Double displacement reactions?

  • #1

Homework Statement


Lead (II) nitrate + manganese (IV) iodide -> _______

2. The attempt at a solution
So I think I know how to do this
2 Pb(NO3)2 + MnI4 -> 2 PbI2 + Mn(NO3)4
I'm just a bit confused because with single displacement you have to check the activity series? And so I'm not sure if I need to use the activity series for double displacement too? If not, how does I check to see if it will take place?
Thanks!
 

Answers and Replies

  • #2
Borek
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You are mixing two things. One is classification of the reaction type as a single or double displacement, the other is classification as a redox, or precipitation, or something else. The reaction you wrote is a double displacement, based on the precipitation of an insoluble salt, it has nothing to do with the redox (in which case activity series could play an important role).
 
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  • #3
Merlin3189
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You're absolutely right.

Except that this reaction does not usually happen! What happens is that both salts are dissolved in water, then the solutions are mixed.
So Pb(NO3)2 + Aq = Pb2+(aq) + 2NO32- (aq)
And MNI4 +Aq = Mn4+(aq) + 4 I-(aq)

Now in solution you have a mixture of 4 type of ion (and H+, OH- and maybe others in some cases.) You do not have either the initial salts or the final salts. In many cases these will happily stay like that, but here there is a salt, lead iodide, which has a very low solubility product, so that the lead and iodide ions cannot remain in solution together in large concentrations. Lead Iodide precipitates, leaving mainly manganese 4+ ions and iodide ions in solution with a very small concentration of lead 2+ and nitrate ions.

So again you are absolutely right to wonder how you know whether the reaction occurs. Here, as I see Borek has now commented, it happens because one of the compounds is much less soluble than the others. That is a common reason for such a reaction to happen. If all of the possible salts are soluble, you can cause the reaction to happen to some extent if one salt has lower solubilty, by evaporating the solution to increase the concentration of ions until the lowest solubility salt starts to crystalise out. But you don't get 100% conversion (which you did not quite get with lead iodide either.)

If one of the possible products is volatile (or unstable with volatile products) this can also priomote a reaction. So for example mixing ammonium sulphate solution and sodium carbonate solution, both fairly stable salts, you get a mixture of sodium, ammonium, sulphate and carbonate ions. The sodium and sulphate ions are goinig nowhere, but carbonate and ammonium ions have other reactions with water and the production od CO2 and NH3, both of which can diffuse into the air and be lost from solution. So if you evaporate this solution, you end up with sodium sulphate.

IMO a similar situation exists with the single displacement reactions. For example when zinc metal displaces copper from copper sulphate solution, the sulphate is largely irrelevant. The actual reaction is, again as Borek pointed out, a redox reaction Zn + Cu2+ ⇒ Zn2- + Cu. The zinc is replacing copper, not from copper sulphate, but zinc metal reacting with copper ions in aqueous solution.

So, again IMO, these descriptions as displacement reactions are empirical views of what happens in some cases. They do not speak in any way about the reaction mechanism, which you need to seek through other ideas.
 
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  • #4
Borek
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Except that this reaction does not usually happen!
Define what you mean by "this reaction". Especially as "this reaction" goes perfectly well.

Lead Iodide precipitates, leaving mainly manganese 4+ ions and iodide ions in solution with a very small concentration of lead 2+ and nitrate ions.
I assume it is kind of a typo, but no, iodide doesn't stay in the solution. It precipitates out. It is nitrate that stays in the solution. A lot depends on initial amounts and what is the limiting reagent, but let's not go this way before we get the basics right, let's assume stoichiometric mixture to keep things simple.

But you don't get 100% conversion (which you did not quite get with lead iodide either.)
Technically you never get 100% conversion, but with well chosen salts you can get as close to that as it is practical in chemistry.
 
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  • #5
Merlin3189
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Define what you mean by "this reaction". Especially as "this reaction" goes perfectly well.
Yes, my comment was probably a bit over the top. I've not checked to see whether and under what conditions it might go.
I was intending to say that this standard reaction, as commonly described in text books and lab exercises, does not take place by mixing lead nitrate crystals with potassium iodide crystals. We dissolve them in water and mix the solutions. Here of course lead iodide has such a low solubility ( Ksp =4.41 x 10−9 ) that vitually all the lead and iodide is removed from a stoichiometricly balanced solution.

