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EDTA Titrations

1

EDTA Titrations

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So now we take a look at E D T. A tight rations. Now we've done acid based situations in the past. Now we're gonna be taking a look at E. D. T. A. Serving as our title. And now we're going to say that the general formula is M. N. Plus where M represents some metal ion with some charge reacting with E D T. A. Remember E D T. A. For the most part we're dealing with its basic form which has a negative four charge when they combine they form are complex here, M Y N minus four. So in our mock Kitrey shin, we're dealing with barium ions. So if were to write this out, we'd have barium B. A. Two plus reacting with E. D. T. A. So the net charge at the end we'd have plus two here minus four here. So the net charge at the end would be minus two. We'd write it as B A Y two minus as our barium E D T. A. Complex. Now we're gonna be dealing with these types of tight rations at a specific ph because of that we'll have to determine what our conditional um constant is. So our conditional formation constant is K. Prime sub f. Remember that is equal to the fraction of E. D. T. A. In the basic form times our formation or stability constant here. Remember to figure out our fraction of basic form of E. D. T. A. Would utilize our chart that we've seen on previous pages. And if we're dealing with whole numbers we can just simply look at that chart from 0 to 14 and see what our alpha value would be. Remember as the ph increases the alpha value will increase as well because the higher the ph becomes, the greater proportion of my total solution will be in the basic form. If we get a ph that is not a whole number, then we'd have to utilize the formula that we've used in the past to calculate what our alpha would be for the basic form of E. D. T. A. When we take a look at our formation or stability constant here, we also utilize the chart that gives us the log of K. F. We do the inverse log to find our K. F. There. We say here because our conditional formation constant which is this is an equilibrium constant. It also equals products. Overreacting. So it's equal our metal E. D. T. A. Complex divided by the concentration of my metal times the concentration of E. D. T. A. Realize that in this process of tight trading our metal ion, we're adding E. D. T. Is our tie Trent. This is going to cause the PM of my metal to gradually increase over time. Remember p here stands for negative log. So this side here represents the negative log of the metal concentration. We're gonna say we can reach an equivalence point because we're dealing with the tight rations at the equivalence point we'd say that our metal ion concentration equals the concentration of E. D T A. And really it's the molds that are equal. Not necessarily the concentrations. So we say the molds of our metal ion equals the moles of R E D T A. At the equivalence point before the equivalence point is reached, we'd have excess metal ion. So we'd say that the amount of metal ion in moles is greater than then the amount of E. D. T. A. After the equivalence point is reached, we'd have an excess of E. D. T. A. So E D T A would be greater than the concentration of my metal ion beyond the equivalence point. Now, with any types of tight rations, it's always important to determine what are equivalent volume for the tightrope will be. This allows us to determine, are we before the equivalence point at the equivalence point or after the equivalence point here we have the titillation of 50 mls of 500.100 moller barium ion it's buffered to a ph of 9.0 and we're gonna say with 0.50 moller E D. T A. So here we say that the polarity of my medal times the volume of my medal equals the polarity of E. D T A times its equivalent volume. So remember E D T A is serving as the tie Trent. So we're looking for the equivalent volume of that type trend. So we plug in what we have. So the polarity of my barium ion metal is 0.100 moller Times its volume, which is 50 MLS equals the polarity of my E D. T. A. Times the equivalent volume Divide both sides here by .050 Moller. So this will cancel with this. The polarities will cancel out. And we see that the equivalent volume from my E. T. A. Would be 100 mL. So we would know that before we get to 100 mls, we'd be dealing with calculations before the equivalence point. Take a look at the next video and see what are the calculations involved for determining the concentration of my metal ion before the equivalence point.

2

EDTA Titrations

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So we're now starting with our tight rations were slowly adding E. D. T. A tight trend to our barium solution. So if we take a look, we have the attrition of 50 mls of 500.100 moller of barium ion. It's buffered to a ph of 9.0. With 80 ml of 800.50 molar of E. D. T. A. Now remember we require 100 mls of E. D. T. A. To get to the equivalence point. Where before the equivalence point at this, at this juncture in our titillation curve here. Because we're before the equivalence point we need to determine what amount of our free floating barium ion exists. As we're slowly adding R E D T. A tightrope here, we're going to say that the concentration of our metal ion equals the equivalent volume that we calculated from before minus the volume of E. D. T. A. Um added, divided by the equivalent volume. This represents the fraction of our free floating metal that remains upon filtration. Now we're gonna then multiply by the initial concentration of our metal ion, times the volume of our metal ion divided by the volume of the solution. This ratio here represents our dilution factor. So taking to heart what we've just seen, we're gonna say that the concentration of my barium ion at this point equals the equivalent volume, which we said was 100 mls minus the volume of E. D. T. A. Which is 80 mls divided by 100 mls. We're gonna multiply by the initial concentration of our metal ion which is 0.100 moller times the volume of our metal ion which is 50 mls divided by the volume of our solution which is 50 mls from my metal ion plus 80 mls from E. D. T. A. So when we plug all that in we're gonna get a concentration of 7.69 times 10 to the -3 moller for our barium ion. Now that we have the concentration of barium ion, we just need to take the negative log to find our final answer. So here we're gonna take the negative log of my barium ion solution and that gives me 2.11. At this point, realize the more E. D. T. We add, the higher the negative log of our metal concentration will go. This is the amount that's remaining. After we started originally with 0.0.100 moller of the metal ion. Remember the E. D. T. A. Is grasping onto any free floating barium ions, which is why we see a drop in the concentration. We'll continue onward in the next video and see what happens when we're at the equivalence point itself? What happens to the negative log of the concentration of barium ion

