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Spectator ions

  • Thread starter crybllrd
  • Start date
  • #1
120
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Homework Statement



The spectator ions in the precipitation equation Pb(NO3)2+ 2KI --> PbI2 + 2KNO3 are ________.

Homework Equations





The Attempt at a Solution



I've been working through a lot of practice problems for next week's final exam, and I seem to be struggling with this concept.
I know the spectator ions 'do nothing', so I am thinking that means they are the ones that are unchanged. That logic got me through most of them, but not this last one.
Here is what I have been doing to solve them (which I know does not yield the correct answer):

I added the charges:

Pb+2 (NO3+)2+ 2K+I- --> Pb+2I-2 + 2K+NO-3

and then I note what each one does:

Pb+2 --> Pb+2

(NO3+)2 --> NO-3

2K+ --> 2K+

2I- --> I-2


So from that logic I would assume that

Pb+2 and 2K+

are the spectator ions (because they did not change), which is not an answer choice.
What did I do wrong?
 

Answers and Replies

  • #2
Borek
Mentor
28,303
2,687
It is not that they "do nothing", it is more about not changing during reaction.

Instead of writing substances the way you did it you should write them as dissociated or solids/gases (using (aq), (s) and (g) designators).

PbI2 is solid, so ions that were precipitated are no longer present in the solution.

Why do you think nitrate in lead nitrate is positive, and in potassium nitrate negative? Do you mean that any lead nitrate molecule has a +4 charge?

Also note - when dissociating things like Pb(NO3)2 they don't dissociate into Pb2+ and (NO3)2whatever charge, but into separate ions:

Pb(NO3)2 -> Pb2+ + 2NO3-

Similarly, what you did to lead iodide is wrong. There is no I2- in PbI2, there are two I-.

I am surprised you were able to get correct answers to other problems, as it seems like you misunderstand almost every concept needed to solve this type of the question.
 
  • #3
164
0

Homework Statement



The spectator ions in the precipitation equation Pb(NO3)2+ 2KI --> PbI2 + 2KNO3 are ________.

Homework Equations





The Attempt at a Solution



I've been working through a lot of practice problems for next week's final exam, and I seem to be struggling with this concept.
I know the spectator ions 'do nothing', so I am thinking that means they are the ones that are unchanged. That logic got me through most of them, but not this last one.
Here is what I have been doing to solve them (which I know does not yield the correct answer):

I added the charges:

Pb+2 (NO3+)2+ 2K+I- --> Pb+2I-2 + 2K+NO-3

and then I note what each one does:

Pb+2 --> Pb+2

(NO3+)2 --> NO-3

2K+ --> 2K+

2I- --> I-2


So from that logic I would assume that

Pb+2 and 2K+

are the spectator ions (because they did not change), which is not an answer choice.
What did I do wrong?

Nope. Pb2+ and 2I- combine and form a solid PbI2 precipitate. The spectator ions would be K+ and NO3- since they don't react and remain in the solution as ions. I don't know if they have you memorize the solubility rules, but you have to memorize the solubility rules. XD Without knowing them, you can't really tell what is soluble and what isn't. Also the charges of the ions must be known.
 

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