In the equilibrium unit you learned about the reaction qoutient Q and we're going to utilize that Q to predict if a precipitation reaction will occur. That'll be the first part of this learning objective. And then we will continue on by taking a mixture of many ions in solution And predict who will precipitate first in that mixture. So if you calculate a Q for a solution you can compare it to K and determine what state that insoluble salt is in. So let's say for example we have the insoluble salt of lead chloride and we know this is the equilibrium process that we would consider. We know that K, which we call Ksp, would be the concentration of lead times the concentration of chloride squared. But you could also talk about a Q And a Q would be what is the concentration of lead right now Initially, what is the concentration of chloride right now? and let's square those, and compare that to the K value the Ksp value of that equilibrium. Let's start with the notion that we have an unsaturated solution. If it's unsaturated that would me we could dissolve more if we had it to dissolve. That would mean that we haven't quite gotten there and that the Q is too small. So any time your concentration that you have in solution of the ions of that salt gives you a value smaller than Ksp then you have an unsaturated solution. Well certainly if it's saturated you're right there at the point where they're equal to each other. When you will get a precipitate to occur is when it is supersaturated. If it is supersaturated it is going to precipitate out. So this would occur when you have two much of the ions present. If there's too much of the ions present and they need to come out of the solution how would Q compare to Ksp? Well they would be too large, so if Q is greater than Ksp You would need a precipitate to occur when you mix solutions together. If it's unsaturated or saturated you would not get a precipitate to occur. So here is an example problem. I want to determine if a precipitate would occur (silver chromate) when sodium chromate and silver nitrate are added. Now what you want to do is start by writing the reaction for Ksp. Ag2CrO4 is the solid. You would have 2Ag+ ions in solution and you would have a CrO4^2- ion in the solution and that would be the equlibria. We know that the Ksp is equal to the silver ion concentration squared times the chromate concentration. And that would also be the Q expression if they were all initialed. And that's what we're going to figure out. What are the initial concentrations of those? Now you will not come up here and take this and this number directly and plug them into those concentrations because we have two separate beakers. We have one beaker that has in it 250ml of 0.33 molar in a Na2Cr04. In other words since it disassociates it has a 0.33 molar solution of the chromate ions in solution. That's one beaker. In the other beaker we have the solution that contains the silver nitrate. And there are 250ml of this solution. It is 0.12 molar AgNO3. What I need to plug into my expression is Ag so it's 0.12 molar AG+. And then we're going to take those two and we're going to mix them togther. So we're going to pout them into a larger beaker. There's going to be in this beaker 500ml of solution. And what that does as we add that together is it's going to dilute down the chromate ion concentration because I'm adding the other solution and it's going from 250ml to 500ml. and I'm going to dilute down the silver ion concentration because I'm adding the other solution and it's going up to 500ml. So let us calculate the silver ion concentration after these have been mixed and before they have had a chance to react. That would be the initial concentration of silver in this beaker down here. I'm going to be considering a dilution so it's going to utilize M1V1=M2V2. So the silver ion concentration would be the N2 of this equation. M1V1=M2V2. M1 is 0.12 molar. I have 250ml of it. M2 is what I'm trying to determine. V2 is 500ml. So M2 which is the silver ion concentration initially after I've added it to that beaker is going to be 0.06 molar. No we'll do the same thing for the chromate in the other beaker. For this I'm going to consider these amounts. And I'm going to realize that the chromate concentration would be M2 in this solution. M1V1=M2V2 where M1 started out as 0.33 molar. V1 is 250ml. M2 is what I'm trying to determine. And V2 would be 500ml. So M2 in this solution would be equal to 0.165. Now I'm carrying an extra sig fig there but that's ok. That's the concentration of the chromate. So now at this point I'm going to take these two amount, the concentration of the silver initially, the concentration of the chromate initially, and I'm going to plug it into my Q expression and determine Q. So Q would be 0.06 and I have to square that number, times 0.165, and that will get me a value for Q equal to 5.9X10^(-4) . NOw we are ready to compare that Q to the K_sp. There is the K_sp. We see that Q is much greater than the K_sp we have ion in solution and whenever Q is greater than K_sp it needs a precipitate to bring that value back down. So will a precipitate K_sp it needs a precipitate to bring that value back down. So will a precipitate form? The answer is yes. Now we are going to utilize this concept of precipitation to talk about selective precipitation. And it is going to use the different solubility to talk about selective precipitation. And it is going to use the different solubility properties of salts to separate out ions from solution. Lets start by considering this. You are going to look at this solution and I have added some chloride ions, bromide ions, and iodine ions now of course there is going to be some kind of cation in there. Maybe it is sodium chlorides, sodium bromide, and sodium iodine that I have dissolved into solution. And in this solution I am going to start adding silver ions. Slowly, drip by drip. And the question is K_sp value that I have give down here which one of those three ions would precipitate first? If you selected AgI then you are correct. Silver iodide will precipitate first. It has the smallest K_sp. And so it would precipitate first, it is the least soluble of the bunch. Now once you have decided what is going to precipitate first, you can then figure out well