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Solubility from the solubility product constant

Video transcript

our goal is to calculate the solubility of copper two hydroxide and we're given the solubility product constant KSP which is equal to two point two times ten to the negative twenty at 25 degrees Celsius so let's conceptualize this problem first let's say we have some copper two hydroxide which is blue so let's say we put some copper two hydroxide some solid copper two hydroxide into a beaker containing water this is a slightly soluble ionic compound so not everything we put into the beaker is going to dissolve let's say only a small portion of this dissolves and take my eraser here I'm going to take off a small bit of our solid there at the top and let's say that small amount turns into ions so what ions would we have in solution copper 2 tells us we have copper 2 plus ions in solution so Cu 2 plus and then we also have hydroxide ions in solution o H minus so we'd have some hydroxide ions in solution to eventually we reach equilibrium right so we have a solubility equilibrium where the rate of dissolution is equal to the rate of precipitation so let's go ahead and write that out what's the chemical formula for a copper two hydroxide we could use this simple little trick here of crossing over of crossing over your charges to figure out the chemical formula is Cu with parenthesis o H and a 2 here all right so that's our solid and we also have our ions so Cu 2 plus or ions in solution we also have hydroxide ions in solution o h minus we need to balance this so we need a 2 in front of the hydroxide ions and everything else here would get a 1 all right so let's set up a nice table so we have our initial concentration our change and then finally our concentration at equilibrium well before that small amount of copper two hydroxide dissolves right so that small amount that I erased earlier we didn't have anything for the concentration of our ions in solution so that's our initial concentration of our ions zero now let's think about this all amount of copper two hydroxide the solid that dissolved all right so let's say that X is equal to the concentration of copper two hydroxide that dissolves all right so we're going to lose a concentration of copper two hydroxide which we'll say is X look at your mole ratios for every one mole of copper two hydroxide that dissolves we get one mole of copper two plus ions in solution so for losing if we're losing X for the concentration of copper two hydroxide we're going to gain X for the concentration of copper two plus ions in solution and for hydroxide ions for every one mole of copper two hydroxide that dissolves we get two moles of hydroxide ions all right so instead of X would be 2x all right so we're going to gain two X for the concentration of hydroxide ions so at equilibrium right our equilibrium concentrations of our ions would be X for copper two-plus and 2x for hydroxide alright let's write our equilibrium expression right so KSP KSP is equal to we look at our products all right so we have Cu two plus we put the concentration of Cu two plus and we raise the concentration to the power of the coefficient and here our coefficient is a 1 so we raise this to the first power next our other product here will be the hydroxide ion so o- and we raise that concentration to the power of the coefficient which in this case is a 2 so we need to put a 2 here and once again we leave this pure solid out of our equilibrium expression alright let's let's plug in for KSP write the solubility product constant was given to us it's 2.2 times 10 to the negative 20 so let's plug that in so this is equal to 2 point 2 times 10 to the negative 20 and this is equal to the concentration of copper two-plus ions at equilibrium which is X so we put that in this is X to the first power times the concentration of hydroxide ions at equilibrium raised to the second power so this would be 2x and then we have 2 where it here and this is where some students get a little bit confused because they say well you're doubling the concentration here right and then you're squaring it aren't you kind of like doing the same thing twice but remember these are two different things right this 2x is because of the mole ratios all right and we raise it to the power of the coefficient because that's what you do in an equilibrium expression all right so those are two different things we're not doing the same thing twice all right when we do our algebra right on the right side we would have x times 4 x squared so that's equal to 4 X cubed and this is equal to 2 point 2 times 10 to the negative 20 so we need to divide that by 4 all right so we need to divide 2.2 times 10 to the negative 20 by 4 so you could you could do that in your head or on the calculator 2.2 times 10 to the negative 20 all right we divide that by 4 and we get five point five times ten to the negative 21st so we have five point five times ten to the negative 21st is equal to X cubed all right so to solve for X we need to take the cube root of five point five times ten to the negative twenty first and unfortunately on this calculator it's a little bit it's a little bit trickier than on most calculators and most calculators it's pretty straightforward it's pretty easy to do so let me show you one way to take the cube root of something on this ti-85 here so we would put in we take the cube root so we put a three in here and then one way to one way to find this is to go to second catalog and then just and then just move upwards here until you see the symbol all right so I don't see it yet and there it is so right there it's a symbol we want right so we're trying to take the cube root of we want five point five times ten to the negative twenty first and that should give us the cube root which is equal to let's go ahead and round that to one point eight times ten to the negative seven so this is equal to X is equal to one point H times 10 to the negative 7 and this would be this would be the concentration right this would be molar this would be the concentration of copper two-plus at equilibrium right let's go back up here so X is equal to the concentration of copper two-plus at equilibrium and notice it's also equal to the molar solubility of copper two hydroxide right that's how much copper two hydroxide dissolved X so we found the molar solubility of copper two hydroxide our question asked us for solubility so maybe they meant molar solubility in which case we're done or maybe they meant solubility in grams per liter so let's go ahead and do that now so this is equal to alright this is equal to the molar solubility this is the molar solubility molar solubility which is moles over liters what if they wanted grams over liters right you would need to have the molar mass of copper two hydroxide alright so why so you could look that up on your periodic table so copper two hydroxide has a molar mass of ninety seven point five seven grams per mole so our answer here for molar solubility this would be moles be moles per liter so if we want to go to solubility in grams per liter let's look at our units and see we would have to do we have 1.8 times ten to the negative seven moles over liters and look at the molar mass if you want to get to grams over liters all we have to do is multiply the molar solubility by the molar mass because the units for the molar mass are grams over moles and if we multiply the moles cancel right and we're end up with grams over liters so let's go ahead and do that calculation here so we have we rounded this to one point eight times ten to the negative seven what's the molar solubility to get to the solubility in grams per liter we multiply that by ninety seven point five seven which is the molar mass of copper two hydroxide and we get we get if we round that 1.8 times 10 to the negative 5 all right so this is equal to 1.8 times 10 to the negative 5 and this would be grams over liters so this is the solubility right in one liter of solution you could only dissolve 1.8 times 10 to the negative 5 grams so copper two hydroxide is not very soluble at all alright so that's how to figure out the solubility if you're given the solubility product constant KSP