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a easy question (1 Viewer)

Accuracy

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SO like NaCl + CaCO3 == > CaCl + NaCO3
is it always going to be the metals switching with the other one?
i don't get why it switches in that manner? could someone explain
 

smilingstoic

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OK so there is a little bit going on here. NaCl(aq) + CaCO3(s) won't react at all.

The equation you are talking about is probably CaCl2(aq) + Na2CO3(aq)--->CaCO3(s)+2 NaCl(aq)

What is happening here is that the reactants are all aqueous, so they are all ions floating freely in solution, ie it is really:

Ca(aq) + 2 Cl(aq)+ 2 Na(aq)+CO3 (aq)

The Calcium and the Carbonate ions react together to form a precipitate because CaCO3 is insoluble. It is not a matter of metals swapping, it is simply that insoluble combinations will precipitate out of the solution.

Was this helpful? Happy to explain further.
 

Accuracy

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is it possible for an insoluble and insoluble to react? so like there's a possibility of 2 precipitates? :S
 

smilingstoic

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Yes and no. You can't react two insolubles - once it has precipitated it will stop reacting. However is it possible to react two soluble compounds to get two precipitates? That's a good question, you can skip to the bottom for one such equation, but the full answer really gets into the guts of precipitation reactions.

Basically there are three types of ions:

Some ions refuse to form precipitates altogether: Group 1 metals (Li+, Na+, K+ etc.), Ammonium (NH4+) Nitrates (NO32-) and Acetates (CH3COO-)

On the other hand, some ions will form precipitates with all corresponding ions: Lead (Pb2+), Carbonate (CO32-) and Phosphate (PO42-). These ones won't form a precipitate with the first type - the ones that refuse altogether.

The final type are ones that form precipitates in some combinations but not others: OH-, I-, SO42-, Cl-, Ag2+, Ca2+, Ba2+, Cu2+, Fe3+.

If any reactant is from the first group, no precipitate will form from it, so there will be at most one precipitate, in most HSC questions one of these is in the mix precisely because it keeps things simpler to know, for instance, that the Nitrate won't form a precipitate.

If any reactant is from the second group, the only way to get it into solution is to combine with a corresponding ion from the first group (Lead Nitrate for instance). If it isn't paired with one from the first group, it is already a solid, so it cant take part in a precipitate reaction. Once again, it is very common for them to slip a Lead Nitrate or Sodium Carbonate into a question because it will conveniently then form a single precipitate with almost anything else that is floating in solution.

The only way to get two precipitates is to combine the right combination of four ions of the third type - the sometimes category. You need to choose two sets of ions that form precipitates despite being initially soluble with their partner. I have charted the combinations, so I can quickly see:

BaI2(aq) + AgSO4(aq) ---> BaSO4(s) + AgI2(s)

Two precipitates. I suspect that the HSC avoids such questions precisely because it is quite complicated and they can test your knowledge of individual solubility rules one at a time with other equations.
 
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