how much silver would be needed to make it precipitate? So I have expanded this question a little bit by giving you silver would be needed to make it precipitate? So I have expanded this question a little bit by giving you concentrations of those three ions. Now we know that the iodide going to precipitate first. Especially since I am starting with the same amount of all those. Now if we were not starting with the same amount of all of those Especially since I am starting with the same amount of all those. Now if we were not starting with the same amount of all of those there could be trouble. But we are starting with the same amount of all of those. And I am adding the silver and the silver iodide will precipitate first. But we are starting with the same amount of all of those. And I am adding the silver and the silver iodide will precipitate first. So I am going to use my K_sp value expression for the silver iodide. It is the silver concentration times the I- concentration. I am going to put in the K_sp for silver iodide it 8.3 x 10^(-17). I am going to put in the K_sp for silver iodide it 8.3 x 10^(-17). I am going to put in the iodide concentration that I have here 0.010 molar. That would give me a silver ion concentration equal 8.3x10^(-15). So that concentration of silver is all I would need to start the precipitation of the silver iodide. At that point, the bromide and the the precipitation of the silver iodide. At that point, the bromide and the Chloride would not precipitate at all. At that point, the bromide and the Chloride would not precipitate at all. That is not very much. Chloride would not precipitate at all. That is not very much. So lets expand this even more. The iodide is precipitating and I keep on adding. Eventually the next ion in line to precipitate which is the silver bromide is going to join the precipitation ranks. So I want to know what would be the concentration is going to join the precipitation ranks. So I want to know what would be the concentration of I-. Would the silver bromide just begins to precipitate. So lets think about what is being asked here. just begins to precipitate. So lets think about what is being asked here. First thing that starts to precipitate is the AgI. It precipitates First thing that starts to precipitate is the AgI. It precipitates by adding a little bit of silver, add a little bit of silver it keeps is the AgI. It precipitates by adding a little bit of silver, add a little bit of silver it keeps precipitating. If I add a little bit of silver it keep precipitating. by adding a little bit of silver, add a little bit of silver it keeps precipitating. If I add a little bit of silver it keep precipitating. Eventually, my concentration of silver is going to increase to the point that the silver bromide is going to precipitate. Is this a point where there is still a lot of iodide in there or have we gotten rid of most of the iodide? We can see this with this problem. So we going to take that second equilibrium. The K_sp for silver bromide, would be the silver concentration times the bromide concentration. The K_sp for silver bromide, would be the silver concentration times the bromide concentration. So K_sp for silver bromide is 7.7 x 10^(-13). That is going to allows us to figure out how much silver is needed to precipitate this bromide. So the silver concentration that is just going to get this silver bromide to precipitate is going to 7.7x10^(-11). OK so as soon as I get the concentration of silver to that amount, finally the silver bromide is going to precipitate. Remember what it was on the other one? to that amount, finally the silver bromide is going to precipitate. Remember what it was on the other one? With the silver iodide it was only 8.3x10^(-15). That is what got the first one to precipitate. Now, I have go to get this concentration of silver up to That is what got the first one to precipitate. Now, I have go to get this concentration of silver up to 7.7x10^(-11) and then it begins to precipitate. Now once it begins to precipitate when can now answer this question what would the I- concentration be? Well, lets go back to our initial K_sp for silver iodide. Lets us not put in this silver concentration that we got from here into this equation. Now the K_sp is still of course 8.3x10(-17). We have upped our silver concentration now to 7.7x10^(-11). That is when the silver bromide starts to precipitate. We can see how much iodide would be in there. That is when the silver bromide starts to precipitate. We can see how much iodide would be in there. So we divide both sides We can see how much iodide would be in there. So we divide both sides by 7.7x10(-11) we will get an I- concentration of 1.1x10^(-4). So we started out with 0.1 molar I- and the silver iodide precipitates. Its concentration will get all the way down to that value and the silver iodide precipitates. Its concentration will get all the way down to that value 0.0001 before Its concentration will get all the way down to that value 0.0001 before the second one starts to precipitate. So we have got 0.0001 before the second one starts to precipitate. So we have got a lot of it out of the solution before the silver bromide precipitates. the second one starts to precipitate. So we have got a lot of it out of the solution before the silver bromide precipitates. So this is the end of our learning objective 13. In which we are looking precipitation reactions will they occur, we are looking at selective precipitation. This is also the end of our aqueous equilibria unit. Historically for students this is a very challenging unit. So make sure you watch, and re-watch the lessons. Make sure you spend time So make sure you watch, and re-watch the lessons. Make sure you spend time examining the practice problems where I am working through problem and really work through those problems we have assigned to you. And every case you have to write your own reaction and I think for most students that is the challenge. Thing about what is occurring in solution what do you have, is it an acid, is it a base, is it a mixture? Do you have an insoluble salt you are considering? How do you write those? So make sure you are really spending the time working problem because there is just no substitute for that, with this unit.
