Aamc cbt 3 ps 4

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

sudo

Señor Member
10+ Year Member
Joined
Sep 12, 2011
Messages
431
Reaction score
34
I hope I don't get in trouble for posting the contents here (CBT 3 is free anyway, so I didn't think it'd be a big issue):

Passage:
LTcKc.png


Question/Answer:
EOYwe.png


I have NO idea how this is the answer. The explanation (which I didn't include) doesn't even make sense to me. How did the rest of you approach this problem?
 
You have to look at the experiments. The more soluble ion is displaced first and dissolves into solution. Since by looking at the experiments we see that ultimately CO3 displaces the other ions, it is the least soluble and will be the first to precipitate because it does not want to be displaced from the compound and dissolve into the solution. In experiment one the insoluble SO4 displaces the soluble NO3 so we have PbSO4 precipitate, but then in expirement 2, I replaces the SO4 so we have PbI precipitate, and finally the CO3 replaces the I making it the most insoluble because it displaces the rest and refuses to dissolve in the solution. Most insoluble= first to precipitate.
 
You have to look at the experiments. The more soluble ion is displaced first and dissolves into solution. Since by looking at the experiments we see that ultimately CO3 displaces the other ions, it is the least soluble and will be the first to precipitate because it does not want to be displaced from the compound and dissolve into the solution. In experiment one the insoluble SO4 displaces the soluble NO3 so we have PbSO4 precipitate, but then in expirement 2, I replaces the SO4 so we have PbI precipitate, and finally the CO3 replaces the I making it the most insoluble because it displaces the rest and refuses to dissolve in the solution. Most insoluble= first to precipitate.

Makes sense, thank you 🙂
 
Top