I understand that we normally accept this equation to represent the usual process and I said that Cheesychese is absolutely right in saying that. Perhaps I was wrong to try to draw a distinction in the way that I did, especially without saying what I just said above.

I assume it is kind of a typo, but no, iodide doesn't stay in the solution. It precipitates out. It is nitrate that stays in the solution. A lot depends on initial amounts and what is the limiting reagent, but let's not go this way before we get the basics right, let's assume stoichiometric mixture to keep things simple.
Again you're right. The amount of Pb2+ and I- left in solution is tiny - negligible. But it is there.(To measure it, as I inadvertently discovered when looking up the SP, they start by dissolving PbI2 to make their solution.)
Nitpicking, irrelevant detail I guess you'd say. But lead iodide is not outstanding in the insolubility stakes, so where do you draw the line? There's no lead left in the solution, but there's 100x more than there would be barium in the sulphate test
I think of gettingn KNO3 by mixing hot NaNO3 and KCl solutions, and cooling in an ice bath. Is that a double displacement, even though there's a substantial amount of K+ and NO3- left in the solution?

The point I was trying to make, maybe foolishly (and maybe wrongly) because of the bee in my bonnet, was that mixing solutions of ionic salts is not generally a reaction. It's just putting those ions into the solution. What happens next, if anything, depends on what the ions get up to, not where they came from. It seemed to me that that was the crux of the OP question - how do we know whether an overall DD reaction occurs. I thought he answer was not simple, but depends on whether and to what extent the relevant ions can leave the solution.

It looks like I was probably wrong to think that. Provoked by this thread I looked up double decomposition and found that it is actually a recognised "thing", rather than just the casual jargon people occasionally use to summarise some precipitation reactions.
I haven't yet found out what the value / utility of this concept is and am therefore in the same boat as the OP, in wondering how I can know whether a particular mix of compounds is a DD reaction or not.
And whether the concentrations of the solutions (assuming we stick to stoichiometric mixes) and temperature enter into the decision.
Or even whether the presence of other ions and the pH of the solution (eg. adding KOH to this mix to change the concentrations of ions already present) affects the judgement or excludes it from consideration.
 
  • #6
Borek
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I was intending to say that this standard reaction, as commonly described in text books and lab exercises, does not take place by mixing lead nitrate crystals with potassium iodide crystals.
Good point, it would be better to mark some salts as (aq) and lead iodide as (s) to make things clear.

I think of gettingn KNO3 by mixing hot NaNO3 and KCl solutions, and cooling in an ice bath. Is that a double displacement, even though there's a substantial amount of K+ and NO3- left in the solution?
You can always filter out the solid, then it is much more clear that the reaction did happened.

The point I was trying to make, maybe foolishly (and maybe wrongly) because of the bee in my bonnet, was that mixing solutions of ionic salts is not generally a reaction. It's just putting those ions into the solution. What happens next, if anything, depends on what the ions get up to, not where they came from. It seemed to me that that was the crux of the OP question - how do we know whether an overall DD reaction occurs. I thought he answer was not simple, but depends on whether and to what extent the relevant ions can leave the solution.
In general you are perfectly right. In practice most chemists would do a mental shortcut and treat the equation as describing reaction taking part in water, especially as it is a textbook example used quite often in this context. Technically every chemist taking the shortcut is wrong in general.

I haven't yet found out what the value / utility of this concept is and am therefore in the same boat as the OP, in wondering how I can know whether a particular mix of compounds is a DD reaction or not.
It is a can of worms, zillions of possibilities even for simple examples. Still, in the case of water solutions it is enough to remember some simple solubility rules to predict the most obvious cases.

And whether the concentrations of the solutions (assuming we stick to stoichiometric mixes) and temperature enter into the decision.
Or even whether the presence of other ions and the pH of the solution (eg. adding KOH to this mix to change the concentrations of ions already present) affects the judgement or excludes it from consideration.
All these can be important, especially when there are substance substances ready to get reduced/oxidized/complexed. No way to predict what will happen without tables (redox potentials, complex stability, solubility, dissociation etc).

As I wrote, the equation that started this thread is a rather trivial case.
 

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