VCE Chemistry Question Thread

[IndefatigableLover]

Probably a really stupid question, but can two different precipitates form when you mix two different solutions?

Yeah of course!

One good example is:



Going by normal solubility rules, all sulphates are soluble with the exception of some including BaSO4 and likewise with hydroxides (that is they are soluble with the exception of a few inclyding Mg(OH)_2)!

[cosine]

Thanks, I was just curious about it for some reason haha

So would the two precipitates be like a different colour or something? I mean how would you identify which one is which? And if all the ions in the solution are formed into an ionic compound, does that mean there is no more solution in the reaction and everything has turned into solid? Or would there still be water in the beaker?

Thanks man!


That sounded weird, let me rephrased

Does this mean that because all the free ions in the solution have reacted, and formed an ionic compound, does this mean that there are no spectator ion in the solution, as all the ions have reacted? So we only have water in the beaker, as well as the two precipitates right?

[IndefatigableLover]

Thanks, I was just curious about it for some reason haha

So would the two precipitates be like a different colour or something? I mean how would you identify which one is which? And if all the ions in the solution are formed into an ionic compound, does that mean there is no more solution in the reaction and everything has turned into solid? Or would there still be water in the beaker?

Thanks man!


That sounded weird, let me rephrased

Does this mean that because all the free ions in the solution have reacted, and formed an ionic compound, does this mean that there are no spectator ion in the solution, as all the ions have reacted? So we only have water in the beaker, as well as the two precipitates right?

I'll answer one of them because I'm not totally sure about the other (probably because it's venturing out of VCE a bit I think LOL) but in terms of identification, I would believe that you could identify which is which by looking at the difference in colour, structure or even odour or you could dissolve it into something else to see whether it's the product which you're looking for!

[bobisnotmyname]

Quick question guys, do we need to know the solubility rules or will we be given that information? Thanks

[cosine]

I'll answer one of them because I'm not totally sure about the other (probably because it's venturing out of VCE a bit I think LOL) but in terms of identification, I would believe that you could identify which is which by looking at the difference in colour, structure or even odour or you could dissolve it into something else to see whether it's the product which you're looking for!

Alright thanks haha

Also say there is only one precipitate, 2NaOH (aq)  + MgCl2 (aq) ---> Mg(OH)2 (s) + 2NaCl (aq)

The precipitate is MgCl2, so only the Mg2+ ions and the OH- ions reacted in the mixing of the two solutions. If only these two react, what's the point of the spectator ions, Na+ and Cl-, what are they used for?

So instead of using 2NaOH in the reactants, why didn't the chemist only mix 2OH ions with Mg ions, but instead the chemist mixed NaOH with MgCl.

If you dont get what im saying, in simpler terms: Instead of mixing two ionic compounds in solution, why dont we mix the only two ions that actually react, so mix OH- ions with Mg2+ ions to get the precipitate.

Thanks

[Rishi97]

Quick question guys, do we need to know the solubility rules or will we be given that information? Thanks

You need to know them

[IndefatigableLover]

Alright thanks haha

Also say there is only one precipitate, 2NaOH (aq)  + MgCl2 (aq) ---> Mg(OH)2 (s) + 2NaCl (aq)

The precipitate is MgCl2, so only the Mg2+ ions and the OH- ions reacted in the mixing of the two solutions. If only these two react, what's the point of the spectator ions, Na+ and Cl-, what are they used for?

So instead of using 2NaOH in the reactants, why didn't the chemist only mix 2OH ions with Mg ions, but instead the chemist mixed NaOH with MgCl.

If you dont get what im saying, in simpler terms: Instead of mixing two ionic compounds in solution, why dont we mix the only two ions that actually react, so mix OH- ions with Mg2+ ions to get the precipitate.

Thanks

Not too sure on this one but I believe that by separating it into the ions that you want it to react with, we have to go back to the basics (that is ions are positive and negative). As OH ions are negative in nature, for them to exist they have to be near positive ions or else they won't exist in nature. This is the same with Mg ions since you can't just simply have two ions that are both positive/negative in nature without it being with it's opposite.

