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I asked my teacher if the reaction

$$\ce{(NH4)2CO3 (aq) + MgBr2 (aq) <=> MgCO3 (s) + 2(NH4)Br (aq)}$$

can be considered a displacement reaction. She answered that it was a precipitation reaction instead.

So what happened in this kind of reaction?

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  • $\begingroup$ Precipitation reaction is just reactions that forms precipitates and is a subset of displacement reaction, $\endgroup$ Commented Dec 23, 2022 at 5:28

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In fact, it is just

$$\ce{\ce{Mg^2+(aq) + CO3^2-(aq) -> MgCO3(s)}},$$

so the double displacement is very formal, as $\ce{NH4+}$ and $\ce{Br-}$ ions are just "spectator ions".

There are no real $\ce{(NH4)2CO3(aq)}$ nor $\ce{MgBr2(aq)}$, it is just a way of inventory of particles.

Precipitation reactions are, intuitively and not surprisingly, reactions forming precipitates.

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  • $\begingroup$ Thanks! However can you explain what do you mean by MgBr2 is not real? $\endgroup$ Commented Dec 21, 2022 at 13:31
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    $\begingroup$ I mean nothing by that as I have not said it. MgBr2(aq) is just a shortcut for Mg^2+(aq) + 2 Br-(aq). $\endgroup$
    – Poutnik
    Commented Dec 21, 2022 at 13:44
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It's a double displacement reaction, but it forms a solid precipitate which is $\ce{MgCO3}$

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When you have a solution where the mathmatical product of the activity of two ions which form a 1:1 solid are greater than the solubility product then a solid will form.

A first approximation can be made that the activity is equal to the concentration.

The combination of ammonium carbonate and calcium bromide can be regarded as being equal to a mixture of ammonium bromide and calcium carbonate. The ionic solids dissociate when they are immersed in water thus forming mobile ions. The ions are then in an "ionic atmosphere" which alters their ability.

We normally write for a ZX salt

log fz = -A sqrt(I)/(1+(1.5*sqrt(I)) + B[X] + C[X^2]

This is a Pitzer equation. We can then use it for a poorly soluble salt to predict the concentration of the salt, when we have a more soluble salt EF present

fz = 10^(Bze[E] + Cze[E^2] - {A sqrt(I)/(1+(1.5*sqrt(I))}

So we can then write

Ksp = [X][Z]fx fz

Which becomes

Ksp = [X][Z] (10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})

We can simply things a little when [X] = [Z]

Ksp = [X]^2 (10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})

Rearrange to give us

1/[X]^2 = {(10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})} / Ksp

Rearrange further to give us

[X]^2 = Ksp / {(10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})}

From which we get

[X] = sqrt (Ksp / {(10^(Bex[E] + Cex[E^2] - {A sqrt(I)/(1+(1.5sqrt(I))})(10^(Bfz[F] + Cfz[F^2] - {A sqrt(I)/(1+(1.5sqrt(I))})})

This might seem like a lot of maths, but it is not so bad. It explains why we should expect the solubility of CaCO3 to increase as the concentration of the NH4Br increases.

This is a rather interesting problem in environmental chemistry, geochemistry and the chemistry of radioactive waste stores. One concern in the future is the release of radium from spent fuels store thousands of years into the future. The solubility of radium sulfate will be higher in salt water than it will be in distilled water.

If we ignore the activity issues then we can write a solubility product which is based on concentrations alone, this is an apparent solubility product. But the apparent and real solubility products will deviate more and more as the ionic strength (I) of the system increases.

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  • $\begingroup$ Why is an answer using algebra so unpopular ? $\endgroup$ Commented Dec 22, 2022 at 19:15
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    $\begingroup$ I guess the issue is formatting, or rather a lack thereof. As a friendly reminder, we use MathJax as markup for literate typesetting of math expressions. Currently I find the math pretty much unreadable. $\endgroup$
    – andselisk
    Commented Dec 23, 2022 at 15:45
  • $\begingroup$ I have got no idea of how to use MathJax, I normally write algebra as if I was writing an expression in excel. How do I use MathJax ? $\endgroup$ Commented Dec 23, 2022 at 15:50
  • $\begingroup$ Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also, see Which symbols are written in roman (upright) font and which are italicized? $\endgroup$
    – andselisk
    Commented Dec 23, 2022 at 16:22

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