as we mentioned in previous video, E. D. T. A. Out of all the Polident, it Liggins is the most prolific and most important that we can use in any types of titrate asians in the determination of the concentration of ions. Now we're gonna say here that E. D. T. A. And it's fully predominated form exists as a hex a protic acid. So it has six acidic hydrogen. Therefore it has six P. K. Values. Now it's fully propagated form would take on H six Y two. Plus here we'd say that when it comes to these six different P. K. S. We have our car box cell groups. And when it comes to car boxes groups, they are more acidic than amine groups. So these car boxes groups would have the lowest P. K. Values since all four car boxes groups are basically equivalent. Um we don't really assign a PK one to, let's say the top left one or two, the bottom right one. They're all four car basilica acids, same type of structure. So those numbers can belong to any of them. The higher P. K. Is the ones that are less acidic belong to these amines here. And remember we've talked about tight rations in the past, we said that if we have a ph that is greater than R. P. K. A. Then that can lead to deep throat a nation where we are able to remove an acidic hydrogen. And then if your ph is less than your P. K. A. Then it leads to proto nation where the acidic hydrogen is not removed off of that particular compound. Now here in this form we have four negative charges. Which is why when we're talking about E. D. T. A. In this form it it has a negative four charge. Again. E. D. T. A. Can have different forms based on the ph of the solution that it's in. So that charge can change over time. Now we're gonna say here in order to form metal complexes, these acidic hydrogen must first be removed. And in fact the complexes that we form when E. D. T. A. Combines with a metal is usually in its basic form. So the basic form is the most important form that we have here and going off of what we just set up above. Remember ph and PK And the relationship between them, we're gonna say at lower phs will exist in the acidic form. And as our ph begins to increase then we'll be we'll have a basic enough solution that we start removing H. Plus ions from the E. D. T. A. Species. So here we've removed the first acidic hydrogen. So now we have a church five Y plus were removed the next one. This structure here is important because this is the only structure that exists in a neutral form in this neutral form. We're able to properly store R. E. D. T. A solution. So this one here, although it's not involved in many of the complex ion um formations that are important in terms of tight rations. This form is important because it's neutral. We're able to store it next. If we continue to remove H plus we now have a three y minus and then H two y minus H H two, Y two minus H Y three minus. And then finally our basic form here at high phs we get closer and closer to our basic form here in this image. This is giving us the fractional composition for a particular E. D. T. A form based on the ph presented. For example, let's say that we're looking at a ph that's slightly above six and we're looking at this point here. What that point is telling me is that we have we trace. So that dot there overlaps these two forms. So what that point there is telling me is it's telling me that the largest percentages of the E. T. A. Form are found in these two forms here. That's what that particular point is telling me. And it'd be like a 5050 composition between those two. And again, as we increase the ph we can see that the basic form will predominate and we see that happens once we get to uh p K A p K six which is 10.37. So basically above p 10.37. And again we got that value from up here. So above ph 10.37. The basic form is the predominant form but as we decrease the ph we have more overlapping between the different forms. And we can see that the most overlapping between majority of the forms happens at a ph less than three. When the ph is less than three. We can see that we have all four of these forms here interacting with each other in some way. So again, when it comes to E. D. T. A. We can talk about the seven different forms that it can take based on the ph of the solution are fully product form where it has all of its acidic hydrogen is H six Y two plus H Y, H four Y is the neutral form. It's the most useful in terms of storage of E D. T A over periods of time. And then why 4 - is the most useful form when talking about the complex structures we can form when EDTA. Is combining with a metal ion. Now, in addition to this, we can calculate the fraction of E D T A. In its basic form when we use the following equation. So here, alpha would just talk about the fraction of E D T. A. Of the basic form. So that's basically equal to the amount of the basic form over E D T A. In this case E D T A is the concentrations of all seven forms all the way from the product form to the basic form and all the other forms in between. Now we could also use an additional formula where it is the fraction of E D. T. A. In the basic form equals each one of the six K. S multiplied together, divided by. So here we have we'd use this version when they give us the ph of a solution and they're asking us to determine the fraction of the basic form. So here if we knew the ph of the solution then we can determine what H. Plus is. Because remember H plus concentration equals 10 to the negative ph we plug that in. So on the bottom which still represents the total amount of E. D. T. A. In terms of this now be um ph of a solution. It equals H plus the six plus H plus to the fifth times K one, H plus two, the fourth times K one and K two and so on and so forth. So you can see that as the exponential power for H plus decreases. We include more of the K. A values. And these K values are coming from up here. R P K values that I gave us in the very beginning. So P K one all the way to P K six. And remember you can find K. If you know P K. A. Because K equals 10 to the negative P K A. Okay, so that's how we're able to determine the K values. We'll see the utilization of these different formulas in order to determine the the fraction of E D T A. In its basic form. And just remember here that would use this form here when they're giving us the concentration of the basic form and the concentrations of each of the seven forms that make up all of E D T. A. And would use this formula here when they give us the ph of the solution. And they want me to find the fraction of E. D. T. A. In its basic form from that given ph value. So both formulas are pretty jam packed with variables. So be careful when you're inputting your values to get the correct answer for the fraction of E D T A. In the basic form. So just remember E D T A. Is a hex of product acid, it has six different p K. S. And based on the ph of the solution, it can exist in one of seven forms.

