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Learn Advanced Placement Chemistry: Solubility Product 2 Video
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 Learn Advanced Placement Chemistry: Solubility Product 2 Video
Learn about Advanced Placement Chemistry, Solubility Product 2, in this comprehensive video by bannanaiscool.
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Rob Lederer: Relative solubility is comparing the Ksp to compounds that are relatively the same, and they have to be in order to compare the values, otherwise you have to do some math in order to figure it out. I will show you what I mean. If you got AgCl, NaI, CaCo3 and the Ksp values for each and somebody says, okay who is the most soluble and who is the least soluble, rank them. Well, because each one of these chemicals disassociates into two ions, Ag positive Cl negative, Na positive Iron negative, Ca2 positive Co32 negative because, they form the same number of ions in solution, those ions when multiplied together in terms of their x values will give you the same type of number. Like for instance, this will disassociate into two ions, both of which would have concentrations x so many moles times  get it as x squared. this has the same and this has the same. Because they disassociate to the same amount of ions, you can compare their Ksp values. And all you have to do is see this. The one with higher Ksp value which is this, this number is actually greater than these two right because, it's got a greater value, its x value would be greater which means greater concentrations before precipitations which means more soluble. Here is the question. We have got a step wise formation of this chemical right here and first of all copper ions are reacting with ammonia and actually forming a very weird complex ion. that is complex ion because, certain cations in solution can attract chemicals to themselves called ligands like ammonia, water, Cn negative, Co negative and they form complex ionic structures like this one. Complex ion and ammonia acts as a ligand, okay. So, here is the first step and that ammonia reacts with this chemical of  in the second step and both of them have K values that have been determined. Now, given that what are the equilibrium concentrations of every one of these ions in solution? When you have a mixture of 100 milliliters or 0.2 mole per litre ammonia reacting with 100 milliliters or 0.001 mole per liter copper two nitrate to get the copper one nitrate to get the copper ions in solution, okay. Here is the first thing you have to understand about this. You see these K values, they are very large. When a K value is very large, it's kind of saying to you we are almost a 100% reaction. So if that's the case, these two look like they are both almost a 100% reaction and you know what happens when you have a step wise reaction mechanism in a 100% arrows here. You can add the two equations together, like we are going to do right now. Now here is that second equation in the mechanism, equation number two. We know the concentration now of that chemical equilibrium and this one. But, we don't know what the concentration of this one is in that mechanism step number two. Well, we can write an expression for that, can't we? And since we know the K value 8.2 times 10 to the 3 and we know the other two concentrations at equilibrium, we can find for x ad x is the concentration of that ion which will in the end give 6.1 times 10 to the negative 8. That is the moles per liter, the concentration of that species in solution and we have now found that third one. Now, remember the copper one ion can't be zero. We kind of estimated to be close to zero in order to solve for everybody else. Now, we went backwards and we go to the first equation. Here is that very first equation in the reaction mechanism. We know the concentration of the ammonia at equilibrium, one mole per liter. We know the concentration of this ion, we just found it as we react into this step but it was the product in that previous reaction with that concentration. So we write the expression. We substitute everything we know into the expression and leave x which is the concentration of that chemical Cu positive at equilibrium. When we solve for x, we get 2.9 times 10 to the negative 11 and we just solve for the concentration of the copper ion, the copper 1 am