3

EDTA Titrations

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So at the equivalence point we have equal moles of our metal ion and moles of E D. T. A. At this point we'd set up a basic ice chart setup in order for us to organize our equilibrium expression. Now here, if we take a look, we'll have to determine what our concentration for a metal E D T. A complex will be. So here would be the initial concentration of my metal ion times the initial volume of the metal divided by the volume of the solution as a whole. If we take a look, barium With EDT would be BY 2 -, It would equal the initial concentration of Barry Mayan, which is .100 times the volume of barium ion Divided by the total volume of the solution, which is 50 plus 100. So that's 1:50. So the concentration of our barium E D. T. A complex at this point would be 0.0.33 molar. So that's the concentration that we would place here within our basic ice chart. Then we'd say that since there is an initial amount of that, we're gonna be decreasing it over time. Initially we don't have any concentrations of our metal ion and R E D. T. A. Because we've converted it into our complex itself. These will be building up over time. So we have them as plus now, remember this again, is our generic equation to show how our metal ion is binding with R E D T A comp E D T a molecule to form our metal E D T a complex because we're buffered at an exact ph we'd have to determine what our conditional formation constant will be, which is K prime sub F. To do that. We multiply the fraction of the basic form of R E D T A times the formation or stability constant here. So if we look that up, we'd see that at 9.0 Alpha Equals .041. And because we're dealing with barium ion, we'd say that the log of K F for barium ion Equals 7.88. And so K F for barry Mayan Equals 10 to the 7.88, Which year equals when we plug all this into here. .041 times 10 to the 7.88 Equals 3.11 times 10 to the six. So that would be our conditional formation constant. Now we would take that formation, that will take that formation constant. And we'd plug it into our equilibrium expression here. So that would equal 0.33 molar. Which is the concentration we calculated for E D T. A complex minus X divided by x squared. We have to solve for X here. So you multiply both sides by X squared. So 3.11 times 10 to the six times X squared equals 60.33 minus X. Bring everything over to the left side. So add plus X. subtract .033 from both sides. So we'd have 3.11 times 10 to the six X squared plus x minus 60.33. You would set up your basic quadratic formula in order to solve for X. And when you set up the quadratic formula we find that X equals 1.03 times 10 to the -4 for the concentration. So that X. There would give us the concentration of the metal ion itself. So we've determined the concentration of barium ion as being this value here and because we know that we can just take the negative log of it to find our final answer. And as we predicted we should see an increase In the negative log of the concentration of our metal. So here it's 3.99 that we get. So remember it's imperative that you are able to calculate the equivalent volume of our titra E. D. T. A. To determine where exactly within the tight tray shin our calculations will take us at the equivalence point. We have a lot more work to do not just simply plugging into one formula. We have to set up our concentration of R. E. D. T. A. Complex from there. Set up our equilibrium expression. You can set up a basic generic ice truck in order to help organize the layout for your equilibrium expression here and remember because we're dealing at a buffered ph we'd have to figure out what our conditional formation constant which is K prime sub F utilizing the expression the concentration calculated and our equilibrium expression helps us to determine what X. Is with X. You know what the concentration of your metal ion will be. And from there we can take the negative log. Finally, we'll go on to our last video in terms of articulation to see what happens after the equivalence point and see what steps need to be taken to determine the concentration of the metal at that point in our titillation curve.