EDTA represents a hexaprotic compound that possesses 7 different forms based on the pH of the solution.

The basic form of EDTA, Y^{4-}, is the predominate form at higher pH values, while the acidic form, H_{6}Y^{2+}, predominates at lower pH values.

The fraction of basic form present in solution can be determined by either given concentrations or by the pH of the solution.

EDTA Calculations

2

example

EDTA Calculations 1

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So here it states the formal concentration of E. D. T. A. Is 1.50 millimeter. Now we're asked, what is the concentration of the basic form at a ph of 5.0. Alright so we need to determine what the concentration of our basic form will be at this given ph and here we have the overall concentration of E. D. T. A. Out of this total concentration, a portion of it belongs to the basic form. So we're gonna say that the concentration of my basic form equals the initial concentration of E. D. T. As a whole times the fraction of E. D. T. A. That exists in the basic form which we designate as alpha. So here we're gonna change our concentration to molar itty, you can keep it in millimeter but I'll just change it to polarity. So remember here that one million moller is equal to 10 to the -3 molar. So here we have 1.50 times 10 to the -3 molar. And here we're dealing with a ph of five. So what we have here are all our ph values set to a temperature of 25 degrees Celsius and an ionic strength of 250.10 Moeller. Remember what do we notice as the ph is increasing as we go from 0 to 14. Look at the fraction that exists in the basic form, we can see that it's increasing and this makes sense because on the previous page we saw that chart with the waves and the lines and we saw that as the ph increased, we got more and more of our basic form. Taking up the um the predominant form of the E. D. T. A solution. And that's all this chart here is telling us as the ph is increasing a greater percentage or a greater fraction of my total E. D. T. A solution exists in the basic form. And what this is telling me is that once I get to phs of 13 and 14 it's so basic that 100% of my total E. D. T. A solution is in the basic form. So all we do now is we just have to look up five for ph and here is my alpha for the basic form at that particular ph so we're just gonna plug that value in. So when we multiply that we get 4.35 times 10 to the -10 moller. So what that answer is telling me, it's telling me that the total amount of E. D. T. A. With all seven forms together totals 1.50 times 10 to negative three molar out of that amount. This is how much of of it exists in the basic form at that ph of five. So just remember this goes in hand with the graph that we saw on the previous page with the waves as the ph is increasing the basic form becomes more and more a predominant part of my overall solution. Now that we've seen this, take a look at example two once you attempted. Or even if you don't attempt it, don't worry, just come back and see how I approach example, too.

