all right in this video we’re going to take a look at a review for the solubility unit test and/or for the solubility portion of your final exam we’re going to look at a handful of questions from the yellow supplementary problem booklet so question number 17 we’ll start with and you’ll notice I’ve made a small change in the question it says what mass of lead iodide pbi – do we need to make 5.00 liters of a saturated solution at 25 degrees Celsius and then it tells us that the KSP is seven point nine times ten to the minus nine at that temperature now if you know what you’re doing then pause the video and go ahead and try that problem for me the plan that I’m foreseeing to find the mass of solid needed to make the saturated solution I’m going to first find the molar saw the ability of that iodide once I know its solubility then I’ll use some you know multipliers with the volume and molar mass to find the answer the question how many grams do I need to make this saturated solution so first of all let’s jump in and find the molar solubility with question 17 the salt would lead iodide so PB I 2 and when you see a KSP problem you see a solubility problem you know that most of the time we write a balanced equation like this with the salt simply dissociating into its two ions in this case PB 2 plus and I minus but because it was PB I – over here we need to write – I – is over here on the right the ions are dissolved in water so they’re equal yes and the solid salt was on the left-hand side of that equation we’re going to set up a little ice table underneath as we do in most equilibrium problems because the PB i2 is a solid we’ll ignore it in the ice table we’ll ask ourselves what are we dissolving this lead iodide in to make the saturated solution well the answer to that is we’re dissolving it in water and the reason it’s important to ask that is because it tells you what to do initially if you’re dissolving in water then initially there’s no lead and there’s no iodide to pure water doesn’t have either one of those as the lead iodide dissolves though the lead concentration will increase and we don’t know how much it increases by so we’re going to use the letter s to represent the concentration that lead increases by but that also reminds us that this will represent the solubility of the salt as well since you get one mole lead for every one mole of lead iodide that dissolves then the concentration of the lead led my leads concentration in moles per liter will be the number of moles per liter of lead iodide that dissolved so in other words lens concentration in the saturated solution is the solubility of the salt as well the iodide concentration also increases but because there’s a – there it will increase by twice as much so two s and at equilibrium we have acid – ass in the saturated solution right the KSP expression the solubility product constant is equal to the concentration of lead times the concentration of iodide squared because the lead iodide was solid we don’t include it in the equilibrium expression the question told us the value for the KSP said it was seven point nine times ten to the minus nine and that’s equal to asks the concentration of lead at at Librium times 2 s squared now when you square 2’s you’ll get 4s squared and then multiplied by s out front you get 4 s cubed now this basic layout works for two different kinds of problems it works for when you’re giving the KSP that you were given here and you’re trying to find with solubility but we could have been given the solubility we might have known what that is and then four times that cubed would give us the KSP so sometimes we’re looking for the KSP in this question S will then equal the KSP value divided by four since it equals four s cubed we’ll divide by four and then to get rid of s cubed will cube root now cube root you can write like

that cube root or we can use brackets and do this seven point nine times ten to the minus nine divided by four and a cube root is the same as 2 raised to the power one-third so now with my calculator let me evaluate that so I’m going to type open a bracket 7.9 times 10 to the power of minus 9/4 close the bracket and now 2 cubed root I need to raise to the power of 1/3 but one-third of the decimal is 0.333 3 repeating so all raised to the power of point 3 3 3 3 3 something like that equals so for me the solubility asks is equal to zero point zero zero 1 3 will keep two significant digits because the KSP had two significant digits seven point nine so point zero zero 1 3 and this is a molar solubility so this is the number of moles per liter of lead iodide which would dissolve in water so that’s why it’s called the molar solubility okay so now the question wants to know if we had to leave or sorry 5 liters 5.00 liters and then we wanted to make of the saturated solution how many grams of lead iodide would we need to answer that we could use one of two approaches we could use the formula C equals n over V followed by the formula moles equals mass or molar mass we’re trying to find the mass trying to find how many grams of lead iodide so to do that we would need to know the molar mass of lead iodide which we can get a periodic table and the moles of lead iodide well how many moles would we need that’s why we use the first formula we know we have 5 liters so the volume is given and we know the solubility is point zero zero one three moles per liter well moles per liter is a unit of molarity which is the unit concentration so where we say see in the formula