Solubility and Net Ionic Equations for Chlorides and Iodides

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Discussion Overview

The discussion revolves around the net ionic equations for the reaction between sodium iodide (NaI) and chlorine gas (Cl2), focusing on the solubility of chlorides and iodides. Participants explore the correct representation of the net ionic equation and the role of spectator ions in the reaction.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes the net ionic equation as 2I- (aq) + Cl2 (g) = 2Cl- (aq) + I2 (s), questioning the solubility implications of the states of matter.
  • Another participant suggests that the reaction simplifies to 2Na + Cl2 = 2NaCl, arguing that iodide ions are spectator ions.
  • A participant expresses uncertainty about the correctness of the previous claim, seeking clarification.
  • Some participants assert that while most chlorides are soluble, chlorine gas is not, which may affect the interpretation of the reaction.
  • One participant emphasizes that net ionic equations should only include ions that change state, noting that Na+ remains in solution and does not participate in the net ionic equation.
  • Another participant suggests that the net ionic equation should ultimately reflect the formation of solid iodine (I2) from iodide ions, while also discussing the cancellation of spectator ions.
  • There is a correction regarding the notation of iodine, with participants clarifying the correct representation as I2.
  • A later reply claims that the initial interpretation was correct, asserting that Cl2(g) does not cancel out the chlorides because it is not aqueous.

Areas of Agreement / Disagreement

Participants express differing views on the correct net ionic equation and the role of spectator ions, indicating that the discussion remains unresolved with multiple competing interpretations.

Contextual Notes

Participants highlight the importance of solubility rules and the states of matter in determining the net ionic equation, but there are unresolved assumptions regarding the interpretation of spectator ions and the implications of the reaction conditions.

ldixon
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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.

Borek
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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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