3

example

EDTA Calculations 1

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7m

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So for example to we need to determine the fraction of R. E. D. T. A. That exists in the basic form when the ph is 8.50. Now from our chart given here, 8.50 would lie somewhere between eight and nine. Of course. So we should expect our answer to be somewhere between 4.2 times 10 to negative three and 30.41. Remember if they're asking us to determine what our fraction of the basic form of E. D. T. A. Is and all they give us is the ph then the formula that we utilize is alpha. Our fraction of the basic form equals K. K. One times K. A. Two times K. Three times K. Four times K. Five times K six. Remember E. D. T. A. And it's fully pro donated form is a hex a protic acid. Therefore it has six K. Values or six P. K. Values divided by the H plus concentration to the six plus H plus to the fifth times K one plus H plus two. The fourth Times K one times K. two. And keep going. So H. Plus to the third times K. One times K. A. Two times K. A. Three H plus two the second. So we just keep adding R. K. S. So H plus times K one times K. Two times K. Three times K. Four times K. A five plus one more K. A one. All the way to K. A. Six A five K. Six. Now they give us ph so with ph we know what the H plus is because remember H plus equals 10 to the negative ph So it'll be 10 to the negative 8.50. And then here we have our P. K. Values that were given to us. This was zero P. K. A. Two was 1.50 P. K. Three. We had was to P K. Four Was 2.69. P. K. A. five Was 6.13. And then p. K six was 10.37. If you know your P. K. Value then you know what your K. Value is. So here we say that K. One equals 10 to the negative P. K. One. So 10 to the zero. So this would be 10 to the -1.50. Okay this will be 10 to the negative 2.00. 10 to the -2.69. 10 to the -6.13. And finally This would be 10 to the negative 10.37. So all we do here is we plug in these values that we have here into each one of the spaces that we have here. So it's a long arduous process. But this is the formula we have to utilize in order to figure out our fraction of the basic form of E. D. T. A. When you plug in all those values. So when we plug in all the K. Values and we multiply them with each other where we would get as a result is 2.04174 times 10 to the -23. Then we could plug in all the values we have for each one of these here. So when we do that for each one of them would get a bunch of different values. So I'll write them here. So we have 1.0 times 10 to the negative 51 plus 3.162 to 8 times 10 to the negative 43 Plone 3.16228 Times 10 to the negative 36 Plus 1.0 times 10 to the negative 29 plus 6.454 on +654 times 10 to the negative 24 Plus 1.51 356 times 10 to the negative 21 plus 2.04174 times 10 to the -23. So when you add up all of that on the bottom you still have what you have on the top, Divided by all of those units added together on the bottom gives us a grand total of 1.54043 times 10 to the -21. So here when we plug that into our calculators to get our final answer we get an answer of .013 three. Which is a reasonable answer. Technically we should only have two sig figs. Uh for the most part because materials numbers have two sig figs. So we say 20.13. Which is a reasonable answer because it is a number that falls in between these two values of 4.2 times 10 to negative three and 30.41. Remember as our ph is increasing the fraction of the basic form of E. D. T. A. Is increasing in amount. And remember we used this version of the formula when we're only given the ph of the solution. We use this form to figure out the fraction of the basic form. If they had given us the form of each of the seven major forms of E. D. T. A. Then we'd use the original equation to figure out the fraction of basic form of E. D. T. A. Like both formulas are pretty complex and long. So make sure you carefully plug in the right numbers. Step by step in order to find the final answer. Now that you've seen this answer. This practice question that's left on the bottom. This one is much more straightforward telegraph which you'll have to utilize or to find your answer. Once you attempt that come back and see if your answer matches up with mine

4

Problem

Problem

Determine the pH where Î±Y4â€“ equals 0.20.