C equals n over V that can be thought of as your solubility so we know the solubility so the moles of salt would be C times V and then the mass would be the moles times the molar mass and we’d be dot I assistant you prefer to use unit multipliers when I do these problems so I’m going to take my two points or sorry 5.00 liters of solution and I’m going to use two unit multipliers to get the answer as many grams of PB I two are needed the first multiplier will convert my leaders to moles of PB I 2 and the second one will convert the moles to grams of PB I to the first multiplier is my solubility the second multiplier is my molar mass of PB I – so the solubility point 0 0 1 3 moles per liter that was our molar solubility and then the molar mass is looking on a periodic table for a moment 207 point two is lead plus two times one twenty six point nine two iodides 460 one point zero grams per mole so now five liters five times point zero zero one three times four hundred sixty one I need two point nine nine six or three point zero grams a two sig figs again because our solubility had two sig figs in it so there’s the basic question so we needed to first find molar solubility then use you know multipliers or we could have used these two formulas to find out that we need three point zero grams of lead iodide to make the saturated solution let’s look at a question which is similar but which approach is this format from the other perspective question number 25 in your booklet the one on the top of the page here says the solubility of calcium hydroxide CA o h2 is 0.93 grams per liter at some temperature what would be the value of the KSP at that temperature so notice they gave us the solubility now it’s not expressed in moles per

liter it’s expressed in grams per liter so that’s a small problem but then they’ve basically given us the solubility and they want to know what the KSP would be so we’re going to approach it in a very similar way but the first thing I’m going to do is take that solubility that they gave us and convert it from grams per liter to moles per liter we need to have the solubility as a molar solubility so we have 0.93 grams of calcium hydroxide per 1 liter that’s our solubility so let’s convert that to moles per liter so we’ll just get rid of the grams then we’ll switch to moles so grams and moles is again molar mass one mole of calcium hydroxide using a periodic table find the molar mass we have one calcium so forty point oh eight we have two oxygens two times sixteen and we have two hydrogen’s of two times one point one seventy four point one zero grams is the mass of one mole so multiplying like that the grams will cancel and we’re left with moles per liter of calcium hydroxide so 0.93 divided by seventy four point one we get a molar solubility with two sig figs of 0.01 three moles per liter there’s our solubility now let’s go and find the KSP and we’ll do it in a very similar way so we’ll start with the balanced equation calcium hydroxide dissociates to calcium 2 plus of two hydroxide ions this is solid there is no calcium in pure water now we know that pure water does have some hydroxide it has one times 10 to the minus 7 molarity hydroxide that number is so small that we almost always ignore it so we will ignore it here noticing that our solubility is actually very large compared to that and now we would normally we could if we wanted to do exactly what we did before we could say plus acid plus 2 s acid 2 s and then we could say the KSP is equal to the concentration of the calcium times the concentration of the hydroxide squared so it equals asked times 2 s squared therefore just like before it equals 4 times s cubed so that finally since we know that s is point 0 1 3 moles per liter we can just put that in a KSP is equal to 4 times point zero 1 3 cubed and so grab the calculator 4 times point zero 1 3 so the power of 3 is 8 point 8 times 10 to the minus 6 there’s the value for my KSP now there was another way you could have approached this just grab a different color we knew that the solubility was point zero 1 3 so that solubility would have to equal the concentration of calcium because the calcium as we know is a one-to-one ratio with the calcium hydroxide so someone may have chosen instead of putting an S in the ice table here they may have said well we know the concentrate the solubility is 0.01 3 so why don’t we say it equals 0.01 3 and then this would have gained 0.013 this would have gained 2 times 0.01 3 and you would have had 0.026 molarity here now that you know those two numbers you could have just put those numbers into the KSP expression and found two SP that way personally I prefer to do it the way we did first with the solubility letter s and then substituting at the end that way the basic setup of doing the problem is the same when we are given the KSP let be we’re back in number 17 we can set it up in exactly the same way to find the solubility or in number 25 where we take the solubility set up the same way to go find the KSP alright let’s look at one more problem with a very standard question of X it’s a question of will a precipitate form when we mix two solutions this is question number forty and it says will a precipitate form if

