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Author Topic: HSC Chemistry Question Thread  (Read 1040857 times)  Share 

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Happy Physics Land

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Re: Chemistry Question Thread
« Reply #150 on: March 15, 2016, 10:13:22 am »
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The information is very much correct.

A weak acid (CH3COOH) yields a weak base conjugate CH3COO-

A strong acid (HCl), specifically, however has an extremely weak conjugate base (Cl-).

It actually isn't, basically.

No pun intended.

But isnt CH3COOH a weak acid?
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RuiAce

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Re: Chemistry Question Thread
« Reply #151 on: March 15, 2016, 10:14:35 am »
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Hi, i was wondering if you have any resources/information of the lead-acid battery. The chemistry of it was quite confusing to me and the way the teacher taught it to us was in a hurry and she didn't really explain much. We also did the silver button cell

Thanks

Note that with such cells, you don't have to listen to your teacher. You can choose yourself to do the dry cell over lead-acid, and pick another cell over the button cell. (However, admittedly the button cell, whilst not the one I did, is probably the easiest to do.)

EasyChem offers some decent notes on the cell. http://www.easychem.com.au/production-of-materials/electrochemical-methods/comparison-of-battery-cells. I didn't study this one (and my HSC was last year) so I can't offer something comprehensive - I will leave that to another user.

Take strong note that the 'chemistry' of a cell specifically refers to the reactions at the anode and cathode (and any electrolyte if applicable) - thus it's mostly the EQUATIONS.

RuiAce

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Re: Chemistry Question Thread
« Reply #152 on: March 15, 2016, 10:14:58 am »
+4
But isnt CH3COOH a weak acid?

Acetic acid is very much a weak acid, yes.

And the acetate ion is ALSO a weak base.

You need to be aware of the fact that the conjugate of only an EXTREMELY weak acid/base is a strong base/acid.

This is because the degree of ionisation of a strong acid/base is virtually 100% (at least 90% but we assume 100% in our course), thus the presence of it's conjugate is NOT going to reform into the original thing. I.e. this reaction can't happen: Cl- + H3O+ → HCl(aq) + H2O(l). Only the reverse reaction can happen.

As opposed to the acetate ion, where we do have the equilibrium happening and both substances can form due to a low degree of ionisation.
CH3COO- + H3O+ ⇌ CH3COOH(aq) + H2O(l)
« Last Edit: March 15, 2016, 10:25:31 am by RuiAce »

katherine123

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Re: Chemistry Question Thread
« Reply #153 on: March 15, 2016, 06:02:10 pm »
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Acetic acid is very much a weak acid, yes.

And the acetate ion is ALSO a weak base.

You need to be aware of the fact that the conjugate of only an EXTREMELY weak acid/base is a strong base/acid.

This is because the degree of ionisation of a strong acid/base is virtually 100% (at least 90% but we assume 100% in our course), thus the presence of it's conjugate is NOT going to reform into the original thing. I.e. this reaction can't happen: Cl- + H3O+ → HCl(aq) + H2O(l). Only the reverse reaction can happen.

As opposed to the acetate ion, where we do have the equilibrium happening and both substances can form due to a low degree of ionisation.
CH3COO- + H3O+ ⇌ CH3COOH(aq) + H2O(l)



so basically
Very strong/weak acid/base produces very weak/strong conjugate base/acid
Moderately strong/weak acid/base produce moderately weak/strong base/acid   ?



Happy Physics Land

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Re: Chemistry Question Thread
« Reply #154 on: March 15, 2016, 06:41:21 pm »
+2


so basically
Very strong/weak acid/base produces very weak/strong conjugate base/acid
Moderately strong/weak acid/base produce moderately weak/strong base/acid   ?

Hey Katherine!

Very strong acid/base would produce very weak conjugate base/acid. Very weak acid/base would produce very strong base/acid. With weak acids and weak bases such as CH3COOH and NH3, their conjugate bases and acids will be strong, but it is only strong when compared to water. When compared to other strong bases or acids they might still be weak. To say "Moderately strong/weak acid/base produce moderately weak/strong base/acid" is justifiable but I think its better to say that these bases and acids are weak/strong compared to water. So yes I would largely agree with your statement here. Sorry for my initial incorrect response on this by the way!

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RuiAce

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Re: Chemistry Question Thread
« Reply #155 on: March 15, 2016, 07:24:42 pm »
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so basically
Very strong/weak acid/base produces very weak/strong conjugate base/acid
Moderately strong/weak acid/base produce moderately weak/strong base/acid   ?

Not exactly the best way to memorise it. Because there's no such thing as very strong or moderately strong.

There's only strong, weak and extremely weak.