A

pH â‰ˆ 11.1

B

pH â‰ˆ 12.5

C

pH â‰ˆ 9.80

D

pH â‰ˆ 9.00

EDTA Complexes

5

concept

EDTA Complexes

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5m

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So with E. D. T. A complexes comes the discussion of our formation or stability constants. So formation or stability constants represents the equilibrium constant for the reaction between the ligand and a medal here. Our formation or stability constant as shown by this variable. Now the lingering question that we're dealing with is E. D. T. A. Now the general formula is our metal ion here. We don't know what it's charges. So we say N. Plus so it could represent a metal with a plus one charge or plus two charge or plus four. Charger plus three charge plus Y. To the negative four. Why do the negative four represents E. D. T. A. This is the basic form of E. D. T. A. And this is the predominant form that we usually use when dealing with a metal ion. But the other forms of E. D. T. A could also form living connections with the metal is just that it's this one in particular. That is the major form that it's usually used now here when they combined together we get our E. D. T. A. Complex. So here it's M. Y. To the N minus four. So for example if you're dealing with calcium which is plus two in its charge and we're dealing with E. D. T. A. So calcium is plus two. E. D. T. A. Will be minus four here. So the overall net charge would be minus two. So we'd say C. A. E. D. T. A. And this would be 2 - overall at the end If we're dealing with let's say iron three with E. D. T. A. So this is plus three. This is still -4. So overall This would be -1. So that's how the overall charge works. And here we're dealing with an equilibrium that's being established because we have the reversible arrows. And just like with any equilibrium constant will say that our formation or stability constant K. F. Equals products. Overreact. It's so it's equal the E. D. T. A. Complex divided by the concentration of my free metal ion times the concentration of E. D. T. A. This formula itself is useful in determining the concentration of free metal ion within the solution. What happens is that not every ounce or every amount of the free metal will connect with E. D. T. A. There will be some free floating metal ions that are around that having combined with every single E. D. T. A molecule. This expression that we have here can be used in order to determine the amount of free metal ion that's existing within the solution. Now we're gonna say the formation constant for metal E. D. T. A complexes are given below. So these are all the values that we have to realize that some of them have superscripts of A. And B. So when it's A. That means the temperature is 20 degrees Celsius and are ionic strength is 200.1 molar. And for the one that's labeled with a subscript of B. That is 20 degrees Celsius still. But now are ionic strength is one Moeller. What we should realize here is that the common trend is, the higher our charge becomes than our positive charge becomes than the greater the affinity between the ligand and the metal. And the higher your KF value will be okay so again the more positive the charge, the stronger the connection between the living and the metal, the higher your K. F. The more products would be favored. So the more your E. D. T. A. Complex would be formed. Now we're gonna say only a portion of E. D. T. A. Exist in its basic form. And the lower the ph the more the other forms predominate. Remember we have to go to phs of 13 and 14 for 100% of my solution to be in the basic form, if my P. H. Is less than that. Then portions of the solution will exist in the other six forms of E. D. T. A. Now here we're gonna say under a fixed ph the fraction of the basic form of A. D. Ta becomes a constant and can be used to determine the conditional formation constant. So our conditional formation constant is here. So it's K. F. Prime. So we have this value here to say prime. Um It represents the formation of our E. D. T. A complex at any ph value. So here are conditional formation constant equals the fraction of your E. D. T. A. In basic form times your formation or stability constant and that still equals products. Overreacting. So it's still the E. D. T. A complex divided by the concentration of free metal ion times the concentration of E D. T. A. We'll have to utilize these different expressions equations as well as values to answer the following questions. When it comes to E D. T. A, complex is so click on to the next video and see how we approach the question where we're asked to find the concentration of free barium ion within R E D T A. Um complex that's formed when it reacts with e D. T. A. Itself.

Each metal ion has its own unique formation constant.

The conditional formation constant can be calculated by the determining the fraction of the basic form and the formation constant of the metal ion.

