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Edexcel - Chemistry Unit 2 - 4 June 2013 Watch

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    I'm struggling with Titrations (sodium thoisulphate) and all the questions on that... can anyone summarize it for me
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    (Original post by Hi, How are you ?)
    Hi guys =), any revision notes?
    There are some pretty good notes on this fairly old thread : http://www.thestudentroom.co.uk/showthread.php?t=101569


    (Original post by daniya12)
    I'm struggling with Titrations (sodium thoisulphate) and all the questions on that... can anyone summarize it for me
    It would be easier if you could give me specific questions you are stuck on

    The sodium thiosulfate is used in reactions which produce iodine, it's your standard solution (which means you know how much sodium thiosulfate there is) which you use to titrate against the iodine to find out how much iodine there is.

    Now in the question they usually tell you at the end what the concentration and mean titre was of the sodium thiosulfate. The mean titre is basically how much concentration was required to reduce all of the iodine into Iodine ions using the sodium thiosulfate in the burette
    (you would know this point as there would be a colour change from blue/black to colourless if you use starch indicator).

    You can find out the amount of moles of thiosulfate required with this data, moles = concentration x volume

    Now that you have that you look at your equation, usually the second one where iodine and sodium thiosulfate are reacting - and see what the stoichiometry (mole ratio) is. Then multiply or divide accordingly.

    This is usually a 2:1 ratio.... that means in the equation there's 2 moles of sodium thiosulfate reacting with 1 mole of iodine.
    Therefore in this case you would half the amount of moles you got for sodium thiosulfate previously, as they are proportional

    Now you go to your previous equation where you have iodine as a product on the right hand side... The moles of iodine you have just found is the same amount of moles of iodine produced in that equation, regardless of any stoichiometry differences !

    Now you do the same thing as previous... you look at the mole ratio of the Iodine and the reactant you are trying to find the amount for. You then multiply or divide accordingly ! This will give you the amount of moles of the specific reactant you were asked to investigate.

    Usually the question will ask you to find in grams how much of this particular reactant there was - you use the equation, mass = moles x molar mass to find this

    I hope that was easy to follow and helps If you are still stuck, then please do post the question here...
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    (Original post by posthumus)
    There are some pretty good notes on this fairly old thread : http://www.thestudentroom.co.uk/showthread.php?t=101569




    It would be easier if you could give me specific questions you are stuck on

    The sodium thiosulfate is used in reactions which produce iodine, it's your standard solution (which means you know how much sodium thiosulfate there is) which you use to titrate against the iodine to find out how much iodine there is.

    Now in the question they usually tell you at the end what the concentration and mean titre was of the sodium thiosulfate. The mean titre is basically how much concentration was required to reduce all of the iodine into Iodine ions using the sodium thiosulfate in the burette
    (you would know this point as there would be a colour change from blue/black to colourless if you use starch indicator).

    You can find out the amount of moles of thiosulfate required with this data, moles = concentration x volume

    Now that you have that you look at your equation, usually the second one where iodine and sodium thiosulfate are reacting - and see what the stoichiometry (mole ratio) is. Then multiply or divide accordingly.

    This is usually a 2:1 ratio.... that means in the equation there's 2 moles of sodium thiosulfate reacting with 1 mole of iodine.
    Therefore in this case you would half the amount of moles you got for sodium thiosulfate previously, as they are proportional

    Now you go to your previous equation where you have iodine as a product on the right hand side... The moles of iodine you have just found is the same amount of moles of iodine produced in that equation, regardless of any stoichiometry differences !

    Now you do the same thing as previous... you look at the mole ratio of the Iodine and the reactant you are trying to find the amount for. You then multiply or divide accordingly ! This will give you the amount of moles of the specific reactant you were asked to investigate.

    Usually the question will ask you to find in grams how much of this particular reactant there was - you use the equation, mass = moles x molar mass to find this

    I hope that was easy to follow and helps If you are still stuck, then please do post the question here...
    Omigosh THANK YOU SO MUCH! I'm sending you telepathic hugs right now This was exactly what I was looking for- thanks heaps
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    (Original post by daniya12)
    Omigosh THANK YOU SO MUCH! I'm sending you telepathic hugs right now This was exactly what I was looking for- thanks heaps
    aha your more than welcome !

