Q vs Ksp

Jules Bruno
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So in this example, it says two flasks containing 7.12 times 10 to negative six Mohler silver nitrate and 8.33 times 10 to the negative for Mueller. Strontium bromide respectively, are mixed together into a larger flask. If the K SP of silver bromide is 5.35 times 10 to the negative 13 what will happen? All right, so they're giving us the K sp of solar bromide. So that's what we're concerned with. And we're gonna say here that it's a solid because we're dealing with PSP, which deals with Ionic solids. We're going to see it breaks up into Silver Ion and Bromide Island. Remember, all those silver's a transition metal, and they usually have more than one charge. Slover does it. Silver is always plus one. Remember, in a nice chart, we ignore solids and liquids. So this silver bromide, which is a solid we're going to ignore it now realize that this is not being created in the solution. That's pure water. What's happening here is we have a bucket, and in this bucket we dumped silver nitrate and strong team bromide, and it's the Silver Ion and the bromide ion connecting with each other to help make possibly make a precipitate. All right, so that means that we're not dealing with pure water. So we're gonna be dealing with double Kalmunai in effect. So again, this is a Q versus K sp question. So we have silver here and bromide here. So what's the difference between this one and the previous one? In this one? We're dealing with polarities and the previous one We're dealing with leaders and polarity which forced us to multiply to find moles, then divide by the total volume here. All we do now since they give us only polarity, is we're gonna say there's only one silver ion in this compound and its concentration would be that number. Okay. And we say here now there are two bromide ions here and because there's two bromide ions would actually have to multiply in this concentration times to to get the concentration of bromide islands. So here this would be 1.67 times 10 to the negative three Moeller. So we have initial concentrations for both islands, so they're both products or making them plus X and plus X bring down everything and remember, just like previous questions. Anytime we have a real number in front of our X variable, that means I can ignore my ex variable. That's because K sp such a small number that X will be very insignificant. The ants will be so small for X it's not gonna really affect my final equilibrium concentration. All right, so now, because we have initial amounts for both were actually looking up cute, not k SP. So we're gonna plug those numbers in and figure out what Q is. So we do that we get as Q 1.19 times 10 to the negative eight. We're gonna do a number line K S P is where we wanna be. Equilibrium way. See that he was a much larger number because it's magnitude is 10 to the negative eight. So which way does you have to shift to get to K and has to shift in the reverse direction? If it shifts to the reverse direction here, that means it will shift in the reverse direction for equation, meaning that this would be increasing and this side will be decreasing. So we're making a precipitate of silver bromide, so precipitate of strong term nitrate. No. Ah solution. Saturate but remain clear. No, the solution is not saturated. No precipitate will form. No precipitate does form. So it's gonna be there either D or E. Now, what we need to realize here is anytime Q is greater than ks P. That means that we're dealing with what's called a super saturated solution. I mean that our solution have dissolved beyond its limit beyond its Max. This could happen if we add heat or increase the pressure. But you can only hold this supersaturated solution state for a small amount of time. Eventually, your be unable to continue dissolving this beyond maximum amount of salute. And then the excess saw you that you were able to dissolve. We're no longer stay dissolved. It will precipitate down. Okay, so eventually gonna make solid, solid serve Rome. I but not all the silver ions and not all the bromide ions in the solution will form a solid will be still portions of them free floating around the solid. So that means that D would be the best answer. You will make us precipitate of silver bromide, but there will be some silver and bromide ions that are still free floating around. They're not all going to combine together to make a solid Okay, So really, what this question is talking about is talking about the transition from a supersaturated solution state to just a saturated state. What, you've dissolved the maximum amount off salute. But any excess salud has precipitated down toe a solid at the bottom of your flask. So that's why Option D is better than Option E. Because although it precipitate does form, it doesn't talk about what happens to the other ions hanging around.