6

example

EDTA Complexes

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7m

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So here we have to basically discover the amount of free barium ion that exists when we've already formed our A. B. T. A. Complex. So we're gonna say we're gonna utilize this equation here within the question and substitute and barium for that unknown. Em so we're gonna have be a two plus plus E. D. T. A. Remember E. D. T. A. We're dealing with its basic form, so its charges minus four, so minus four plus two from the very um so the net charge at the end will be minus two. So that's how we came up with B. A. Y two minus now. We're gonna have our initial our change and our amount of equilibrium here, they give me an initial concentration for the barium E. D. T. A. Complex as being 0.10 moller. We don't have any initial amounts for the E. D. T. A. And the barium two plus ion, Since we have an initial amount for complex, it's going to be decreasing over time. So this is -1. These two would be plus x. Bring down everything. So this is plus X plus x 0.10 minus x. Alright, so with any ice chart setups or which we have to figure out equilibrium amounts. We need to know what are our formation constant will be in this case and not just our formation constant but are conditional formation constant. So we're gonna have to utilize this portion here. So our conditional formation constant equals the fraction of E. D. T. A. That exists in its basic form times our formation or stability constant. So they're telling us that the ph is 10. Remember on our previous videos we said that when the ph is 10, if we look on the chart from a few pages ago, we know that the value of our basic form of E D. T. A would be 100.30. Now we need to figure out what our K. F is. Well for barium ion here it is. Right here The log of K. F. of burying a 7.88 salah log of K. F. For barry Mayan Equals 7.88. But I want I don't want the log of K. F. I just want K. F. So I take the inverse of the log function. So now K. F of barium ion equals 10 to 7.88. So that is what my K. F would equal When I multiply those two together. That gives me 2.28 times 10 to the seven. Now that I have my conditional formation constant which is based on the ph given we're gonna say that this constant like all other constant equals products. Overreact mints. So my barium complex divided by my barium ion times E. D. T. A. All we're gonna do now we're gonna plug in the values that we know. So we know that our conditional formation constant. We calculated it as 2.28 times 10 to the seven equals at equilibrium. This is 70.10 minus X. Divided by both of these. Multiplying each other on the bottom is X squared. So we're gonna have to solve for X. So we're gonna multiply both sides now by X squared. So 2.28 times 10 to the seven X squared equals 70.10 minus X. So this X variable has the largest power so it's the lead term. So everything has to be moved over to its side. So my equation now is 2.28 times 10 to the seven X squared Plus X -110. Next we're gonna use the quadratic formula in order to isolate our X variable. So here it will be negative one plus or minus one squared -4 times 2.28 times 10 to the seven times negative .10 Divided by two a which is two times 2.28 times 10 to the seven. Alright, so here when we solve, remember this is going to be plus or minus whatever all of this comes out to be. So when we do the math we're gonna say X equals negative one plus. So when I multiply when I do all the functions inside and then I take the square root, That's gonna give me 301 9.93 divided by 4.56 times 10 to the seven or X equals negative one minus 303019.93 Divided by 4.56 times 10 to the seven. So we get two possible values for X. So here for the first one, X. Would be 6.62 times 10 to the negative five molar. Or X would be negative 6.62 times 10 to the negative five moller. Remember at equilibrium your final concentrations cannot ever be negative. It's not possible. So this cannot be the answer for X. This would be my answer for X. And at equilibrium X. Represents the concentration of both E. D. T. A. And barium ion. Here we don't care about the E. D. T. A. We're focused on the barium ion free floating barium ion concentration. So this would be our answer. So when it comes to determining the amount of free floating metal ion, this is the approach we must take. We have to set up a basic ice chart. Look at the store geometric relationship between E. D. T. A. And the free floating ion in the formation of R. E. D. T. A. Complex. Remember we have a Polident E. D. T. A. That forms a 1 to 1 ratio with the free floating metal ion. So for every one mole of the free floating metal ion, we have one mole of E. D. T. A. You also have to determine the conditional formation constant. And use that at the particular ph given to determine your final concentration at the end. So keep in mind the approaches that we used here to find the answer for free floating metal ions.

EDTA Complexes Calculations

7

example

EDTA Complexes Calculations 1

Video duration:

