If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:11:45

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 the only point to two 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 this is a solid in our beaker some of the lead two chloride does dissolve a small amount only 0.22 grams so we're going to have some ions in solution we're going to have lead two plus ions in solution PB 2 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 right we have a saturated we have a saturated solution so we have ions in solution and we have a large amount of undissolved lead two chloride alright in Part A our job is to calculate the solubility of lead two chloride in water at 25 degrees Celsius at first we're going to find the solubility in grams per liter all right so in grams per liter 0.2 two grams dissolved so 0.2 two grams dissolved in 50 milliliters of water if we move our decimal place one two three that's point zero five liters so this is point zero five liters 0.2 two divided by point zero five is four point four 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 four point four grams of lead two chloride in that one liter of solution so this is why this is why is only a slightly soluble ionic compound a small amount dissolves 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 0.2 - grams so you point to two grams here to find moles we need to know the molar mass so for lead two chloride right we have lead with a molar mass of 207 point two so we have two hundred and seven point two two that we'd have to add two times thirty five point four five because we have two chlorines right P bcl-2 so two times thirty five point four five two times thirty five point four five is 70 point nine alright if we add that to two hundred and seven point two we get two hundred and seventy eight point one grams per moles that's the molar mass of P bcl-2 so if we divide the grams by the grams per mole if we divide grams by two hundred and seventy eight point one grams per mole our grams would cancel we 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 point two two grams divided by two hundred and seventy eight eight point one and that gives us point zero zero zero seven nine moles so if I round that we would get we would get at point zero zero zero seven nine moles we're trying to find the molar solubility so we need to divide moles by liters and we already saw the leaders was point zero five so we need to divide that by point zero five liters all right we'll go ahead and use the use the rounded number here so point zero zero zero seven nine divided by point zero five gives us a molar solubility of point zero one five eight which I'm going to round two point zero one six so this is equal to point zero one six and this would be molar right moles over liters as 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 let's think about 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 PB cl2 led to chloride is our solid and our ions are PB 2 plus in solution and Cl minus we need to balance this alright so we need a 2 here in front of our chloride anion and everything else would get a 1 so if we're trying to find our equilibrium constant KSP we need to start with a nice table all right so we're going to start with an initial concentration all right 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 we haven't made our solution our saturated solution yet so our initial concentrations would be 0 for our products all right next we need to think about how much of our lead two chloride dissolves alright so we did that in Part A right point zero zero zero seven nine moles of lead two chloride dissolved in our water and our molar solubility was therefore point zero six molar so that's the concentration of lead two chloride that we're going to lose here so we're going to lose point zero one six molar right concentration of lead two chloride and we're going to assume that all of the lead two chloride that dissolves 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 for losing 0.016 for the concentration of lead two chloride we're gaining point zero one six for the concentration of lead two plus and for the chloride anion right this time our mole ratio is one to two so we need to multiply this number by two so point zero one six times two is equal to point zero three two so we're gaining or gaining point zero three two molar for the concentration of chloride anions when when the PB cl2 that soluble and water dissolves so therefore at equilibrium we should have a concentration of lead two plus ions is point zero one six molar and our concentration of chloride ions should be point zero three two molar and now we are ready to write our equilibrium expression all right so we write K and since this is a solubility equilibrium we're going to write KSP so k SP 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 2 plus so we have the concentration of PB 2 plus and we're going to raise the concentration to the power of the coefficient and here our coefficient is a 1 so we're going to raise this to the first power and 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 2 so we're going to raise this to the second power all right this is all over the concentration of your reactants but here here we have a pure solid 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 right we can plug these numbers in so it was point zero one six for lead to plus and it was point zero three to four CL minus so we can now solve for KSP KSP is equal to point zero one six to the first power times point zero three two to the second power so we can check out the calculate calculator and go ahead and do this point zero three two squared all right times point zero one six gives us one point six times ten to the negative five so this is equal to one point six times ten 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 of your products of your ions and finally let's just talk about what would happen if we had tried to dissolve what hundred grams instead instead of our original ten grams of lead two chloride in the same volume of water and at the same temperature so let's go all the way back up to the beginning right so we try to do a hundred grams instead of ten this times all the way back up to here all right so instead of a instead of ten let's talk about a hundred well still only point two two grams would dissolve in our 50 milliliters of water so we would have a bigger pile of undissolved led to chloride but since we're still only able to dissolve point to 2 grams right 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 to understand that it's it's really it's the concentration of the P bcl-2 that dissolves that determines your KSP right it's not the undissolved part