Conjugate of a strong acid/base is an extremely weak base/acid.
Conjugate of a weak acid/base is a weak base/acid.

katherine123

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Re: Chemistry Question Thread
« Reply #156 on: March 15, 2016, 10:20:06 pm »
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Is this okay?

Discuss factors that must be considered when using neutralization reactions to safely minimise damage in chemical spills 4 marks

Acid can be neutralised by adding solid amphiprotic substance eg. NaHCO3. Since it is a weak acid/base it will not constitute a hazard when it is in excess.
as an acid neutralise base: HCO3- (aq)+ OH- (aq)< -> CO3(2-)(aq) + H2O (l)
as a base neutralise acid: HCO3(-)(aq) + H30+(aq) <-> H2CO3 (aq) + H2O(l)
As a powder, it provides a great surface area for quick absorption of liquid acid. Moreover, it produces effervescence in acid therefore complete neutralisation of concentrated acid can be known when fizzing stops.   HCO3- + H3O + --> CO2 + 2H2O
Its disadvantages has minimal adverse impact as CO2 produced has negligible contribution to atmospheric CO2 and calcification of water pipes that is causes can be easily managed

RuiAce

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Re: Chemistry Question Thread
« Reply #157 on: March 16, 2016, 10:00:11 am »
+2
Is this okay?

Discuss factors that must be considered when using neutralization reactions to safely minimise damage in chemical spills 4 marks

Acid can be neutralised by adding solid amphiprotic substance eg. NaHCO3. Since it is a weak acid/base it will not constitute a hazard when it is in excess.
as an acid neutralise base: HCO3- (aq)+ OH- (aq)< -> CO3(2-)(aq) + H2O (l)
as a base neutralise acid: HCO3(-)(aq) + H30+(aq) <-> H2CO3 (aq) + H2O(l)
As a powder, it provides a great surface area for quick absorption of liquid acid. Moreover, it produces effervescence in acid therefore complete neutralisation of concentrated acid can be known when fizzing stops.   HCO3- + H3O + --> CO2 + 2H2O
Its disadvantages has minimal adverse impact as CO2 produced has negligible contribution to atmospheric CO2 and calcification of water pipes that is causes can be easily managed

On the note of NaHCO3 being a weak acid/base, you should contrast it to the disadvantage of using a strong acid/base. One of them is yes, causing a hazard of it's own when used in excess, but the other is that the process of neutralisation is highly exothermic and it's only worse when a strong acid/base is used. This occurs due to the reaction H+ + OH- -> H2O(l) having ∆H = approx -57 kJ mol-1, which is reduced to about -42 kJ mol-1 when a weak acid/base is present.

Another suggestion when neutralising the acid, however this being more or so on larger scale spills, is to soak the acid up into sand first. Have the sand absorb the acid, and then still using NaHCO3, neutralise it at a much safer location.
« Last Edit: March 16, 2016, 10:01:53 am by RuiAce »

katherine123

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Re: Chemistry Question Thread
« Reply #158 on: March 16, 2016, 03:14:02 pm »
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how do you answer this ques

explain why a solution of CH3COONa is basic, while a sodium sulfate solution of the same concentration has a pH of 7. Write ionic equations to describe any reactions

RuiAce

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Re: Chemistry Question Thread
« Reply #159 on: March 16, 2016, 06:04:53 pm »
+2
how do you answer this ques

explain why a solution of CH3COONa is basic, while a sodium sulfate solution of the same concentration has a pH of 7. Write ionic equations to describe any reactions

Preface. We know that sulfuric acid is a strong acid. It's first proton will always ionise no matter what.
H2SO4 + H2O(l) → HSO4- + H3O+

But we obviously know that sulfuric acid is diprotic. It has two protons that it's willing to donate. However, the second hydrogen atom does not always get fully ionised - it goes into equilibrium
HSO4- + H2O(l) ⇌ SO42- + H3O+

Except, under normal circumstances this equilibrium lies well to the right. The amount of sulfate ions present will always be appreciably greater than those of hydrogen sulfate. In fact, it almost ionises to completion as well.

The consequence is that whilst H2SO4 is obviously strongly acidic, HSO4- is reasonably acidic and SO42- is only a tiny bit basic.

In Na2SO4, obviously the sodium ion Na+ is neutral. But because the sulfate ion is only somewhat basic, it doesn't adjust the pH of the solution much at all. It is essentially neutral, just not exactly (which the question mistakenly implied).

By contrast, for sodium acetate, the acetate ion is definitely basic. (Obviously, it's a weak base. But it's not as weak as the sulfate ion.)
CH3COO- + H3O+ ⇌ H2Ol + CH3COOH(aq)
A simple analysis of that equation clearly shows why a solution of sodium acetate will be basic.

