represent 100% of all possible um molecules and ions present within my solution at any given ph now that you see in terms of a week mono product acid, click on the next video to look at it in terms of a di protic acid.

2

concept

Fractional Composition

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with die products systems, we need to take into account that there are more forms that exist based on ph with a dia product system, we have our acidic form where we still possess both of our acidic hydrogen. Once we lose that first acidic hydrogen, we now have the intermediate form. Then finally we have our basic form with these different forms for our diabetic system. We have to take into account more than one K value. So remember we're losing the first acidic hydrogen means we're dealing with K. One, which then gives us this equilibrium expression we're arranging. It gives us our concentration of the intermediate form as the acid forum times K one divided by the concentration of H plus ions produced. When we lose that second acidic hydrogen, we have the formation of again additional H plus as well as the basic form here. Once we rearrange our first equilibrium expression, we can figure out what our concentration of a two minus is. Now remember H a minus. Our intermediate form equals this entire expression here. Which you can then substitute in for this H a minus here. So, by substituting in that portion, we get this with only one H plus getting multiplied by K. A two plus one H. Plus on the bottom, which is why this is squared here from that, we get our formal concentration as being equal to the concentration of our acid form times this large expression here. Now the fractions of the three major diabetic forms can be seen as we have our fraction of the acid form, the fraction of our intermediate form and the fraction of our basic form, each with their own equation. Now, if we take a look here on the left side, we can see how the fraction of each of the forms exists at any given ph. We can see here that this darker line here represents the acid form. This dotted form here represents the intermediate form and then this light line right here, gray line here represents the basic form. As the ph is increasing, we're going from left to right. And so the amount of the asset form is decreasing as well as the intermediate form and the basic form is increasing. We're gonna say here at this particular ph we have the intersection between the asset form and the intermediate form. So we say here, P. H equals P K. One at this intersection at that intersection. We can also say that the concentration of my asset form is equal to the concentration of my intermediate form. Then we have this intersection here at a higher ph At this point we'd say that ph is equal to P. K two. So at that point we'd say that the intermediate form concentration is equal to the basic form. So again, we can utilize this chart here to determine what the fraction of two forms are in terms of one another at any given ph So remember when it comes to die products systems, there are more intermediate forms involved, which leads to more complex equations to represent the fractions of each of those forms. And remember your fractional composition is all based on the ph. The higher the ph, the more the basic forms exist and predominate. And at lower ph is, the acid forms will predominate.

3

example

Fractional Composition Calculations 1

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So here we're told we have a die basic compound B. That has a PKB one value of five and a PKb to value of eight. Find the fraction of the acidic form when the ph equals 9.0. All right. So, we're looking for the fraction of the acid form of a die basic species. So that means we're dealing also with the dye product species, looking at it from the opposite direction. So here that would be the fraction of the acid form equals the concentration of H plus squared divided by the concentration of H plus squared plus the concentration of H plus times K one plus K one times K. A two. So, what we have to do here is determine what our concentration of H plus says and determine what my acid dissociation constants. K one and K two are we're gonna say here that H plus equals 10 to the negative P H. So H plus is equal to 10 to the negative 9.0. So, we have that pushing figured out so far. So, plug that in next. We need to determine what my K. One and K two will be. But here we're dealing with P. K. B one and P K. B two. What we need to remember is that the relationship between your acid dissociation constant and your base association constant is P. K. A. One plus PKB two equals 14 and P K A two plus PKB one equals 14. So K. One and K. B. Two are connected and K. Two and K B one are connected. We're gonna say here that PK one equals 14 minus P K B two. So it'd be 14 -8 Which gives us six Here,. A two equals 14 minus P K B one. So that's 14 -5.00. So that's nine. Now that we know P K one and P K. B two, we know what K. One and K two will be. That's because we're gonna say that K1 equals 10 to the negative P K. A. One and K. A two equals 10 to the negative P. K. Two. So plug those values in so we're gonna get here 10 to the negative 6.00 and 10 to the negative 9.00. Take those values and plug it in. So that's gonna be 10 to the negative six And 10 to the -9. So when we punch that in the top portion will be 10 to the negative 18 Divided by. When we add up everything on the bottom, that gives us 2.001 times 10 to the -15. So at the end of fraction in my asset form, What Equal .0005. This makes sense because we'd say that at PK. A one RPK one we were told is 6.00. When our p was equal to RPK one, we'd have an equal amount of the acid form and the conjugate base form. But as the ph gets higher and higher, we're gonna have less and less of the acid form. The ph is three units higher than my P. K. One value of six. So that means we're gonna see a decrease in the amount of my asset for being three units different is a huge difference in magnitude for concentrations. So it makes sense that my asset form would be such a small value. But remember, die basic alludes to die product. Here we're talking about finding the acid form. If they're asking for the intermediate form of the basic form, all we have to do is recall the equations associated with those different forms of die product. Die basic species, utilizing them helps us determine the concentration of any one of those in terms of their fractions. Now that you've seen this example, attempt to do the next one we're talking about, the fraction of tyrosine. So remember the equations that we've seen on previous pages, utilize them to find the fraction of that particular amino acid. Once you do come back and see if your answer matches up with mine

