Titrations and Titration Curves

So we've talked about titrate asians as well as titillation curves in the past. Now we're dealing with redox tray shin curves, we're gonna say here a redox titrate curve follows the change in either the analytics which is coined your tie Trent or the title itself concentration as a function of the titans volume. So as we're adding are tightened volume by volume. We're gonna see a change in the concentration of either species. Now with all tight rations, whether they be acid based tight rations or redox tight rations, we should always first determine what are equivalent volume of our tie Trent will be. So here we have the tight rations of 50 mls of 500.100 moller sodium chloride with 0.100 moller silver nitrate here would produce our precipitation reaction. What happens here is that Any plus would combine with the n. 0. 3 - here. But based on sai ability rules that would give us an acquis compound here, we're concerned with the silver ion combining with the chloride ion to give us silver chloride. Remember when dealing with a precipitation reaction, we're really talking about K. S. P. Of the ionic compound. So remember K. S. P deals with the solid ionic compound breaking up into its ions here in this equation, what I've done is we've reversed the reaction. So now my products are reacting and my reacting tier is now a product to show the formation of the solid here. This K represents our formation constant. It is actually the inverse of my K. S. P. So Kay here is actually one over K. S. P. Remember here when we reverse the reaction we get the inverse of our original equilibrium constant. So for this original breakdown of my ionic solid, we have K. S. P. Because I reverse the reaction now becomes one over K. S. P. Which is represented by this formation constant here. The fact that it is a number much greater than one tells me that the formation of this solid is highly favorable. Which makes sense because based on solid ability rules silver when it combines with chloride ion definitely forms a solid, precipitate. Remember in terms of calculating the equivalent volume here we say the equivalent volume to determine it, we'd say molar itty of my an elite which will say is a times volume of my an elite equals polarity of my tie, Trent times equivalent volume of my tie trend. And when we're calculating the equivalent volume we're looking for the volume of the tightrope. So you plug in 0.100 moller of my an elite times its volume equals. We know that the first compound is the an elite because we're seeing the titillation of this with this. So we're adding this to it. So we have 0.100 moller times the equivalent volume of my tie trend, Divide both sides by .1 Molar. So my equivalent volume of my tie Trent equals 50 MS. So that's the first step in terms of our redox situation. Gathering all this information overall will help us to determine what are our titillation curve would look like once we've calculated all different points, in terms of the thai tray shin, now that we've covered equivalent volume, move over to the next video, where we take a look at calculations before we reached the equivalence point.

So before we reach the equivalence point, we have an excess of an elite. So at this point we have an excess of an allied, we needed 50 mls of our tight. Trying to get to the equivalence point here, we're only using 20 mls at this point we're dealing with an excess of an elite. So we're we're calculating the concentration of our an elite. We're actually calculating the amount that is left a NRI acted. That's what we're calculating at this point. Now here, cl minus happens to be our an elite A G L. Positive is our tie Trent. It is the ion that's going to connect with the cl minus to form the precipitate. So here we say that this represents our concentration of an elite. Here we say its concentration equals its initial moles minus. The moles of titrate added, divided by the total volume. Now I wrote moles here and remember moles equals leaders times polarity. But you could just leave it in milliliters if you want to remember the word of in between two numbers means multiply. So I could have kept it in many moles, which would just be milliliters, times more clarity. If we're using leaders times polarity. Then at the end we have moles on top and we'd have to use leaders on the bottom and that would give me more clarity. If I choose to keep the mls would be milliliters times more clarity would give me millie moles. So you have many moles on top milliliters on the bottom, which ratio wise doesn't change anything proportionately. So this would still give you more clarity at the end. He will just keep the mls. So we're gonna do 50 mls times 500.100 moller of N A C L. So that's going to give me five million moles of my an elite minus 20 mls times 200.100 moller to millie moles of my tie Trent divided by the total volume. So that's 50 mls plus 20 mls. Okay, so that's gonna give me my new concentration of a NRI acted an elite that comes out to .042857 moller here, we're talking about it in terms of a tin tray shin curve. So we're gonna do P of cl remember P here just means negative log. So we're taking the negative log of the chloride ion concentration. So plugging what we just got Here that give me 137,. So before we reach the equivalence point, at this exact moment, based on the amount of titrate used, this would be the negative log of my analyzed concentration as we continue to add more and more tight trend. We should expect this value to increase. Now that we've talked about before, the equivalence point. Move over to the next question and take a look at what happens when we're at the equivalence point. So click over to the next video and see what we do in terms of that situation

So at the equivalence point we have equal moles of our analysts and our tight trend here, we've formed our precipitate. It exists in equilibrium with the ions itself. To figure out the concentration of our particular an elite, we actually have to use the cell viability product constant of the precipitate. So we're gonna have to use the K. S. P. Of my silver chloride solid. So here, remember how would how do we do that? We have our silver chloride solid. It breaks up into ions, it breaks up into silver ion acquis plus cord ion acquis here, solids are ignored. We're gonna say at equilibrium, we only have one silver ion here, which is X. And one chloride ion here, which is X. If there were two here, we'd have to put a two here and that would have an effect on my equilibrium expression. Remember K S P is just like any other equilibrium constant equals products over react ints. But here my reactant is a solid. So we're going to ignore it. So K S P. Which is equal products, so equal a G positive times cl minus. So we'd say K S P. Which is this number equals X times X. So it equals X squared. Let's say that there were two here than Ks P would equal actually bring the two down here. It would equal X times two X squared because remember, whatever the coefficient is that becomes the power. So just in those cases, that's what we would do for those situations. Now we need to do is we need to just isolate X. Because once we know what X. Is, that gives me actually the concentration of chloride ion. So take the square root of both sides here, that gives me X equals 1.34 times 10 to the -5 Moller. That equals the concentration of my chloride ion. Now we just say again P. Of C. L. Equals negative log of my chloride ion concentration, So that equals 4.87. So as we expected, we see an increase in the negative log of my chloride ion as I add more and more titrate. Now, finally, we're gonna go to the point where we are after the equivalence point and we need to determine what our final p of chlorine would be in that situation. So click on to the last video and see what we do in this situation.

