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Solubility and the common-ion effect

Video transcript

our goal is to calculate the molar solubility of lead two chloride which is a slightly soluble compound and we're given the solubility product constant KSP which is equal to 1.6 times 10 to the negative 5 at 25 degrees this time now we're starting out with a point 1 molar solution of potassium chloride so if we have a beaker here and we have a point 1 molar solution of potassium chloride that means the concentration of potassium ions in solution is point 1 molar and the concentration of chloride anions in solution is also point 1 molar so we have some solid lead two chloride right so this is supposed to represent our solid lead two chloride and what our problem wants us to do is to figure out how much of our lead two chloride will dissolve if we add it to our solution of potassium chloride so some of the lead two chloride is going to dissolve and eventually reach equilibrium so let's go ahead and write that out here we have solid lead two chloride and some of it's going to dissolve and give us lead two plus ions in solution and we would also get chloride anions in solutions we would get CL minus when you balance this you need a 2 in front of our chloride anions and everything else would get a 1 so we first think about the initial concentration of our products so what's the initial concentration of PB 2 plus so before we've added our lead two chloride right nothing has dissolved yet so we don't have anything for our concentration of lead two plus right so nothing's dissolved before the lead two chloride has dissolved we don't have anything for the concentration of chloride anions from lead two chloride but we do have a concentration of chloride anions from our point one molar solution of KCl so we're starting with an initial concentration of 0.1 molar for our chloride anions next we think about the change so some of our lead two chloride is going to dissolve and let's make that concentration X so we're going to lose a concentration of lead two chloride and for every one mole of lead two chloride that does solves we get one mole of lead two plus ions in solution so if we lose X for lead two chloride we gain X for lead two plus and for every one mole of lead two chloride that dissolves we get two moles of chloride anions so we're going to gain two X here for our chloride anions so our common ion for this problem is the chloride anion because we have two sources right one source was from our potassium chloride and one source was from our lead two chloride so at equilibrium all right so at equilibrium our concentration of our products would be zero plus X for lead two plus or X and point one plus 2 X for chloride anions so this is equal to point one plus two X next we write our equilibrium expression right so our solubility product constant KSP is equal to concentration of our products so concentration of PB two plus raised to the power of the coefficient so our coefficient here is 1 so this is to the first power and we multiply that by the concentration of chloride anions in solution CL minus and we raise that to the power of the coefficient our coefficient is a 2 so we raise that to the second power next we can go ahead and plug in our solubility product constant so KSP is equal to one point six times 10 to the negative five so one point six times 10 to the negative five is equal to the concentration of lead two plus ions at equilibrium which we already figured out was X so this is X to the first power times the concentration of chloride anions at equilibrium which is 0.1 plus two x squared so we have point one plus two x squared so next we have to solve for X all right so on the left side we have one point six times 10 to the negative 5 is equal to X to the first and over here Oh point one plus two X we could do that math but to make our lives easier we usually make the assumption that is really small compared to point 1 and if X is really small compared to point 1 point 1 plus 2x is approximately the same as point 1 so let's go ahead and put in point 1 so we have point 1 instead of 0.1 plus 2x and point 1 squared is point 0 1 so on the right side we have x times point 0 1 and this is equal to 1 point 6 times 10 to the negative 5 so when you solve for X X is equal to point zero zero 1 6 molar and so this this would be the concentration of lead two plus ions in solution right if we go back up here we can see that lead the concentration of lead two plus ions is X so that's our concentration and that's also the molar solubility of lead two chloride right that's how much dissolved so let's go ahead and write that in here so our goal is to calculate the molar solubility and it was equal to point zero zero one six molar now keep in mind this is the molar solubility for lead two chloride in our solution of potassium chloride alright so this is this is in potassium chloride in the video on solubility product constant we found the molar solubility when it was just pure water and if you look at that video we calculate the molar solubility to be point zero one six molar so this was in pure water in pure water so we didn't have any potassium chloride here and notice what the addition of the potassium chloride did to the molar solubility right it decreased the solubility by a factor of 10 right if you compare these values here so the solubility in pure water was Oh point zero one six and now the solubility has decreased it's point zero zero one six alright so one tenth the original solubility and this is due to the presence of a common ion and our common ion was the chloride anion so according to le chatelier's principle right a system that's disturbed from equilibrium will shift its equilibrium position to leave the applied stress so we can use that idea in this situation so if we think about our reaction being an equilibrium and you increase the concentration of your product here if you increase the concentration of your chloride anion is right that's our stress our stresses increase concentration of a product the equilibrium shifts to decrease the stress so the equilibrium shifts to the left and the excess chloride ions that we have combined with lead two plus ions to form more of our solid therefore therefore we've decreased the solubility of lead two chloride due to the presence of our common ion notice that KSP doesn't change right KSP is still 1.6 times 10 to the negative 5 but the molar solubility has been affected by the presence of our common ion so a common ion decreases the solubility of our slightly soluble compound in this case by a factor of 10 so we can use this concept in a laboratory separation so let's say that we had let's say we had some solid lead two chloride and our goal was to isolate all of the solid all right so before we filtered it we could add a common ion we could add a source of chloride anions right and due to the common ion effect that decreases the solubility of lead two chloride which means which means we're going to get more of our solid because our goal is to isolate as much of our solid as possible so that's one use for the common ion effect in a laboratory separation