you mix twenty five point zero milliliters of a very dilute point zero zero four zero molar solution of sodium fluoride with seventy five point zero milliliters of point zero one six zero molar magnesium nitrate and then as a clue that tells us the KSP of magnesium fluoride and we can see it’s very small so will o precipitate form since the magnesium fluoride KSP is so small really we’re wondering will magnesium fluoride precipitate so because magnesium fluoride is the thing that will precipitate possibly what we’re gonna do first this is basically a three step question what we’re going to do first is find the concentration of the magnesium ion and of the fluoride ion since magnesium fluoride is the thing that may precipitate and to do that we realize that by mixing the two solutions we were actually doing a dilution when you mix two solutions C 1 B 1 has to equal C 2 B 2 the moles before must equal the moles after when you dilute so the concentration after we mix which is C 2 would equal C 1 V 1 divided by V 2 and that’s true for both ions looking back at the question we were told that the volume of naf the volume of fluoride that’s where the fluoride comes from was 25 milliliters so that’s the V 1 and this is a C 1 point zero zero 4 notice that there’s one fluoride in sodium fluoride so if you have point zero zero four molar sodium fluoride you have point zero zero 4 molar fluoride so we know its initial volume we know its initial concentration the V 2 the final volume is the total volume after mixing so 25 plus 75 it’ll be a hundred similarly when we look at the magnesium ions we notice that there’s one magnesium in the formula for magnesium nitrate therefore when we say that there’s a point zero one six molar solution of magnesium nitrate there has to be point zero one six molar magnesium ions as well so that’ll be the initial concentration of magnesium 75 mils the initial volume and the total volume McManus 100 so putting these numbers in the magnesium’s initial concentration was point zero one six molarity we were using 75 milliliters of that and the total volume was a hundred milliliters at the end similarily for fluoride dipped initial concentration was 0.0 0.8 storable volume is still 100 milliliters at the end so the magnesium concentration after we’ve mixed them is point zero one six times 75 over 100 so it’s going to be 3/4 of what it used to be point zero one two molarity similarly the fluoride concentration is gonna be one quarter 25 over 100 is one quarter of what it used to be the point zero zero four times 25 over 100 is point zero zero one molarity okay so there’s the concentration of magnesium and a fluoride now will a precipitate form when we mix these solutions to decide that we’re going to calculate the value of QSP the salts formula is mg f2 so QSP which is the reaction quotient is equal to the concentration of the magnesium ions times the concentration of fluoride squared and we know that this countless expression is based on this reaction right mg 2 plus to floor eyes when we remember from our equilibrium unit that if excu at the value of the reaction quotient Q or in this case Q SP if it’s bigger than the equilibrium constant for the reaction so the Q is bigger than K we know that the reaction shifts backwards that’s because the equilibrium is trying to reduce the value of the cue it’s too big and to do that it’s going to shift backwards to lower the concentrations of those ions notice that in shifting backwards we’d be forming solid solid magnesium fluoride so we would see a precipitate PPT II precipitate if the

reaction shifts backwards and the reaction will shift backwards if Q s P is bigger than KSP so we’re going to calculate Q s P and that’s why we did these two concentrations so magnesium point zero one two molarity and the fluoride point zero zero one zero molarity squared okay and so grabbing a calculator point zero one two times point zero zero one squared is one point two times ten to the minus eight there’s the value of QSP now the question told us the KSP for magnesium fluoride is six point four times ten to the minus nine therefore our QSP is bigger than KSP and therefore the precipitate will form if the QSP were equal to the KSP then the solution would have been perfectly saturated it would have been adequate Librium and there would have not been a precipitate if qsp were less than KSP then the solution would be unsaturated more ions could have been present and then again there would not have been a precipitate so you only see the precipitate if the equilibrium shifts backwards and that would be when Q is larger than K so the basic steps in this problem are three of them the first step was to find the concentrations of the two ions that might precipitate and we knew they were magnesium and fluoride because the question told us the KSP of magnesium fluoride we found those concentrations after diluting because we were mixing two solutions so that we could plug them into the QSP expression calculating QSP our second step that led us to the third and final step which is simply to compare the QSP to the KSP and we found that it’s larger and therefore the precipitate does form so those are three standard questions for solubility equilibria they’re not by any means all that you need to know for this unit but they do form if you really understand as well then they form the heart of this unit you should do relatively well on the unit if you understand be sure that you’re also using your data booklets checklist for the unit in it this is at the beginning of your data but it’s also on your Edmodo page there are a list of things you need to know for solving ability and illiberal I think you’ll agree once we’ve once you look them over that we have done all of those problems in class so good luck in comparing