after the equivalence point but we now have is an excess of our titrate. And in order to figure out the concentration of our analytics, we first need to determine what the concentration of our titrate would be at this point. So we're gonna say here, silver ion represents are tightened so it's moles of the toy train added or millie molds to make it easier. So multiply 70 mls times 700.100 Mohler. That gives us seven million moles - million moles of are an elite. So multiply these two gives me five million moles divided by the total concentration, which would be 50 mls plus 70 mls. So when we plug that in that gives me .0167 moller. Now that we have the concentration of our access an elite, we're gonna use that value with K. S. P. In order to figure out the concentration of my an elite. So we're gonna take this equilibrium expression here that we've seen earlier, we're gonna say K. S. P. Equals the concentration of my tie, Trent times the concentration of my an elite plug in. The value that we have for K. S. P. Which was given as 1.8 times 10 to the negative 10. Here, this will be .0167 moller. And we're looking for the analyzed concentration Divide both sides by .0167 Moller, cancel, cancel. We'd get our concentration here For Claude Iron as being 1.07 784 times 10 to the -8 moller take the negative log of that number would give me P of C. O. So here that would give me 7.97 when I take the negative log of this concentration. So we can see that taking the negative log of my concentration has increased as I go through the steps for the redox detraction. Now here, if we were to construct basically a chart we'd say here, we have the volume of our tight Trent, which would be in this case the silver nitrate. Here we have the negative log of our an elite which in this case is the chloride chloride ion. And we would see from our calculations that we saw that PCL was increasing and what we would expect to see is if we did a bunch of points, we'd see it gradually increase and then you'd see it hit a volume where there's a sharp increase, meaning that we've reached the equivalent volume where there's an equal moles of our tightrope and are an elite reacting with one another and then eventually level off. If we were to use an indicator, that indicator would change colors at a precise moment as it reacts with the titrate and that be somewhere close to the middle or so of this redox titillation curve. Sometimes it may not be the exact middle, but again you'd use an indicator to find more or less where the approximate range of the endpoint would be. Another thing that we could use is we could use uh podium metric tit rations in which we can basically use readings in terms of voltage to give us information on the different types of solution used within this redox filtration. From this information we could uh possibly identify are unknown an elite if this were done in the lab and just realize very similar to things that we've done in the past in terms of asset based tight rations. But now we're using it in terms of redox tight rations. So just remember we apply the same principles that we've done before in the past, where we're trying to determine what our equivalent volume is to start. And then from there, determining are we at the equivalence point before the equivalence point? After the equivalence point, to know what set of equations to utilize to find our final answer?

So here it says to calculate the floor and ion concentration from the tight rations 130 mls of 1300.1 to 0 molar potassium fluoride with 100 and 50 mls of 500.100 moller. Barium chloride here we're told that the saw liability product constant of barium fluoride is 1.5 times 10 to the negative six. Alright, so were tight trading potassium fluoride with with barium chloride. So what we need to do first is determine what are equivalent volume will be. This will help me determine am I at the equivalence point before the equivalence point or after the equivalence point? So M an elite times V. An elite equals M titrate times V titrate We have here .1-0 moller Times 130 M mi equals 0.100 Mueller times the equivalent volume of my tie trend Divide both sides by 0.100 Moller. So my equivalent volume is 156 mL. So I need 100 and 56 mL of the thai trend to reach the equivalence point here, I only have 100 and 50 mls. So that means we're dealing with calculations before the equivalence point. And if we're dealing with the calculations before the equivalence point, that means that we have an excess of an elite. So we're gonna see the concentration of our an elite. How do we know who's the analytics? The analytics in this case would be fluoride ion because we're tight trading with barium chloride. Bury him is part of the ionic solid with the K. S. P. So barium is the tight trend which means my fluoride ion is the analytic. It's what's being tight traded by the barium chloride. So here concentration of my an elite here would equal moles or milli moles of an elite minus millie moles of titan divided by total volume. So multiply milliliters times polarity to get the million miles of each. So that's 15.6 million moles of florida ion minus 15 million moles of barry mayan Divided by 1 30 ml plus 150 ml. So here that would give me a concentration of 2.14 times 10 to the -3 molar of fluoride ions. So that would represent the concentration of my fluoride ion that's remaining before I reached the equivalence point. If we wanted to go a step further, you could have just done after that negative log of this concentration to find P. F. Here, we're not being asked to do that. We just have to figure out the concentration which is this original value here. Now that we've seen this attempted the practice question that's left on the bottom of the page and come back and see if your answer matches up with mine

Calculate the pCN from the titration of 40.0 mL of 0.060 M NaCN with 20.0 mL of 0.050 M AgC₂H_­3O₂. The solubility product constant of AgCN is 2.2 x 10^-16.

• A. 1.53
• B. 0.023
• C. 1.45
• D. 1.63
• E. 0.029