Solubility and Net Ionic Equations for Chlorides and Iodides

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The discussion centers on the net ionic equation for the reaction between sodium iodide and chlorine gas, where participants debate the correct representation of the ions involved. The initial net ionic equation proposed is questioned, with clarification that chlorine gas (Cl2) is not aqueous and thus does not cancel out with the chloride ions. It is emphasized that net ionic equations should only include ions that change states, with spectator ions excluded. Participants agree that the correct net ionic equation reflects the formation of solid iodine (I2) from iodide ions, while sodium ions remain unchanged. The conversation highlights the importance of understanding solubility rules and the role of spectator ions in chemical reactions.
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Ok here is the problem:
2NaI (aq) + Cl2 (g) ----- 2NaCl (aq) + I2 (s)

For the net ionic I got:
2I- (aq) + Cl2 (g) ------- 2Cl- (aq) + I2 (s)


Is this correct? Most chlorides and iodides are soluble, but one states (g) and the other states (s) so does that mean they are not soluble?
 
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Im pretty sure that it is just 2Na + Cl_2 ------> 2NaCl because the Iodine ions are the spectators.
 
Please someone can you clairify this problem for me? I don't think Stevedye56 is correct on this?
 
You are right, Stevedye56 is wrong. Note that while most chloriDes are soluble, chloriNe is not (well, it is soluble to some extent, but you may assume here it is not). Same goes for iodiNe and iodiDes.
 
Borek said:
You are right, Stevedye56 is wrong. Note that while most chloriDes are soluble, chloriNe is not (well, it is soluble to some extent, but you may assume here it is not). Same goes for iodiNe and iodiDes.

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Ah sorry about that, for some reason the past week i have not been reading things right even for simple problems in chemistry. Sorry again about any confusion i may have caused.
 
Net ionic equations only involve ions that are (aq) on one side of the equation and a different form on the other side, i.e. (s),(l),(g). For example, your Na+ is dissolved on both sides. It never comes out of solution, so it just stays as ionic Na+.

However, this doesn't mean that nothing has changed about the Na+. If you filtered out the I2(s) and then distilled the salt, you would be left with NaCl, not NaI.
 
dtx said:
However, this doesn't mean that nothing has changed about the Na+. If you filtered out the I2(s) and then distilled the salt, you would be left with NaCl, not NaI.

How does it change Na+? :wink:
 
ldixon said:
Ok here is the problem:
2NaI (aq) + Cl2 (g) ----- 2NaCl (aq) + I2 (s)

For the net ionic I got:
2I- (aq) + Cl2 (g) ------- 2Cl- (aq) + I2 (s)


Is this correct? Most chlorides and iodides are soluble, but one states (g) and the other states (s) so does that mean they are not soluble?

answer should be

2I- --- 2I (s) because the Chloride cancel out (they're spectator ions)

correct me if I am wrong
 
cheechnchong said:
answer should be

2I- --- 2I (s) because the Chloride cancel out (they're spectator ions)

correct me if I am wrong

Dont you mean I_2. They both would cancel it comes down to the solubility rules. The higher one always prevails
 
  • #10
Stevedye56 said:
Dont you mean I_2. They both would cancel it comes down to the solubility rules. The higher one always prevails

my bad haha i didn't notice typing that out :redface:
 
  • #11
cheechnchong said:
my bad haha i didn't notice typing that out :redface:


Didnt mean any offense i just wasnt sure if you mistyped o:) :wink:
 
  • #12
Stevedye56 said:
Didnt mean any offense i just wasnt sure if you mistyped o:) :wink:

it's no problem...exact answers are key haha
 
  • #13
cheechnchong said:
it's no problem...exact answers are key haha

Yes, the definitely are in Chemistry.:smile:
 
  • #14
Actually according to my chemistry professor My answer was correct, becasue the Cl2(g) is not aqueous so the chloides do not cancel out. Thanks Borek for the correct answer.
 
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