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Question1ptsAsolutlon is 0.021 COj?- (carbonate) anlons znd 001 MS2- (sulfide) anlons If salid PbINO-lz addedto the solutlon selectivelyprecipitate one of the anion...

Question

Question1ptsAsolutlon is 0.021 COj?- (carbonate) anlons znd 001 MS2- (sulfide) anlons If salid PbINO-lz addedto the solutlon selectivelyprecipitate one of the anions while leaving the other Jnom solution Seley would precipitate first andthereason that 6ect| sulnor Aond CATED demonstration tor the process shoknbelow K 0f PLCO_ Ilead (Il) carbonate) [s 74 * 10 Kipof PbS (lead(II) sulndel is 90PDINO hAschutkni002MCO " lexrbonste} anons and 001MS - [el = AnleteaIt PbINO J;ls addedtothesolution

Question 1pts Asolutlon is 0.021 COj?- (carbonate) anlons znd 001 MS2- (sulfide) anlons If salid PbINO-lz addedto the solutlon selectivelyprecipitate one of the anions while leaving the other Jnom solution Seley would precipitate first andthereason that 6ect| sulnor Aond CATED demonstration tor the process shoknbelow K 0f PLCO_ Ilead (Il) carbonate) [s 74 * 10 Kipof PbS (lead(II) sulndel is 90 PDINO h Aschutkni002MCO " lexrbonste} anons and 001MS - [el = AnleteaIt PbINO J;ls addedtothesolution to selertiet pecotreorect the *tons leaving the other Jion splutkn IScea| precicitate nelanttnanerronkthat Ademonstrationlor thr Anloc enna A Feont nctat Cinatoeneent TTEAatn ( Rennale #t4 Jaa 10-*Kdpus Iler ran icabo eandollraua Qukttkn6 Sold PbINOsh " aoned soluton ma cotans COy" anions. Which bes ortabes Ihc corstion when PbCOspicole0.d 5e sollion? Kap of PbCOs 674 10 # ocpocon 0i Poco- POco UnLco Jceco KenlFnco



Answers

A solution is 0.022 M in Fe2+ and 0.014 M in Mg2+.

a. If potassium carbonate is used to selectively precipitate one of
the cations while leaving the other cation in solution, which
cation precipitates first? What minimum concentration of
K2CO3 will trigger the precipitation of the cation that
precipitates first?

b. What is the remaining concentration of the cation that
precipitates first, when the other cation begins to precipitate?

Okay. So during the experiment one we saw that there was no reaction. Yeah. And then during experiment too, your reaction, is there silver ion Plus Chrome eight Ion. So we'll need two of these Form a precipitate of silver chrome eight. Okay, that's our red precipitate that forms Okay. An experiment three. Again, there was no reaction. Okay, so those things are all soluble. The experiment for. We've got our silver ion reacting with reacting with oxalate ion to make silver oxalate. Okay, that was a white precipitate that we saw in that experiment. So experiment five we've got some calcium ion reacting with some oxalate ion and they're going to form calcium oxalate. You just need your sigh. Ability rules to see where your precipitates are coming from. And this is the white precipitate that was formed there, the calcium oxalate. And then finally in experiments six, our silver ion is reacting with chloride ion to make a white precipitate of silver chloride. Okay, so we've written our net ionic equations and we've identified all of the precipitates and then we want to know which one is more soluble salts of chrome eight or oxalate. So you'll see that there's a number of oxalate precipitates, But only one chrome 8 precipitate. So it seems like that with our limited evidence that Crow mate salts are more soluble than oxalate salts

We're going to look at some double displacement reactions. Okay, so the first one is with sodium chrome eight, Miss smith ammonium C 204 which is known as oxalate, obsolete. Okay. So those who will react together in a double displacement reaction where they basically search cat ions and and ions to then form sodium oxalate. Okay, so as you can see here it has a two plus charge or two minus charge. Sorry? So the sodium, we'll need two of them. Okay then we get ammonium chrome eight. And as you can see, chromite is also die violent. So will be quite similar. Okay, so what we're at now is to decide which one is the precipitate. Okay, So if we look at it here, this reaction had no precipitate. So everything here is Equus and this was the only mixture that no precipitates. Now let's move on to the next one. It's again with the sodium crewmates and we'll be reactively reacting it with silver nitrate. Okay, what will form here again is double displacement switching of Catalans and and ions to get sodium nitrate with silver chrome eight. Okay. And what we see here is a red precipitate. So we know all nitrates are soluble. But so we would suggest that the silver crow mates. This is a solid. Okay. And silent. Put sub scripts in front of the other ones. That just means they're acquis. Everything else is Aquarius. Next up will be sodium chromium again and this time it will be with calcium chloride. Okay. And double displacement again to make some salt which is definitely soluble. This is the sodium chloride used to season food and then we have some calcium chromite. Okay, So then we know that this one is a solid here because we know that you know chlorides are soluble. Next up it will be B and C. So we have ammonia oxalate with silver nitrate. Okay, so we know the old form. Another nitrate which will definitely be soluble in this case. That's ammonium nitrate. Okay then the silver crow mates, as we've seen. Even in um example too is not that's soluble so it's just a solid. Again, this is a silver oxalate. Next up will be be in D. So we'll take this ammonium oxalate with calcium chloride. And what we get again, we'll be ammonium chloride, which we know will be valuable cards are soluble. Then we have calcium oxalate which must then be the solid. Okay then the last one is C. And D. So silver nitrate with calcium. All right. What will form is a G. Is a silver chloride. It's a calcium nitrate. So, that's when it's If the trick is here Well, we know nitrates are actually more soluble than quad. So required. Here is the solid. All right, So we just used the rules for the comment that we didn't know would be would be um soluble, which would be the nitrates and most of the chlorides, all the nitrates and most of the clients. So if we had to choose between a cooler and nitrate acquired would be be solved and the solid one.