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Re: Chemistry Question Thread
« Reply #160 on: March 16, 2016, 06:49:29 pm »
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Hey Jake! For HSC chem, in the Production of Materials module, would we need to know anything about Hydrobromous acid in terms of the Bromine water test used to distinguish between alkenes and alkanes?

Happy Physics Land

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Re: Chemistry Question Thread
« Reply #161 on: March 16, 2016, 07:09:24 pm »
+1
Hey Jake! For HSC chem, in the Production of Materials module, would we need to know anything about Hydrobromous acid in terms of the Bromine water test used to distinguish between alkenes and alkanes?

Hey King_sanj:

Jake is temporarily away for a period of time, so I can answer the question here if you wouldnt mind. In Production of Materials alone, we wouldnt need to know anything about Hydrobromous acids (HBr), because when we are doing the Bromine water test we are using only Br2(aq). Because the reaction between bromine water and alkenes is an addition process, it doesnt form hydrobromous acid and hence it is not necessary to know about HBr for the module. However, what you do need to know is the final product between the alkene and the Br2(aq), because oftentimes you would be asked to provide with an equation of the reaction between the alkene and bromine water. For module 2 Acidic Environment you WOULD need to know about HBr and how it is a weak acid compared to other ones such as H2SO4 and HCl.

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Re: Chemistry Question Thread
« Reply #162 on: March 16, 2016, 07:43:44 pm »
+2
Hey King_sanj:

Jake is temporarily away for a period of time, so I can answer the question here if you wouldnt mind. In Production of Materials alone, we wouldnt need to know anything about Hydrobromous acids (HBr), because when we are doing the Bromine water test we are using only Br2(aq). Because the reaction between bromine water and alkenes is an addition process, it doesnt form hydrobromous acid and hence it is not necessary to know about HBr for the module. However, what you do need to know is the final product between the alkene and the Br2(aq), because oftentimes you would be asked to provide with an equation of the reaction between the alkene and bromine water. For module 2 Acidic Environment you WOULD need to know about HBr and how it is a weak acid compared to other ones such as H2SO4 and HCl.

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Happy Physics Land

I know what he's referring to.

In truth, whilst we prepare it using bromine itself, Br2, what we actually get and use for the practical is what is called bromine water - HOBr.
Observe the following equation:
Br2(l) + H2O(l) -> HOBr(aq) + HBr(aq)

Now, hydrobromic acid will not affect the experiment in any way whatsoever. It may be worth noting that it's a strong acid, but it just sits there.
We are only interested in bromine water - HOBr, which actually has the same molecular formula as hyprobromous acid (HBrO) - not hydrobromous acid. Hydrobromous acid isn't actually a thing. Technically, the experiment used to distinguish between the alkane and alkene series actually uses bromine water. This is where the colloquial name "bromine water experiment" is derived from.

For the sake of having an alkene, I will use hex-1-ene, with formula C6H12(l).

The HSC (generally) accepts your usage of bromine itself in the equation you use:
Alkene: C6H12 + Br2 -> C6H12Br2

But the ACTUAL equation of what goes on, is believe it or not, that with the bromine water
Alkene:  C6H12 + HOBr -> C6H12BrOH

Realistically speaking, the second equation would be more accurate.

However, in terms of what bromine water HOBr actually is? That is beyond the scope of the syllabus. It is merely a tool for us to use in our Production of Materials topic.

Aside:
Naturally, we would use hexane as our contrast for the experiment:
Alkane: C6H14 + Br2 -(UV)-> C6H12Br2 + H2
Alkane: C6H14 + HOBr -(UV)-> C6H12BrOH + H2

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Re: Chemistry Question Thread
« Reply #163 on: March 18, 2016, 04:08:44 pm »
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Hey qwerty222!

Jacaranda, definitely jacaranda.
It has the most detailed information on each dotpoint, a whole load of content, if you make notes out of the textbook and be familiar with them, you will be very familiarised with all your modules. The two disadvantages are its cost (60-70 dollars l remember) and the overwhelming amount of content which may bore you. If you want to obtain a fast grasp onto the chemistry knowledge I will recommend Excel chemistry textbook because it has all the information you would need for each dotpoint and present it in very succint manner. But of course it is not as extensive as jacaranda. I havent really used any other textbooks as just yet, but l do recommend these two textbooks, especially jacaranda.

Best Regards
Happy Physics Land

Thank you! :D

katherine123

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Re: Chemistry Question Thread
« Reply #164 on: March 18, 2016, 05:46:38 pm »
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I have asked my teacher about moderately weak/strong acids/base producing moderately conjugates and she told me that I would not come across those except the strong and weak acids/base. Is this true? She also told me to assume CH3COOH is a weak acid and produces strong conjugate in the exam.