    It was good recap for me too
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    I'm struggling a bit with learning the organic stuff. Do we need to know how to prepare say, a halogenoalkane, in its entirety?
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    (Original post by samsimmons)
    I'm struggling a bit with learning the organic stuff. Do we need to know how to prepare say, a halogenoalkane, in its entirety?
    I'm afraid so
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    (Original post by posthumus)
    I'm afraid so
    Ahhhh man. Could someone go through the process for me? And also how would you carry out the reaction of a halogenoalkane with alcoholic alkali to collect the alkene, plus with alcoholic ammonia?

    Meh so confused x_x
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    (Original post by samsimmons)
    Ahhhh man. Could someone go through the process for me? And also how would you carry out the reaction of a halogenoalkane with alcoholic alkali to collect the alkene, plus with alcoholic ammonia?

    Meh so confused x_x
    Okay the common method to make the halogenoalkane chloroalkane is to add phosphorus pentachloride, example:

    CH3CH2OH + PCl5 -------> CH3CH2Cl + POCl3 + HCl

    Another method is adding HCl, example:

    CH3CH2OH + HCl --------> CH3CHClCH3 + H2O

    This reaction is less efficient however also you have to form the HCl, you can do this by adding concentrated sulfuric acid to some potassium chloride:
    KCl + H2SO4 -----> HCl + KHSO4

    Moving on to Bromoalkane:
    Here obviously you can't use PCl5
    You use the second method but do not use concentrated sulfuric acid to form your HBr from KBr + sulfuric acid, remember you use 50% concentration....

    Iodoalkanes:
    A completely different method... you form phosphorus triiodide:
    2P + 3I2 -----> 2PI3
    3CH2H5OH + PI3 -----> 3C2H5I + H3PO3
    (These are all slightly heated together)
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    (Original post by samsimmons)
    Ahhhh man. Could someone go through the process for me? And also how would you carry out the reaction of a halogenoalkane with alcoholic alkali to collect the alkene, plus with alcoholic ammonia?

    Meh so confused x_x
    Okay do you know how heat under reflux apparatus works ?
    You 'reflux' the halogenoalkane, and here's what is required to do turn this into an alkene, the is done via elimination:

    Reagent: concentrated potassium hydroxide
    Conditions: Heat under reflux in solution of ethanol
    Product: Alkene!


    Okay now to form an amine you can't use heat under reflux as the ammonia gas would be released and free to escape!
    Therefore your apparatus needs to be sealed and heated.... or you could just add some concentrated ammonia and leave the mixture at room temperature for a very long time, that's if your feeling patient !
    Make sure the ammonia is excess.

    The halogenoalkane goes through nucleophilic substitution with the ammonia basically, here's an example:

    CH3CH2Cl + 2NH3 ------> CH3CH2NH2 + NH4Cl
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    (Original post by posthumus)
    okay do you know how heat under reflux apparatus works ?
    you 'reflux' the halogenoalkane, and here's what is required to do turn this into an alkene, the is done via elimination:

    reagent: concentrated potassium hydroxide
    conditions: heat under reflux in solution of ethanol
    product: alkene!


    Okay now to form an amine you can't use heat under reflux as the ammonia gas would be released and free to escape!
    Therefore your apparatus needs to be sealed and heated.... Or you could just add some concentrated ammonia and leave the mixture at room temperature for a very long time, that's if your feeling patient !
    make sure the ammonia is excess.

    The halogenoalkane goes through nucleophilic substitution with the ammonia basically, here's an example:

    Ch3ch2cl + 2nh3 ------> ch3ch2nh2 + nh4cl
    thank you so much! ;o;
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    (Original post by geor)
    If this paper is like the Jan 2013 unit 1 paper.... tears will be shed.
    Can someone pllllzzzzz post jan 2013 3b asapp, i need it plzzz, thanks

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    Resitting this one

    Must start actually revising it soon I guess... Thankfully most of the key concepts overlap with U5, so maybe I'll improve on my D grade -_-
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    (Original post by Gnome :))
    Resitting this one

    Must start actually revising it soon I guess... Thankfully most of the key concepts overlap with U5, so maybe I'll improve on my D grade -_-
    Hey your right I am sure you will definitely improve on that !

