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Learn Advanced Placement Chemistry: Solubility Product 1 Video
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 Learn Advanced Placement Chemistry: Solubility Product 1 Video
Learn about Advanced Placement Chemistry, Solubility Product 1, in this comprehensive video by bannanaiscool.
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Male Speaker: Here is our old friend copper 2 sulfate being dissolved in solution, okay make it a nice pretty blue solution out of this we put in some more copper 2 sulfate and then we put it that's the more cooper 2 sulphate you know that the solution is unsaturated until that time where we can dissolve, no more crystals at the bottom, they all entered in to solution as much as they are going to and we have a saturated solution, but here something else right. That we know that those crystals down below are not just sitting there but some are dissolving at the same rate that the microscopic level that, some of them are re crystalizing. So what we've got there is equilibrium, now if you said equilibrium, that means then that this concentration of copper ions and sulphate ions remaining constant over time, we can actually plug those in to an equilibrium expression and come up with a constant that will describe this solution at a given temperature. So an equilibrium a product can actually been determined. So how do we do that, well just pick any kind of crystal like say calcium chloride solid, and you'll always write the solid first, when you do these solubility type equations and the reason is, is because we want to make sure that the, number we get in the end are when we do the equilibrium expression is the concentration of products over, the number 1, so when we write the equation for the dissociation of Calcium Chloride in two times, there is Calcium 2 positive and Chloride its a two negatives, one negative charges so it has to be two of those, when you write the equilibrium expressions the concentration in this time this divided by oh that's a solid so we just put 1, see, and so K, is going to equal the concentration in Ca2 positive times concentration of Cl negative() squared right. And that equals and value called the KSP, the solubility product constant. So here's what we know, we know them the KSP or what these concentrations in solution, reaches certain value here, that then that is the point of saturation, so that's also the plus point where it just starts the form of precipitate so if we alter certain concentrations, we can still determinate if you know the KSP value when they are going to precipitate, or that's very cool. Here is an example, somebody says okay here is a compound Copper 2 bromide and we are going to take that Copper 2 bromide with it's KSPO 1.5 times ten to negative seven and that's at 25 degree celsius, when somebody says, what concentration is that relatively I mean what's the concentration of that chemical and solution before it starts to precipitate, at that temperature. So what you do is you just take that equation, there always putting the salt, first break it down to its ions and remember this doesn't have a concentration it's a salt, but we can say that it will dissociate initially in to nothing but eventually it will change remember the ICE parts, ICE or Initial Change Equilibrium, it will change in to Xconcentration here in one to one ratio, but 2x here, so at equilibrium we have an X here and 2x here as concentrations. So we know then  then if we take those and put them in to the expression and here is the expression remember its concentration this time this square. Put those numbers in, we will be able to do the math knowing the KSP goes in for here, and we'll get X as 0.0033 moles per liter. What is that, okay its this concentration here but its also we going to  we see, that's actually the concentration of this chemical X and then X would be here, the concentration of here, X will be the concentration of this chemical at the point of saturation, so this is the concentration which you could have before precipitation occurs of this chemical, if you have more than this concentration your X values going to be higher and you are going to get a Q value not a K but a Q that is going to be greater than this, and when you've got in this case, Q greater than K you are going to have a shift