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I was just wondering about the following situation; if we had a solution containing 0.1M bromide and 0.1M chloride ions, and treated it with excess silver nitrate, what would happen and would quantitative precipitation of both silver salts be possible?

If I understood correctly, as AgBr is less soluble, it precipitates first. I know that the solubility product of AgBr is lower than AgCl, but why does it selectively precipitate first?

Anyway, it precipitates until the point where the remaining bromide (about 0.48% of 0.1M) remains; at this point, AgCl is more insoluble and starts precipitating, until AgBr is more insoluble again, that’s how I understood it.

So basically, am I correct in the assumption that quantitative precipitation of both salts in same solution is not possible because at some point both salts precipitate simultaneously although there is still a significant amount of bromide in solution?

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    In practice, the very first drop of AgNO3 solution will create a huge oversaturation in both AgBr and AgCl, so both will start precipitating chaotically. – Ivan Neretin Nov 19 '21 at 08:55
  • It can be said that precipitation of chlorides may be preferred kinetically even before it is preferred thermodynamically. – Poutnik Nov 19 '21 at 09:54
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    @KarstenTheis It is even her own question...... – Poutnik Nov 19 '21 at 12:23

2 Answers2

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All depends on the particular scenario.

Let assume the concentrations of both halogenides and silver nitrate are low enough, so only the solubility product of silver bromide is reached. Then silver bromide is precipitated selectively ( if we neglect coprecipitation effects ).

Similar effect is achieved, if nitrate is being added so slowly the reprecipitation is faster than primary precipitation.

Otherwise, both salts are precipitated simultaneously. Saying this, be aware there will be ongoing fast or slow recrystallization/reprecipitation.

More soluble chloride will gradually dissolve in favour of bromide precipitation until chloride/bromide concentration ratio is equal to ratio of the respective solubility products.

Poutnik
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  • And if we add excess we are basically precipitating both of them at the same time, regardless of relative solubility? –  Nov 19 '21 at 09:08
  • If you think otherwise I am curious about your reasons. // Who asks too much, thinks too little. – Poutnik Nov 19 '21 at 09:11
  • Well, in my lab experiment (measuring iodide and chloride) I conducted potentiometry and we measured the potential change of the solution. There were two equivalence points after 1.82 and 3ml of AgNO3 0.1M. That would mean that first the iodide, then the chloride precipitates, we also only gave amounts of 0.1ml of AgNO3 0.1M into it.

    In my handbook it says that measuring AgBr with this method wouldn’t be possible because the solubility of AgBr and AgCl are closer together. But I didn’t really understand if they meant giving in AgNO3 as excess or how you put it, as infinitely small additions

     –  Nov 19 '21 at 09:37
  • Yes, potentiometry is preferred way here. 2 equivalence points do not mean there is no AgCl(s) before the first one. All depends on the scenario details. The point just means I- are done, Generally both nitrate and hologenides are too concentrated. – Poutnik Nov 19 '21 at 09:41
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If some $\ce{AgCl(s)}$ is introduced into a bromide solution, it reacts with $\ce{Br-}$ ions according to : $$\ce{AgCl(s) + Br- -> AgBr(s) + Cl-}$$ so that $\ce{AgCl}$ is quickly transformed into $\ce{AgBr}$ which is nearly insoluble in, for example, $\ce{NH3}$ solution. So if some $\ce{AgNO3}$ is added to a solution containing both $\ce{Br-}$ and $\ce{Cl-}$ ions, both may react in the beginning with $\ce{Ag+}$ ions, but after some time the precipitate is only made of pure $\ce{AgBr}$, because the $\ce{AgCl}$ precipitate (formed in the beginning) has all reacted with bromide ions. $\ce{AgCl}$ will be obtained only when the $\ce{Br-}$ ions have all been consumed.

Maurice
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