I guess one way of visualising it would be a galvanic cell with the salt bridge providing ions which split into the cathode/anode and the metal strips where the metal ions interact with the ions provided there's a current or a continuous flow of electrons from one side to the other (this works because there's a continuous flow of electrons from the anode to the cathode).

[cosine]

HCl (aq) means the H+ ions and Cl- ions have been dissociated in water and hence HCl has been dissolved in water. But how do we dissolve HCl in water, is HCl initially a liquid, solid or gas? Basically, before it becomes HCl(aq) it mustve been HCl(??) right?


Like we know the NaCl(s) + H2O(l) --> NaCl(aq) so the solid form of NaCl is mixed with water to dissociate the ions into Na+ and Cl- ions.


Thanks and sorry if  this is a stupid question

[Kel9901]

HCl (aq) means the H+ ions and Cl- ions have been dissociated in water and hence HCl has been dissolved in water. But how do we dissolve HCl in water, is HCl initially a liquid, solid or gas? Basically, before it becomes HCl(aq) it mustve been HCl(??) right?


Like we know the NaCl(s) + H2O(l) --> NaCl(aq) so the solid form of NaCl is mixed with water to dissociate the ions into Na+ and Cl- ions.


Thanks and sorry if  this is a stupid question

HCl is a gas at room temp, and a gas for pretty much all VCE purposes

[chocolate.cake.1]

Hello 

Just wondering if this would be classified as a redox reaction:   2Cu+  -->  Cu2+   +   Cu   
(sorry I have no idea how to type equations on computers  )
There is a change in oxidation numbers, but is it possible for a reactant to be both an oxidant and a reductant?

[alchemy]

Hello 

Just wondering if this would be classified as a redox reaction:   2Cu+  -->  Cu2+   +   Cu   
(sorry I have no idea how to type equations on computers  )
There is a change in oxidation numbers, but is it possible for a reactant to be both an oxidant and a reductant?

I don't think that is a possible reaction...the equation you wrote just doesn't seem to make sense.
But in general, equations like Cu --> Cu2+ + 2e- aren't redox reactions by themselves, they are half equations.
Only when you have the gain and loss of oxygen then it's redox.

[IndefatigableLover]

I don't think that is a possible reaction...the equation you wrote just doesn't seem to make sense.
But in general, equations like Cu --> Cu2+ + 2e- aren't redox reactions by themselves, they are half equations.
Only when you have the gain and loss of oxygen then it's redox.

Hmm interesting! Maybe not use a half equation but say a full equation such as:



If we're looking at sulphur, it undergoes both oxidation and reduction since the oxidation numbers change from +3 to +2 and +4 respectively however I'm pretty sure they have to be separate for it to be a redox reaction as well (could you clarify that for me alchemy or anybody else)?

[lzxnl]

Hello 

Just wondering if this would be classified as a redox reaction:   2Cu+  -->  Cu2+   +   Cu   
(sorry I have no idea how to type equations on computers  )
There is a change in oxidation numbers, but is it possible for a reactant to be both an oxidant and a reductant?

I don't think that is a possible reaction...the equation you wrote just doesn't seem to make sense.
But in general, equations like Cu --> Cu2+ + 2e- aren't redox reactions by themselves, they are half equations.
Only when you have the gain and loss of oxygen then it's redox.

Actually, that reaction does occur and shows why you don't ever see Cu(I) ions in solution; they react with themselves to form Cu metal and Cu(II).
It's called a disproportionation reaction in which a reactant is simultaneously an oxidant and a reductant and is a special case of redox reaction.

Another example is the decomposition of hydrogen peroxide to form oxygen and water.

[alchemy]

Actually, that reaction does occur and shows why you don't ever see Cu(I) ions in solution; they react with themselves to form Cu metal and Cu(II).
It's called a disproportionation reaction in which a reactant is simultaneously an oxidant and a reductant and is a special case of redox reaction.

Another example is the decomposition of hydrogen peroxide to form oxygen and water.

Interesting, I didn't know about that reaction. Do we have to familiarise ourselves with disproportionation reactions for VCE chem though?

[Kel9901]

Interesting, I didn't know about that reaction. Do we have to familiarise ourselves with disproportionation reactions for VCE chem though?

well you just need to be able to construct them from the data book, or given half equations