4

example

Fractional Composition Calculations 1

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So here it states what fraction of tyrosine which has two PK values exist in all of its forms at ph equals 10.0 here. Because we're dealing with two P. K values. That means that tyrosine exists as a dye protic acid. Here, Tyrosine is one of the amino acids has two acidic K values. Remember for dia product species, we have the acid form, we have the intermediate form and we have the basic form halfway. Here we have P. H. Equals P. K. A one. At that point we should have 50% of the acid form and 50% of the basic form or intermediate form actually. And then here halfway P. H equals P. K. Two. So we should have 50% of the intermediate form and 50% of the basic form. Now, in actuality we won't really have um a flat 50% of both of these at this P. H. Equals P. K. One. There will be some weak acid and small minute amounts uh that we're dealing with realize here that are P. K. A. Two is 8.67. And our ph is above that. What that tells me is it tells me that we've gone beyond this line here. So the predominant form of tyrosine would be its basic form. What we're gonna do now is we're gonna use the equations for the fractions of acid form, intermediate form of basic form. To figure out how much we really have of these three different forms. So for dia product species the fraction in the acidic form the equation is the concentration of H plus squared divided by the concentration of H plus squared plus the concentration of H plus times K one plus K one times K. A two. The fraction of the intermediate form equals K. One times the concentration of H plus divided by the same denominator as the fraction for the asset form. So it's the same exact setup. And then for the fraction of the basic form it be K one times K two. Again we'd have the same denominator again. So H plus squared plus H plus times K. A. One plus K. A one plus times K two. All we have to do is we have to plug in the numbers. So H plus we get that from the ph because remember h plus concentration equals 10 to the negative ph that's 10 to the negative 10. Then our K one, we'd say K one is equal to 10 to the negative P. K. One. So it's 10 to the negative 2.37 K two is equal to 10 to the negative P. K. A two. So that's 10 to the negative 8.67. When we plug these values in we'll get the fraction of each one of these three different forms for tyrosine. So here when we plug them in we'd get uh 10 to the negative 20 because I put 10 to the negative 10 here and I'd square it would give me 10 to the negative 20 divided by when I input each one of these values here into each one of these slots. What I would get at the end would be 9.54669 times 10 to the -12. Because all three equations have the same exact denominator. All of them would have that same value for the bottom. When I plug this into my calculator, I get 1.05 times 10 to the -19. So, that'd be the fraction in the acid form. And look at how small that value is. It makes absolute sense that it would be an incredibly small number because again, where are we in terms of these three different forms were around here somewhere because the ph is equal to 10. Our ph is high enough that we're really talking about. There being a larger amount of the basic form in the intermediate form. We've blown past this first line here which represents our ph of 2.37 or PK of 2.37. Because the ph is so much higher, we're gonna have very little of this acid form present within our solution at that ph. And what this really shows us is that no matter what the ph is, we're always gonna have some small amount of acid form present within my solution. Now here, when we plug in the values we're going to get 4.2658 times 10 to the negative 13 on top divided by the same denominator. So that gives me zero four 47 as the fraction for the intermediate form. And then finally here we're gonna get 9.12, 011 times 10 to the negative 12. Again divided by the same denominator. So that gives me .955. So, what this is showing me, it's it's showing me that. The basic form of course, is the portion that's largest an amount which we expected because the ph is at 10. We've gone beyond this line here. So, the major form would be the basic form and then the next highest form would be my intermediate form with very very, very small amounts of the acid form here. If you were to multiply these values by 100 you get the percentages of each one. So that would mean that we have 95.5% in the basic form. If we're talking about the percentages of the three forms for this dye product, tyrosine compound. So now that we've seen this example, attempt to do the practice question that's left on the bottom of the page. Once you do come back and see if your answer matches up with mine

5

Problem

Problem

Calculate the fraction of the intermediate for sulfurous acid, H _{2}SO_{3}, at pH = 11.00? pK_{a1} = 1.80, pK_{a2} = 7.19.