4

EDTA Titrations

5m

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we have finally reached the point where we have to consider calculations after the equivalence point in R E D T. A tight rations curve. At this point we utilize our equilibrium expression where are conditional formation constant. K. Prime sub F equals our metal E D T. A complex divided by the concentration of the metal times the concentration of E D T. A. At this point we have enough information. We've already calculated our conditional formation constant from previous part in our titrate in curve. We'll be able to calculate our concentration of metal E. D. T. A complex and we'll be able to determine what R. E. D. T. A excess will be in terms of its concentration. With these, with these three pieces of information, we can finally determine what the remaining concentration of our metal ion will be after the equivalence point. So here we have the attrition of 50 mls of 500.100 moller barry Mayan still buffered to nine for ph with 112 mls of 1120.50 moller E D T A. So what we need to do is we need to determine what the concentration of my E. D. T. A. Is and the concentration of my metal E. D. T. A complex. So we're gonna say here that the concentration of E. D. T. A equals the initial concentration of E D. T. A, Which is .050 moller times the excess volume of E D T. A. We only needed 100 mls to get to the equivalence point. We are 12 mls beyond that point. So 12 mls is our excess volume of E. D. T. A. And then we're gonna divided by the total volume. So that's 50 mls plus 100 and 12 mls. So when we do that that gives us a concentration of .00370 moller. Next we need to determine what the concentration of our metal E. D. T. A complex will be, which in this case is B. Y two minus. So we have the initial concentration of our metal which is 0.100 moller times the initial the volume of the metal ion. So 50 mls again divided by the total volume which is 50 mls plus 112 mls. So that gives me .0309 moller. So we're gonna use the conditional formation constant we found previously. Remember we said that that was 3.11 times 10 to the six equals. So the concentration of my metal E. D. T. A complex which is 0.309 moller divided by the concentration of the metal ion which we're looking for which is barium ion times the concentration of E. D. T. At this point. So there we go. And all we have to do is solve for that missing variable there. So when we rearrange this expression we see that the metal ion concentration which is barium equals the concentration of my Baron E. D. T. A complex divided by my conditional formation constant times the concentration of my ex. S. E. D. T. A. So plug in the values that we found for each one. And we see that we get a concentration for barium of 2.69 times 10 to the -6. So when we take the negative log of that we'll have our answer. So that gives us 5.57 at this point. So again it's all based on first determining what are equivalent volume for E. D. T. A. Titrate titrate will be. Once we do that we'll know where exactly and articulation curve. Our calculations will take us are we before the equivalence point? At the equivalence point? After the equivalence point. So just remember for the equivalence point we'll have an excess of barium ion at the equivalence point all of it has been uh reacted with R. E. D. T. A. Where they're equal moles of both. So we just have to determine um what are remaining barium ion will be that's free floating. And then finally after the equivalence point we have an excess of E. D. T. A. But we still have some free floating metal ions in that case we have to determine what that final amount would be. So just remember we're in the equivalents were well in the tight rations curve our calculations take us and then what formulas are necessary to determine our concentration of our metal ion

5

EDTA Titration Calculations

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So here we need to calculate the negative log of manganese three ion. For the titillation of 30 ml of 300.100 Mueller E. D. T. A. With 50 ml of 500.200 molar of manganese, three phosphate. When the ph is 10.0. All right. So what we need to do first is we need to determine is this tie trey shin that's occurring before at or after the equivalence point. So we're gonna need to determine what are equivalent volume for R E D. T. A tight Trent will be. So we're going to say here that the polarity of our metal Which is Magnus three in this case times the volume of magazines. three equals the polarity of E D T A. Times the equivalent volume of E D T A, plug in what we know. So the polarity of manganese three ion is 30.200 moller, The volume is 50 MLS Equals. We're gonna have .0100 Molar. And we don't know what the equivalent volume is yet Divide both sides here by .0100 Moller. So the polarities cancel out and we'll see that our equivalent volume equals 100 mls of E D. T A. Well, the volume in the question itself is only 30 mls. So this is dealing with calculations before the equivalence point. And since we're dealing with calculations before the equivalence point, that means we're gonna have an excess of our manganese three ion. So we have to do here is calculate what that excess amount of manganese three ion will be. So we're gonna say the concentration of manganese three ion equals your equivalent volume minus the volume of E D. T. A. Divided by the equivalent volume. This will give us the amount of the magnets three ion remaining, the fraction remaining times the initial concentration of magnets, three ion Times its dilution factor. So that would be the volume of magnets three ion divided by volume of the solution. So plug in what we know so are equivalent volume that we calculated for E. D. T. A. Is 100 mls minus the 30 mls that we're actually using divided by 100 mls. Next multiplied by the concentration of magnets three ion times the volume of magnets three ion divided by 80 mls which is the volume of our solution which is 30 mls plus 50 mls. When we work all that out, we get a concentration for magnus three ion as 8.75 times 10 to the minus three moller. Now that we have that concentration, we can take the negative log of it To find our final solution. So negative log of 8.75 Times 10 to the -3 Gives me 2.06 as my final answer. So remember for questions like this we need to determine where exactly within articulation curve. Our calculations um will take us here, we're dealing with calculations before the equivalence point. So we don't have to worry about things such as the conditional formation constant. The use of equilibrium expressions. All that stuff occurs at and after the equivalence point here. Since we have an excess of our metal ion, just determine what that concentration is. Take the negative log and you'll have your final answer now that you've seen this example attempted to the practice question that's left here here. It's a bit trickier because if we do need to calculate our conditional formation constant, we're not dealing with a whole integer for our ph That means that you'll have to utilize some of the formulas we've seen previously to determine what our alpha of our basic form of E. D. T. A will be. Once you do that, then you'll be able to find your formation constant or your conditional formation constant. So, good luck and attempt to do this one. Once you do come back and see how your answer matches up with mine.

6

Problem

Calculate the pNi** ^{2+}** for the titration of 50.0 mL of 0.120 M EDTA with 15.0 mL of 0.100 M NiCl

A

1.90x10^{-17}

B

17.7

C

16.7

D

17.4

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