Edexcel A2 Chemistry 6ch04/05 JUNE 2015 Watch

laurenjjj
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What do we need to know about recrystallisation? (for unit 4)
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cbfks
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(Original post by laurenjjj)
What do we need to know about recrystallisation? (for unit 4)
We would need to know about the recrystallisation of the precipitate produced from the Brady's reagent reaction (and then the melting point determination).

The precipitate is filtered and washed with, for example, methanol and then recrystallised from a suitable solvent which will vary depending on the nature of the aldehyde or ketone. For example, you can recrystallise the products from the small aldehydes and ketones from a mixture of ethanol and water.

The crystals are dissolved in the minimum quantity of hot solvent (removing impurities that are insoluble at high temperatures). When the solution cools, the crystals are re-precipitated and can be filtered (removing impurities that are soluble at low temperatures), washed with a small amount of solvent and dried. They should then be pure.

A minimum quantity is used to prevent loss of product that could remain in solution.
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rasil23
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Could someone explain that question please
Its from jan 2015 [International].
CH3COOH + CH3CH2OH = CH3COOCH2CH3 + H2O
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AnishaJayne
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(Original post by FMSN)
H2O = 1x10-7
1. [H+] = 10/1000 x 1 = 0.01 mol
2. [OH-] = 990/1000 x 1x10-7 = 9.9x10-8
3. Excess = 0.01 - 9.99x10-8 = 9.99x10-3
4. Conc= 9.99x10-3 x 1000/(990+10) = 9.99x10-3
5. -log(9.99x10-3) = 2
I know its a long method but it got me there!
Brill this is the most logical thank you


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laurenjjj
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(Original post by cbfks)
We would need to know about the recrystallisation of the precipitate produced from the Brady's reagent reaction (and then the melting point determination).

The precipitate is filtered and washed with, for example, methanol and then recrystallised from a suitable solvent which will vary depending on the nature of the aldehyde or ketone. For example, you can recrystallise the products from the small aldehydes and ketones from a mixture of ethanol and water.

The crystals are dissolved in the minimum quantity of hot solvent (removing impurities that are insoluble at high temperatures). When the solution cools, the crystals are re-precipitated and can be filtered (removing impurities that are soluble at low temperatures), washed with a small amount of solvent and dried. They should then be pure.

A minimum quantity is used to prevent loss of product that could remain in solution.
So helpful! Thank you very much :-)
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cbfks
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(Original post by rasil23)


Could someone explain that question please
Its from jan 2015 [International].
Firstly, calculate moles of HCl: (5/1000) x 1 = 5 x10-3 moles HCl

Then we need to find the moles of NaOH = (45/1000) x 1 = 0.045 moles NaOH

To find the moles of ethanoic acid at eqm we can do this: 0.045 moles - 5 x10-3 moles = 0.04 moles ethanoic acid

This is because NaOH reacts with both acids, therefore the rest of the NaOH required will be due to the NaOH reacting with ethanoic acid as we have determined the max moles of HCl it can react with.

So initially we had: 0.120 moles ethanoic acid, 0.220 moles ethanol, 0 moles of the ester and 0.278 moles of water.

As 0.04 moles of ethanoic acid are present at eqm (we know this from titre), this has decreased by 0.08 moles so it can be deduced that at eqm: 0.04 moles ethanoic acid, 0.220-0.08=0.14 moles ethanol, 0.08 moles of ester and 0.08+0.278=0.358 moles of water.

Now plug moles into Kc= 0.08 x 0.358 / 0.04 x 0.14 = 5.1 with no units as these cancel.

The only reason we do not use concentrations and can use moles is because the volumes will also cancel.

P.S. I would say it is best to either leave this kind of calculation to the last thing or avoid spending too much time on simple Q's to leave time for more demanding calculations.
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thymolphthalein
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(Original post by cbfks)
Firstly, calculate moles of HCl: (5/1000) x 1 = 5 x10-3 moles HCl

Then we need to find the moles of NaOH = (45/1000) x 1 = 0.045 moles NaOH

To find the moles of ethanoic acid at eqm we can do this: 0.045 moles - 5 x10-3 moles = 0.04 moles ethanoic acid

This is because NaOH reacts with both acids, therefore the rest of the NaOH required will be due to the NaOH reacting with ethanoic acid as we have determined the max moles of HCl it can react with.

So initially we had: 0.120 moles ethanoic acid, 0.220 moles ethanol, 0 moles of the ester and 0 moles of water.

As 0.04 moles of ethanoic acid are present at eqm (we know this from titre), this has decreased by 0.08 moles so it can be deduced that at eqm: 0.04 moles ethanoic acid, 0.220-0.08=0.14 moles ethanol, 0.08 moles of ester and 0.08+0.278=0.358 moles of water.

Now plug moles into Kc= 0.08 x 0.358 / 0.04 x 0.14 = 5.1 with no units as these cancel.

The only reason we do not use concentrations and can use moles is because the volumes will also cancel.

P.S. I would say it is best to either leave this kind of calculation to the last thing or avoid spending too much time on simple Q's to leave time for more demanding calculations.
Wouldn't 0.278 be considered the initial amount of water. It wouldn't make any difference to the calculation, just asking.
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cerlohee
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(Original post by thymolphthalein)
Basically there's two types of transesterification.

Reaction with another organic acid:
Acid part of an ester being replaced by an acid reactant in the presence of an acid catalyst.
Ethyl ethanoate and methanoic acid react to form ethyl methanoate and ethanoic acid.
This is the type used in manafacture of low-fat margarine.