    I am resitting to improve my D too (& my D in unit 4 for that matter )
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    (Original post by posthumus)
    Hey your right I am sure you will definitely improve on that !

    I am resitting to improve my D too (& my D in unit 4 for that matter )
    Ah, we'll get there eventually!

    The organic stuff seems to overlap quite a bit, and the core understanding is simply built on in U5 I feel. Need to go over the group tends and intermolecular forces etc, but apart from that I'm just going to do past papers.

    Unit 4 aswell?! Ouch!

    Going to uni?
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    (Original post by Gnome :))
    Ah, we'll get there eventually!

    The organic stuff seems to overlap quite a bit, and the core understanding is simply built on in U5 I feel. Need to go over the group tends and intermolecular forces etc, but apart from that I'm just going to do past papers.

    Unit 4 aswell?! Ouch!

    Going to uni?
    I'm going to go over the chapters that aren't covered in unit 4 & 5, and do past papers ! (hopefully I can accomplish that within a week )

    aha, I'm sitting unit 1 as well ! I have 13 exams

    I sure hope so ! Applied for Computer Science, and yourself ?
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    (Original post by posthumus)
    I'm going to go over the chapters that aren't covered in unit 4 & 5, and do past papers ! (hopefully I can accomplish that within a week )

    aha, I'm sitting unit 1 as well ! I have 13 exams

    I sure hope so ! Applied for Computer Science, and yourself ?
    Fun times eh? Ah well, not long left now, just a few final weeks of torture

    13? Wow. Good luck with that!

    Awesome, confident about meeting your offer? Yeah I'm hopefully studying nursing, but my firm wants AAB which I am on track for but only just :/
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    PS Reviewer
    I'm doing this exam too!

    What ums do you lot need in this to meet your target grade?
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    (Original post by Gnome :))
    Fun times eh? Ah well, not long left now, just a few final weeks of torture

    13? Wow. Good luck with that!

    Awesome, confident about meeting your offer? Yeah I'm hopefully studying nursing, but my firm wants AAB which I am on track for but only just :/
    Thanks you too !

    Erm not so much, I was confident - but January exams went very badly ! At the moment I am on B in both Math & Physics and D in Chemistry You ?


    (Original post by James A)
    I'm doing this exam too!

    What ums do you lot need in this to meet your target grade?
    I need roughly 120 UMS (for a B grade) from my 3 retakes + Unit 5

    I got 70 & 68 UMS in unit 2 & 4 ... I hope to get at least 50 UMS from those to be honest

    Unit 4 I worked bloody hard knew everything inside out and just blanked out in the exam ! ergh

    how about yourself ?
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    (Original post by posthumus)
    Thanks you too !

    Erm not so much, I was confident - but January exams went very badly ! At the moment I am on B in both Math & Physics and D in Chemistry You ?




    I need roughly 120 UMS (for a B grade) from my 3 retakes + Unit 5

    I got 70 & 68 UMS in unit 2 & 4 ... I hope to get at least 50 UMS from those to be honest

    Unit 4 I worked bloody hard knew everything inside out and just blanked out in the exam ! ergh

    how about yourself ?
    Well, as you know I'm doing this and Unit 5.



    I need 15 ums in total from either unit 2 or 5 for my A overall. So if I gain 10 ums in Unit 2 and 5 in unit 5, then I have my A, overall. It's still gonna be hard, defo not easy.

    I got 90/120 for Unit 2 and 88/120 for Unit 5.
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    (Original post by James A)
    Well, as you know I'm doing this and Unit 5.



    I need 15 ums in total from either unit 2 or 5 for my A overall. So if I gain 10 ums in Unit 2 and 5 in unit 5, then I have my A, overall. It's still gonna be hard, defo not easy.

    I got 90/120 for Unit 2 and 88/120 for Unit 5.
    Wow ! Good luck

    I am finding it so difficult to study at the moment
    Have you gone through all the topics in unit 5 yet, and moved onto past papers maybe ?
 
 
 
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