A

6.31x10^{-11}

B

9.77x10^{-14}

C

0.000155

D

0.00977

E

1.00

Fractional Concentrations

6

concept

Fractional Concentrations

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So recall when it comes to mono product systems that the fraction of the acid molecules and the consequent based molecules were represented by alpha sub H. A. And alpha sub A minus here. When we're talking about the fraction of acid molecules, it would equal the H plus concentration divided by H plus concentration plus K. A. And the fraction of the base concentration was equal to K. A. Divided by H plus plus K. A. As we look at this chart here on the left, as our ph increased, going from left to right, the amount of the acid molecules would decrease as the amount of contribution based molecules would increase. Here at this intersection is where both the concentration of the H. A molecules and a minus molecules would be equal to one another. Also remember that that happens when ph is equal to the P. K. A. Now when we try to determine what our concentrations are for these fractions of these molecules, we can restructure our equations to now represent, the concentration of my acid is equal to the fraction of the acid molecules times the formal concentration of the acid molecules using this expression now inputs the formal concentration of my weak acid solution. And if we're looking for the overall concentration of my conjugate base, it's equal to the fraction of my conjugate based molecules times the formal concentration of my solution asset solution, so that B. K. A. One times formal concentration of acid again divided by H plus plus K. A. So remember the fraction can be determined by comparing the ph of the solution to the P. K. A. Value to see which species is greater in amount. We can utilize these equations themselves to find either the exact fraction of the weak acid or the exact fraction of the conjugate base. From there, we can attach formal concentration to find the new concentrations of the weak acid and the conjugate base, click onto the next video and see what we do when dealing with dia product systems.

7

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Fractional Concentrations

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So as we stated before in the past, when it comes to die protic systems we have three possible forms that are compound can take. We have the acidic form, We have the intermediate form after it's lost its first acidic hydrogen. Then we have the basic form here with these three different forms. We have the introduction of K. one and K two. When removing the first acidic hydrogen, we produce your hydro ni um ion plus some intermediate K one would be involved. Which would give us this expression of products. Overreacting by rearranging it. We'd be able to isolate the concentration of our intermediate as the concentration of your asset form. Times K one divided by H plus hear from the intermediate form we could lose our second acidic hydrogen to give us more hydro name, ion created Plus now the basic form of our compound by substituting in this expression for intermediate. We can get the concentration of our basic form as being H. Two a times K one times K two over H plus squared. Now our formal concentration comes about through the use of our mass balance here. We have our formal concentration equal to the concentration of my asset form, times the expression within here. Now remember the formal concentrations now of the three major dia product forms can be restructured as the following. We've seen the equations related to the fractions of the acid form. The intermediate form and the basic form. With the introduction of our formal concentration here, we'll be able to calculate the new concentrations of each one of these compounds by doing that We just have the insertion of the formal concentration into each one of these compounds. And remember all three forms, although different possess the same denominator on the bottom. Also recall that when we're taking a look at the fraction versus ph chart that this line here represents the amount of concentration of my acid form. This here represents the concentration of my intermediate form and then this one here represents the concentration of my basic form at this intersection we have P. H. Equal to P. K. One Because the concentration of H. two a would be equal to the concentration of H. A. -. And here at our second intersection we'd have ph equal to P. K. A. Two because the intermediate form would be equal to the basic form. This just shows us as the ph is increasing, going from left to right, the proportion of the conjugate base or the basic form is increasing. And remember even at high enough phs we're still gonna have some amount of our acidic form remaining. It'll just be a very very very small amount. The higher the ph gets. And remember all three forms exist in some proportion. The higher the ph gets. So we've seen these equations in terms of the fractions now we're introducing our formal concentrations to find our new conditional concentration of our asset form, our intermediate form and the basic form