Then there's reaction with another alcohol.
The alcohol part of the ester is replaced by the alcohol reactant in the presence of an acid catalyst.
Ethyl ethanoate and methanol react to form methyl ethanoate and ethanol.
This is used in the manafacture of biodiesel where the reactant alcohol is usually methanol/ethanol which is generally reacted with a triglyceride(an ester) to form a 1,2,3-triol.

That's pretty much it.
Do you have the equation for the first transesterification? And does it ever come up?
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rasil23
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(Original post by cbfks)
Firstly, calculate moles of HCl: (5/1000) x 1 = 5 x10-3 moles HCl

Then we need to find the moles of NaOH = (45/1000) x 1 = 0.045 moles NaOH

To find the moles of ethanoic acid at eqm we can do this: 0.045 moles - 5 x10-3 moles = 0.04 moles ethanoic acid

This is because NaOH reacts with both acids, therefore the rest of the NaOH required will be due to the NaOH reacting with ethanoic acid as we have determined the max moles of HCl it can react with.

So initially we had: 0.120 moles ethanoic acid, 0.220 moles ethanol, 0 moles of the ester and 0 moles of water.

As 0.04 moles of ethanoic acid are present at eqm (we know this from titre), this has decreased by 0.08 moles so it can be deduced that at eqm: 0.04 moles ethanoic acid, 0.220-0.08=0.14 moles ethanol, 0.08 moles of ester and 0.08+0.278=0.358 moles of water.

Now plug moles into Kc= 0.08 x 0.358 / 0.04 x 0.14 = 5.1 with no units as these cancel.

The only reason we do not use concentrations and can use moles is because the volumes will also cancel.

P.S. I would say it is best to either leave this kind of calculation to the last thing or avoid spending too much time on simple Q's to leave time for more demanding calculations.
Thank you so much, the bit that was confusing me was the NaOH ,but I get it now,
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Pilz
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(Original post by cbfks)
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holy crap, that should be more than 6 marks
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Pilz
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I got a question, why does the alkylation of benzene (Friedel-Crafts) need to take place in dry ether? Does water react with RCl? Does it react with the catalyst, AlCl3?
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thymolphthalein
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(Original post by cerlohee)
Do you have the equation for the first transesterification? And does it ever come up?
We don't have to know specific equations, it's just given in our notes as an example.

We have to know the basic principle that when an ester reacts with an acid: it replaces the acid part of the ester to form a transester and an acid.

But for the example, the equation is just:

CH₃COOCH₂CH₃ +HCOOH -----> HCOOCH₂CH₃ + CH₃COOH

In the paper, they could give you a different ester and acid combination. Although, I don't really recall huge transesterification questions in the papers anyway.
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cerlohee
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(Original post by Pilz)
I got a question, why does the alkylation of benzene (Friedel-Crafts) need to take place in dry ether? Does water react with RCl? Does it react with the catalyst, AlCl3?
Unit 5 thread is here http://www.thestudentroom.co.uk/show....php?t=3247585
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Pilz
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oh, cheers!
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cerlohee
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(Original post by thymolphthalein)
We don't have to know specific equations, it's just given in our notes as an example.

We have to know the basic principle that when an ester reacts with an acid: it replaces the acid part of the ester to form a transester and an acid.

But for the example, the equation is just:

CH₃COOCH₂CH₃ +HCOOH -----> HCOOCH₂CH₃ + CH₃COOH

In the paper, they could give you a different ester and acid combination. Although, I don't really recall huge transesterification questions in the papers anyway.
Thankyou!!
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cbfks
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(Original post by thymolphthalein)
Wouldn't 0.278 be considered the initial amount of water. It wouldn't make any difference to the calculation, just asking.
Sorry! yes of course, my mistake
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HaydenMoussa
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The equation for the reaction between a weak acid, HX, and sodium hydroxide isHX(aq) + NaOH(aq) ---> NaX(aq) + H2O(l)
The pH of the solution of the salt NaX is most likely to be:
A 5.5
B 7.0
C 8.5
D 13.0

Does anyone know the answer??
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aersh8
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(Original post by HaydenMoussa)
The equation for the reaction between a weak acid, HX, and sodium hydroxide isHX(aq) + NaOH(aq) ---> NaX(aq) + H2O(l)
The pH of the solution of the salt NaX is most likely to be:
A 5.5
B 7.0
C 8.5
D 13.0

Does anyone know the answer??
X- will react with H+ from water dissociation to reform HX so the concentration of H+ will be slightly less than OH- so the salt will be slightly alkaline ie C 8.5

Please someone correct me if I'm wrong, I haven't started revising this yet

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Ripper Phoenix
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(Original post by HaydenMoussa)
The equation for the reaction between a weak acid, HX, and sodium hydroxide isHX(aq) + NaOH(aq) ---> NaX(aq) + H2O(l)
The pH of the solution of the salt NaX is most likely to be:
A 5.5
B 7.0
C 8.5
D 13.0

Does anyone know the answer??
its a weak acid strong base... most likely has to be C
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Raghadabuali
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(Original post by thymolphthalein)
We don't have to know specific equations, it's just given in our notes as an example.

We have to know the basic principle that when an ester reacts with an acid: it replaces the acid part of the ester to form a transester and an acid.

But for the example, the equation is just:

CH₃COOCH₂CH₃ +HCOOH -----> HCOOCH₂CH₃ + CH₃COOH

In the paper, they could give you a different ester and acid combination. Although, I don't really recall huge transesterification questions in the papers anyway.
it only came in jan 2011 and it was very easy
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