On this problem. For the first one there's no precipitate. So there's no net ionic because all the items are going to cancel. So for the next one There is a precipitate. Nitrous are always soluble. So the precipitate must be the silver chrome eight. So basically to write the net ionic, you're just going to write two moles of silver because there's a sub script of two on the silver added to one mole of chrome eight because there's no self grip on the chrome eight, we get silver chrome eight solid. So that's basically what the net ionic shows is just showing the ions turning into the precipitate. So for the next one and A C. L. Is table salt we know it's soluble. So therefore this must be the precipitate. So we're going to write the non ionic for that For the next one, the nitrate is always soluble. So therefore this has to be the precipitate. And we'll have to write an ionic for that one and just be sure you're getting the correct charge too on this. So this would be the charges on these ions and these will just be something you would have to memorize. So be sure that you know the charges on the poly atomic ions and the common ions. So for this one the ammonium chloride is soluble because that's also assault. So this one has to precipitate. So we're going to just write none on it for that For the next one the nitrate is soluble. So we know this is a precipitate and we're going to write an ionic for this one we can get rid of the twos in front of everything because we're just dividing everything by two, and we'll just get that.

Okay, So in this video, we're answering question 108 from Dr 17 which tells us we have a solution that is 170.2 to Mueller and F people. It's too ions and 0.1 for Mueller and MG plus two ions. In part they were asked if potassium carbonate is used to selectively precipitate one of the cat ions while leaving the other cat eye. And in solution which cat and will precipitate first. What minimum concentration of potassium carbonate will trigger participation? Precipitation. Sorry of the cat eye on the precipitates first, um so the possible products we conform with our carbonate eye on our iron carbonate and magnesium carbonate and to figure out which one will precipitate first, all we need to do is to compare the scalability product. Constance those equilibrium, Constance, that air calculated from our amounts of ions and solution. So for iron carbonate, we have a lower liability product constant than for my museum carbonate, which means that iron carbonate is the least soluble in the most likely of department precipitate. While magnesium carbonate is the most soluble in the least likely to form precipitate eso, we're going to form a precipitate from iron carbonate before magnesium carbonate. And we want to know what concentration of que Tu Sio three will trigger the precipitation of r F E C 03 s. So, basically, if there are more ions that Aaron the solution, then there would be equilibrium than some of the solid reactant will have to form in order to reach equilibrium. And we're going to form a precipitate. So basically, when our reaction question Q is greater than our Selya bility product, Constant K um, so are our reaction quotient. Q is equal to our iron ions. Times are carbonate I arms and we know that that needs to be greater than our KSB. So basically, we need our concentration of carbonate ions to be greater than the K S p divided by our concentration of iron ions which were given. So we have 3.7 times 10 to the negative 11 for our ke sp divided by 110.2 To moller, that gives us 1.4 times 10 to the negative nine Moeller That's the concentration of potassium carbonate we need in order to initiate this precipitation of iron carbonate out of solution um, Andy were as what is the remaining concentration of the cat eye on the precipitates first, when the other cat eye on begins to precipitate. So what's the remaining concentration of iron ions wants? Magnesium carbonate begins to precipitate. So when is the other cat eye on going to start precipitating on? When are we going to start forming magnesium carbonate? We need to check when Q is greater than K for magnesium carbonate. So Q is just equal to our concentration of my museum ions, times their concentration of carbonate ions. And that needs to be greater than our K S P, which is 6.82 times 10 to the negative six for my museum carbonate. Um, so are carbonate ions. I need to be greater than 6.82 times in the neck of six, divided by our point No. One for Moller from our concentration of my museum ions, that gives us 4.87 times 10 to the negative for Moeller. So the magnesium I owns air going to start precipitating when our carbonate concentration is 4.87 times 10 to the negative for Mueller. Um, so what is our iron concentration when we have that concentration of carbonate ions are so basically all we need to do is plug that carbonate concentration and the K s P of iron carbonate into, um into our equilibrium expression. And what we end up with is 6.3 times 10 to the negative eight Moeller for our iron ion concentration.


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