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ემზადებით გამოცდისთვის? მოემზადეთ ამ 3 გაკვეთილის დახმარებით შემდეგ თემაზე: Buffers, titrations, and solubility equilibria
იხილეთ 3 გაკვეთილი
ვიდეოს აღწერა
- [Voiceover] There are many ionic compounds that are only slightly soluble in water. And lead two chloride is one of those ionic compounds. So let's say that we added 10 grams of lead two chloride to 50 milliliters of water at a temperature of 25 degrees Celsius. And let's say that only .22 grams of the lead two chloride dissolve. All right, so most of the lead two chloride is undissolved and we can go ahead and show that in our beaker over here. So let's say that this represents our undissolved lead two chloride, right? So this is a solid in our beaker. Some of the lead two chloride does dissolve, a small amount. Only .22 grams, so we're going to have some ions in solution. We're going to have lead two plus ions in solution, PB two plus and chloride anions, Cl minus. So this is just a little bit of a picture of what's happening, right? We now have a saturated solution of lead two chloride. We have a saturated solution. So we have ions in solution and we have a large amount of undissolved lead two chlorides. All right, in part A, our job is to calculate the solubility of lead two chloride in water at 25 degrees Celsius. And first, we're going to find the solubility in grams per liter. All right, so in grams per liter, .22 grams dissolved. So .22 grams dissolved in 50 milliliters of water. If we move our decimal place one, two, three that's .05 liters. So this is .05 liters. .22 divide by .05 is 4.4 grams per liter. So that is the solubility. And the solubility is the number of grams of solute in one liter of a saturated solution. So if you had one liter of water, you could only dissolve about 4.4 grams of lead two chloride in that one liter of solution. So this is why this is only a slightly soluble ionic compound. A small amount dissolves. All right, we could also find the solubility in moles per liter, which would be the molar solubility. So we know grams, we know the grams is .22 grams. So we have .22 grams here. To find moles, we need to know the molar mass. So for lead two chloride we have lead with a molar mass of 207.2. So we have 207.2. To that we'd have to add two times 35.45 because we have two chlorines, right, PbCl2. So two times 35.45. Two times 35.45 is 70.9. If we add that to 207.2, we get 278.1 grams per moles. That's the molar mass of PbCl2. So if we divide the grams by the grams per mole, we divide grams by 278.1 grams per mole, our grams would cancel. We'd get one over one over moles. So this would tell us how many moles. So let's get out the calculator and let's do this. So we have .22 grams divided by 278.1 and that gives us .00079 moles. So if I round that, we would get .00079 moles. We're trying to find the molar solubility, so we need to divide moles by liters. And we already saw the liters was .05, so we need to divide that by .05 liters. We'll go ahead and use the rounded number here. So .00079 divided by .05 gives us a molar solubility of .0158, which I'm going to round to .016. So this is equal to .016 and this would be molar, right? Moles over liters is molarity. So this is the molar solubility of lead two chloride in water at 25 degrees Celsius. You have to make sure to specify the temperature because, obviously, if you change the temperature, you change how much can dissolve in the water. All right, so that's the idea of solubility and molar solubility. In part B our goal is to calculate the solubility product constant, Ksp, at 25 degrees Celsius for lead two chloride. Ksp is really just an equilibrium constant. So let's think about a solubility equilibrium. Let's think about this picture right up here. So we have a saturated solution of lead two chloride and our solution is in contact with our solid, lead two chloride, here. And at equilibrium the rate of dissolution is equal to the rate of precipitation. So the rate at which the solid turns into ions is the same as the rate in which the ions turn back into the solid. So let's go ahead and represent that here. PbCl2, lead two chloride is our solid. And our ions are Pb two plus in solution and Cl minus. We need to balance this, so we need a two here in front of our chloride anion, and everything else would get a one. So if we're trying to find our equilibrium constant, Ksp, we need to start with an ice table. So we're going to start with an initial concentration. So an initial concentration, then we need to think about the change, and finally, we can find equilibrium concentrations. So let's pretend like nothing has dissolved yet. So let's pretend like we haven't made our solution, our saturated solution yet. So our initial concentrations would be zero for our products. All right, next, we need to think about how much of our lead two chloride dissolves. All right, so we did that in part A. .00079 moles of lead two chloride dissolved in our water and our molar solubility was, therefore, .016 molar. So that's the concentration of lead two chloride that we're going to lose here. So we're going to lose .016 molar concentration of lead two chlorides. And we're going to assume that all of the lead two chloride that dissolved dissociates completely into ions. So for every one mole of lead two chloride that dissolves we get one mole of lead two plus ions in solution. So if we're losing .016 for the concentration of lead two chloride, we're gaining .016 for the concentration of lead two plus. And for the chloride anion, this time our mole ratio was one to two, so we need to multiply this number by two. So .016 times two is equal to .032. So we're gaining .032 molar for the concentration of chloride anions when the PbCl2 that's soluble in water dissolves. So therefore, at equilibrium we should have a concentration of lead two plus ions is .016 molar and our concentration of chloride ions should be .032 molar. And now we are ready to write our equilibrium expression. So we write K, and since this is a solubility equilibrium, we're going to write Ksp. So Ksp is equal to-- remember concentration of products over reactants and for these we also need to think about the coefficients. So let's think about our products first. Pb two plus, so we have the concentration of Pb two plus and we're going to raise the concentration to the power of the coefficient and here our coefficient is a one. So we're going to raise this to the first power. Then we're going to multiply this by the concentration of chloride anions, so Cl minus, and then we're going to raise the concentration to the power of the coefficient. So here our coefficient is a two, so we're going to raise this to the second power. All right, this is all over the concentration of your reactants, but here we have a pure solid. Remember, we leave out pure liquids and pure solids out of equilibrium expressions, so this is our equilibrium expression. The solubility product constant, Ksp, is equal to the concentration of lead two plus ions to the first power times the concentration of chloride anions to the second power. And so now we can solve for Ksp because we know the equilibrium concentrations of our ions. We can plug these numbers in. So it was .016 for lead two plus and it was .032 for Cl minus. So we can now solve for Ksp. Ksp is equal to .016 to the first power times .032 to the second power. We can take out the calculator and go ahead and do this. .032 squared times .016 gives us 1.6 times 10 to the negative five. So this is equal to 1.6 times 10 to the negative five. So that is the solubility product constant, Ksp, of lead two chloride at 25 degrees Celsius. Now, I should say I have seen different values for Ksp, for lead two chloride at this temperature. And so you might see a different one if you're looking in a different textbook, but to me that's not the important thing. To me the important thing is understanding how to calculate Ksp by thinking about-- by writing an equilibrium expression and thinking about the equilibrium concentrations of your products, of your ions. And finally, let's just talk about what would happen if we had tried to dissolve 100 grams instead of our original 10 grams of lead two chloride in the same volume of water and at the same temperature. Let's go all the way back up to the beginning. So we try to do 100 grams instead of 10 this time. So all the way back up to here. All right, so instead of 10, let's talk about 100. Well, still only .22 grams would dissolve in our 50 milliliters of water. So we would have a bigger pile of undissolved lead two chloride, but since we're still only able to dissolve .22 grams the molar solubility would stay the same. And if the molar solubility stays the same, that means that our equilibrium concentrations would stay the same and so Ksp is the exact same value. And so, hopefully, that helps you understand that it's really-- it's the concentration of the PbCl2 that dissolves that determines your Ksp. It's not the undissolved part.