8

example

Fractional Concentrations Calculations 1

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here it states a di basic compound B has a PKB one value of 4.0. And a PKB to value of 6.0. We're asked to find the concentration of the intermediate form. When the formal or total concentration of the dye protic acid equals 0.150 moller. And the P H equals 8.0. All right. So here we need to determine the concentration of our intermediate. So that's so we're gonna say the concentration of our intermediate form equals the fraction of the intermediate form times the formal concentration of the dia protic acid. When we expand this out, it gives me a K A one times H plus times the formal concentration of my dia protic acid divided by H plus squared plus H positive times K one plus K one times K. A. Two. Alright, so, we know what H plus is equal to because H plus concentration equals 10 to the negative ph and in this case we're dealing with the ph of 8.0. Next, we need to figure out what K. One and K two will be. We're gonna say that P K. A one plus PKB two equals 14. And that P K. A two plus P K B one equals 14. So, P K one here equals 14 minus P K B two, which is six. So, this comes out to being eight because we now know what P K one is. We know what K one is. K one equals 10 to the negative P K one. So that's 10 to the negative 8.00. P. K two will equal 14 minus P. K. B. One. So it equals 10. So K. A two equals 10 to the negative P. K. Two. So it's 10 to the -10. We now have all the units that we need. So we're just gonna plug them into the formula up above. So K one is 10 to the negative eight Times 10 to the -8 Times the formal concentration of .150 moller divided now by 10 to the negative 8.00 Squared plus eight H. Plus. So 10 to the negative 8.00 times 10 to the negative 8.00 plus the two K. Is multiplying each other. So 10 to the negative 8.0 times 10 to the negative 10. When we figure out what's on the top and on the bottom we're gonna get 1.5 times 10 to the negative 17, divided by 2.1 times 10 to the negative 16. So at a ph of 8.0 the amount of it that exists in the intermediate form concentration wise equals .075 molar. So that would be our amount of the intermediate form. Remember when we're dealing with our die product or die basic form. We have three things to worry about. We have the acid form the acidic form. The intermediate form. And the basic form here P. H equals P K. Two, which equals six. But our ph is above six. R. P. H. S. Eight. Which means that a majority of this formal concentration of my dia protic acid would actually exist in the basic form. A majority of it exists in that form, and a portion of it exists in the intermediate form and the acidic form. What we just found was the amount that exists in the intermediate form from again from the initial formal concentration of 0.150 moller. As the ph would increase. This number would decrease even more because the higher the ph gets, the more the basic form will become the predominant concentration involved in my solution. And if we're to find the concentration of my acidic form, it would be even less than this value here. Now that you've seen this example in terms of determining the concentration. Look to see if you can answer example two that's given down below. Once you do come back and see if your answer matches up with mine.

9

example

Fractional Concentrations Calculations 1

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So here it tells us to calculate the concentration of the acidic form for 0.230 moller tart acid When the ph is 6.0 were given to p. K. Values. So P K one equals three and p K two equals 4.34. Since they're telling us to calculate the concentration will assume that this number here must be the formal concentration of my dia protic acid. So we're gonna say here that the concentration of my dia protic acid equals the fraction of my dia protic acid times the formal concentration of my dia protic acid. This translates to H plus squared times the formal concentration of my acid divided by H plus squared plus H plus times K. One plus K. Two times K one times K two. Remember here that H plus is equal to 10 to the negative ph So equals 10 to the negative six K. A one equals 10 to the negative p. K. One So equals 10 to the negative three and K two equals 10 to the negative p. K. Two. So it equals 10 to the negative 4.34. All we do now is we insert our values so H plus squared so that's 10 to the negative six squared times the formal concentration divided by H plus squared plus H plus times K. One which is 10 to the negative three plus 10 to the negative three Times 10 to the -4.34. When we figure out what the values for top and bottom we get 2.30 times 10 to the negative 13 divided by 4.67098 times 10 to the negative eight. So that gives me a fraction of or concentration of 4.92 times 10 to the negative six moller. So our concentration here is incredibly small because with our die protic acid, we'd have the acidic form. We'd have the intermediate form and we have the basic form Here, Ph would equal PK one And here PHOPK two Here at PK two. It is equal to 4.34. That's when we would have approximately 50% of the basic form and 50% of the intermediate form. Our ph is even higher than that. It's at six. So, we'd have a majority of my solution composed of the basic form. The intermediate form will form the second most amount. And because the acidic form is at 50% around a ph of three, we'd expect very little of it when the ph is all the way up to six. So, what this shows us is that although the ph is incredibly high, we never totally get rid of all of our acidic form, they'll still be small amounts of it present within our solution. They'll just be incredibly small. Small amounts. That explains why our concentration here is times 10 to the negative six Mohler. The other two would be much greater in amount because the ph again is at six. And at ph 4.34, we'd have 50% of both of these. Since the ph is greater than six, the majority of our solution exists in the basic form. So remember the formulas that are associated with each one of the forms for Dia product system. And remember what our peaches telling us? It's giving us basically the fraction distribution between the three different forms for my di product species.