Within this learning objective we are going to learn how to utilize what we learned in the previous one which was to write the reactions for these insoluble salts and write the equilibrium constant expressions for the salts. We going to be able to use that and convert between what is called molar solubility and the solubility product we learned about there. And we going to learn how to rank salts and the solubility product we learned about there. And we going to learn how to rank salts according to their solubility. Which one is more soluble, which one is less soluble. By looking at either molar solubility or K_sp. So as you look at these boxes down below in this slide here. We see that we can convert between K_sp. So as you look at these boxes down below in this slide here. We see that we can convert between what is called K_sp and molar solubility. So the K_sp is one-way to express how soluble salt is and in general the smaller the K_sp the less soluble the salt. Now that comparison really only works well when you are comparing salts that have the same general formula. Now that comparison really only works well when you are comparing salts that have the same general formula. So you might compare a salt with a formula like MX to MY. They have a 1 to 1 ratio but it might not be good for comparing MX to another substance of MX_2. Where there is two anions for every one cation. So you have to be a little careful when comparing K_sp's for these insolubles salts and say who is more soluble and who is less soluble. But the other way of expressing solubility is what is called molar solubility and here is the definition. It is abbreviated with a lower case 's' and it is the number of moles of solute in one liter of saturated solution. It is in unit of moles per liter, in other words it is molarity. And it is basically the molarity of the amount of the salt that dissolved. So if you are given one of these two. Either the K_sp or the molar solubility you can calculate the other. By way of determining the concentration you can calculate the other. By way of determining the concentration of the ions in solution. In our first example problem we are given the K_sp of this lead chloride and we are asked to determine the molar solubility. So as with any equilibrium problem, you want to start with a reaction we are not giving you the reaction, we are expecting you to know the reaction. And you put the solids on the left, and the ions on the right. I will leave room for an ICE table. I am going to write the K_sp expression. We have the lead. We have the chloride and it is squared. So what is happening here? We have a beaker of water and we are going to dump into that water some lead chloride. Lead chloride is not very soluble so most of the lead chloride will sit at the bottom. We are going to dump and excess of lead chloride in there. And some people do that the say excess. But that looks too much to me like X - 5, and writing out excess I just don't do. So what I general write is I put some of the lead chloride into solution. I put a bunch. And I will put zero because none of it has dissolved yet. I will allow instead of using X's, I am going to use S's. in this change line because S and I always use a script s because my s looks too much like a 5. Is moral solubility. And where I have written this s here that is how much lead chloride dissolves. So by definition that s or what sits right here in my change line by definition is molar solubility. And so that is why in any text book you look at they use S's here. Well we will follow that S change across to the product side. A S and 2S. And only a little bit of that lead chloride dissolves so there is still some in the E line. So we will have S and 2S. We come to our expression and plug in what we know. We know the K_sp, that was given to us that was 2.4 x 10 ^-4. For lead there was an S, so we will put an S here. For Chloride there was a 2S. For lead there was an S, so we will put an S here. For Chloride there was a 2S. So we will put a 2S here and we will square it. That will give me, 2.4 x 10 ^ -4 Is equal to 4 do you see where the 4 came from? The two square, and then S cubed. Is equal to 4 do you see where the 4 came from? The two square, and then S cubed. See why it is cubed? You have squared this S and you have multiplied by another S. So now we can divide by 4 on both sides. And give me and S cubed So now we can divide by 4 on both sides. And give me and S cubed equal to 6.0 x 10 ^-4 Then we can take the cube root of both sides and that will give me S equal to 3.9 x 10 ^-2. Now it has a unit, it is how much has dissolved in mole per liter so it is in molarity. Now it has a unit, it is how much has dissolved in mole per liter so it is in molarity. And that is the value for the lower solubility of lead chloride. OK in this problem we have molar solubility given to us and we are trying to determine K_sp so it is the reverse of the previous problem. We begin by writing the reaction. Same as we did before we have the solid on the left the ions it breaks apart into on the right. I will do an ICE table. Before I do that I write my K_sp expression. Manganese 2 plus Before I do that I write my K_sp expression. Manganese 2 plus carbonate so far nothing is different. We are going to dumb some of the solid into our water before anything happens, we don't any of those ions present. Now here is where it varies. We could put minus s here, but we know what the solubility is. Remember S stands for the molar solubility. And that was given to me. the solubility is. Remember S stands for the molar solubility. And that was given to me. So i know how much of this is dissolving. Not very much, but some is dissolving. That quantity. We carry the change line across so we are going to produce 4.2 x 10 ^ -6 of this substance and produce 4.2 x 10 ^ -6 of the carbonate. So there is a large excess of the solid, so there is still some sitting on the bottom. But we now know the quantity at equilibrium of these two substances. So K_sp's value would simply be 4.2 x 10 ^-6 which is the manganese's concentration times 4.2 x 10 ^-6 which is the carbonate's concentration. When I multiply those two value I will get 1.8 x 10 ^-11. which is the carbonate's concentration. When I multiply those two value I will get 1.8 x 10 ^-11. and that is the K_sp of Manganese 2 Carbonate. We are going to work this one together but before we work it together I want you to figure out We are going to work this one together but before we work it together I want you to figure out the molar solubility of this salt, so lets describe what is happening here. You were given the solubility not is molar per liter but in grams per liter. We cannot use the K_sp expression unless we know things in molarity. So you job is to take this information and determine which one of those answer is the molar solubility of this compound. Well did you choose B? If so that is the correct answer. To two decimal places. Or two significant figures. Now we are going to complete this problem together. New we have space to work the problem and we are going to start by writing the reaction the solid is M_2X_3. And it is turning into the ions in aqueous solution, there are two of the metal ion. It probably has a 3 plus charge. And 3 of th anion, it probably have a 2 minus charge. Its charge is not important but that is probably the charges that they are I will leave room for an ICE table and I will write the expression I will leave room for an ICE table and I will write the expression K_sp expression down here. K_sp is the metal ion's concentration squared times the X concentration cubed. So we will fill in our table we have some solid that we put into the beaker. There is none of the ions to begin with. We figured out with your previous There is none of the ions to begin with. We figured out with your previous question that you answered what the molar We figured out with your previous question that you answered what the molar solubility of this salt is. I carrying just a little bit extra sig. fig here just to carry extra until the end. It was actually 1.25 just to carry the extra. This is the moles of solubility. We carry the change line across, so it has to be twice 1.25 x 10 ^-19 here. And 3 x 1.25 x 10 ^-19 here. 1.25 x 10 ^-19 here. And 3 x 1.25 x 10 ^-19 here. So there is still going to be some of the solid on the bottom of my flask. But now we are going to have 2.5 x 10 ^ -19 and 3.75 x 10 ^ -19. Now we are ready to obtain the K_sp value. For the metal we put 2.5 x 10 ^ -19 and that is squared. For the none metal we put 3.75 x 10 ^-19 and that is cubed. We you plug these into your calculator and obtain a value, you will get 3.3 x 10 ^ -93. So that is an absurdly small K_sp value for this hypothetical compound. But I wanted to show you that these coefficients come into play twice. They come into place in the change line and they come into play in a power. So that is the end of this learning objective where we have learned to covert between K_sp value and molar solubility.