3m

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So for this question, it asked us to determine the conditional formation constant for this E. D. T. A complex at a ph of 8.0. Now remember our conditional formation constant is K prime sub F. It is equal to the fraction of the basic form of E. D. T. A. Which will designate his alpha times my formation or stability constant. Now here, if we take a look remember et itself is -4 in terms of its charge. When dealing with the basic form, the overall charge is minus one. For that to occur, Cobalt here would have to have a charge of plus three. So we're dealing with cobalt three ion interacting with E. D. T. A. To produce this E D. T. A complex. Now here they're telling us that our ph is 8.0. So remember in the chart that we saw a few pages ago when the ph is 8.0, then the fraction of the basic form of E D. T. A gives us a value of 4.2 times 10 to the minus three. Here, when we're dealing with whole integers in terms of ph so 0 to 14, we can simply look at that chart and pick the value that we see for the fraction of the basic form of E. D. T. A. Now remember if our ph is a value that is not a whole imager. So let's say we had uh 6.15 instead of 8.0, we have to utilize the equation that we've used before to figure out the fraction of the basic form. Remember that was the one where we had. K. A. One all the way to K. Six on top divided by H. Plus to the six plus all the other portion of portions of the denominator. So that one would be a longer process in order to figure out the the fraction of the basic form of R. E. D. T. A. Species. But again, we're dealing with whole numbers so we can simply look at the chart and pick the value that we see. Alright, so now we have that then when we're dealing with cobalt three while the log of K. F. For cobalt three ion equals 41.4. But we don't want the log of K. F. We just want K. F. So we're gonna take the inverse log function. So that's K. F. Of cobalt three ion equals 10 to 41.4. So we'll put 10-41.4 here And then multiply those two values together gives me an answer of 1.05 times 10 to the 39 as the conditional formation constant at that exact ph Now remember your K value is greater than one. That means that products are highly favored. Which would mean that when cobalt three ion comes into contact with E. T. T. A. There is a high propensity or high desire to form this complex here. So that's what our KF value was telling us here. And remember the more positive the charge of the metal ion, the greater the affinity the e d t, a molecule or Ligon has for the positive ion, the more they're gonna combine together to help make this complex here. Now that we've seen this simple example, move on to example to remember and utilize some of the techniques that we use in the previous page to help figure out the amount of free tend to ion within this example, given so, good luck guys.

8

example

EDTA Complexes Calculations 1

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5m

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So here we need to find the concentration of free 10 2 ions in 20.20 Moeller of this E D. T. A complex at a ph of nine. Now, what we have here is we have potassium ion involved. But then this right here has to be S N E D T A two minus. And it's this complex E D T A complex that we're focused on Here for to have a negative two charge overall. That means that 10 Had to have been plus two. was -4. And that's how we got that charge of -2 at the end. We won't pay attention to the potassium which is acting as a spectator ion. So we'll have our initial amount the change and then what we have at equilibrium. So we have 0.20 moller of REDT. A complex. And initially we have zero of these to react ints we're losing some of our E D T A complex in order to create some of these free floating metal ions and E D T. A. We bring down everything. Now, we need to determine what our conditional formation constant will be. So K prime sub F equals alpha. So the portion or fraction of our basic form of E D T A times our formation or stability constant. So we're working here with 9.00 as our Ph. So if we look at our table, we find that the fraction of the basic form of EDTA at a ph of nine comes out to .041. Then if we look at our chart which looks at the metal E. D. T. A complexes and the formation constants connected to them, we'll see that log of K F for sn two plus Is 18.3. So that means K f of sn two plus equals 10 to 18.3. So 10 to 18.3 multiply those two together gives me 18 8.18 times 10 to the six as my conditional formation constant. Now that we have that we can set up our equilibrium expression and solve our missing X variable. So here 8.18 times 10 to the six equals products. Overreact ints. So 60.20 minus X divided by X squared. We're gonna multiply both sides by X squared. So we'll get here 8.18 times 10 to the six X squared equals 60.20 minus x, add everything over to the left side. So here we're gonna have 8.18 times 10 to the six x times X squared Plus X -120. We use the quadratic formula now at this point. So we're gonna say here negative one plus or minus one squared minus four A. C. So four times 8.18 times 10 to the six times negative 60.20 Divided by two times a. So two times 8.18 times 10 to the six. So we're gonna get two possible values for X. But remember it's the positive answer that we're focused on so that's negative one. Plus when I do the functions in here and I take the square root I get 2.558 times 10 to the eight Divided by 1.636 times 10 here to the So that's 16 to the 17. And that should be 16 here as well. All he should be 16. Alright so here so just remember your formation constant actually is to the 16 not to the six. So make that small little correction guys. Alright so then When we plug that in we'll get our answer of X. Being equal to 1.56 times 10 to the -9 moller. This here represents our ex which gives me the concentration of my metal ion which in this case is tend to ion. So just remember these are the steps that we need to take in order to figure out the amount of free floating uh 10 2 ions within this particular question from the examples that we've seen to attempt a practice question that left that's left on the bottom of the page. Once you do come back and see if your final answer matches up with mine

9

Problem

Problem

Find the concentration of free Na ^{+} in 0.15 M Li _{3}[Na(EDTA)] at pH = 10.00.