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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread Watch

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    (Original post by C0balt)
    Which one's the answer?
    I think it's B?
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    (Original post by Funky_Giraffe)
    Yes but that also gives OH- ions in solution. NH3 is not an acid.

    Yeah I know it isn't but since I thought it's in a solution then it would dissociate and and CH3COONa would dissociate as well to give OH- ions so that's why I am confused
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    can someone explain why C is the answer?

    multiply?
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    In this question volumes are different but I still used moles to get the correct answer :s, I've been trying using moles for a lot of questions now and they give the same answer even when volume us different.

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    (Original post by PlayerBB)
    B (3,2,1)
    Am I being stupid
    Why is NH3 stronger base than ethanoate ion
    I mean my gut feeling does tell me that it has a higher pH but I thought ammonia was a weak base but ethanoate ions were strong base (conjugate base of a weak acid is a strong base?)
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    (Original post by _H_V)
    In this question volumes are different but I still used moles to get the correct answer :s, I've been trying using moles for a lot of questions now and they give the same answer even when volume us different.

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    Volumes are different initially but when you mix to give final solution you get the same volume. Adding 20 and 30 give 50 in the end and both the moles are in 50
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    (Original post by C0balt)
    Am I being stupid
    Why is NH3 stronger base than ethanoate ion
    I mean my gut feeling does tell me that it has a higher pH but I thought ammonia was a weak base but ethanoate ions were strong base (conjugate base of a weak acid is a strong base?)
    Yeah same here!!! I am feeling really stupid like my gut feeling tells me that NH3 is definitely a stronger base but then I thought ethanoate ions are the conjucate base and that they;re stronger and also if they dissociate they would give CH3COOH while ammonia will give NH4+ (NH3 + H+) so I can't figure it out!
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    (Original post by PlayerBB)
    Yeah I know it isn't but since I thought it's in a solution then it would dissociate and and CH3COONa would dissociate as well to give OH- ions so that's why I am confused
    Ohh I see what you mean. That's got me very confused!
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    (Original post by ihaspotato)
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    can someone explain why C is the answer?

    multiply?
    General form for straight line is y=mx+c
    y=lnrate
    m=-Ea/R
    x=1/T
    c=constant

    So you've got gradient=-Ea/R and rearranging this gives Ea=-( gradient) R
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    (Original post by PlayerBB)
    Yeah same here!!! I am feeling really stupid like my gut feeling tells me that NH3 is definitely a stronger base but then I thought ethanoate ions are the conjucate base and that they;re stronger and also if they dissociate they would give CH3COOH while ammonia will give NH4+ (NH3 + H+) so I can't figure it out!
    The salt of a weak acid and strong base is slightly basic. This is because ethanoate ions react with water CH3COO- + H2O --> CH3COOH + OH-.The salt of a strong acid and weak base is slightly acidic. Again because the salt is hydrolysed NH4+ + H2O --> NH3 + H3O+.We know ammonia is a weak base so it has to have the highest pH out of the 3 as it would dissociate the most. Hence the answer is B (Copied this from my earlier post)
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    Ahhh finally okay that makes so much sense, thankss

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    (Original post by MacroniCheese)
    The salt of a weak acid and strong base is slightly basic. This is because ethanoate ions react with water CH3COO- + H2O --> CH3COOH + OH-.The salt of a strong acid and weak base is slightly acidic. Again because the salt is hydrolysed NH4+ + H2O --> NH3 + H3O+.We know ammonia is a weak base so it has to have the highest pH out of the 3 as it would dissociate the most. Hence the answer is B (Copied this from my earlier post)
    AH thanks!
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    (Original post by C0balt)
    General form for straight line is y=mx+c
    y=lnrate
    m=-Ea/R
    x=1/T
    c=constant

    So you've got gradient=-Ea/R and rearranging this gives Ea=-( gradient) R
    thanks!
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    (Original post by ihaspotato)
    thanks!
    No probs
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    I do not understand part b at all, would be really helpful if anyone could explain it if they have time. I know its super late!
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    I bet you we're going to get a question like this tomorrow:

    https://www.youtube.com/watch?v=mb9v7NIaSoo

    You heard it here first peeps!!!1onetwo!!
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    (Original post by Dren)
    I do not understand part b at all, would be really helpful if anyone could explain it if they have time. I know its super late!
    Okay I'm kinda of going to answer this all together instead of in the separate parts because I think it'll make more sense that way.

    Basically, the job of sodium thiosulfate is to reduce any iodine formed back into iodide ions:

    2S2O32- + I2 --> S4O62- + 2I-

    So, the solution initially will be colourless, since the persulfate ions oxidise iodide ions to iodine, but the sodium thiosulfate reduces the iodine immediately back to iodide ions, so a black colour doesn't form straight away (which is what would happen if sodium thiosulfate wasn't used --> answer to part ii). The final colour change occurs when all of the sodium thiosulfate has reacted, so any iodine formed will no longer be reduced back down to iodide ions. Therefore, the final colour change is colourless to black (answer to part i). In relation to part iii), the iodide concentration initially remains constant due to persulfate ions oxidising iodide ions to iodine but then sodium thiosulfate reducing iodine immediately back down to iodide ions again.

    Hope that makes da sense
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    Can anyone help?
    Find the pH of the mixture formed when 25 cm3
    of 2 mol dm–3 sodium hydroxide solution is
    added to 50 cm3
    of 2 mol dm–3 ethanoic acid, for which Ka
    = 1.7 × 10–5 mol dm–3.
    pH=
    A 2.2
    B 2.5
    C 4.5
    D 4.8

    Thanks
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    (Original post by Don Pedro K.)
    I bet you we're going to get a question like this tomorrow:

    https://www.youtube.com/watch?v=mb9v7NIaSoo

    You heard it here first peeps!!!1onetwo!!
    You are a beautiful person! I've never seen that type of calculation before, thank you so much for posting that! Would have definitely got it wrong tomorrow if it came up! 😅
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    (Original post by Don Pedro K.)
    Okay I'm kinda of going to answer this all together instead of in the separate parts because I think it'll make more sense that way.

    Basically, the job of sodium thiosulfate is to reduce any iodine formed back into iodide ions:

    2S2O32- + I2 --> S4O62- + 2I-

    So, the solution initially will be colourless, since the persulfate ions oxidise iodide ions to iodine, but the sodium thiosulfate reduces the iodine immediately back to iodide ions, so a black colour doesn't form straight away (which is what would happen if sodium thiosulfate wasn't used --> answer to part ii). The final colour change occurs when all of the sodium thiosulfate has reacted, so any iodine formed will no longer be reduced back down to iodide ions. Therefore, the final colour change is colourless to black (answer to part i). In relation to part iii), the iodide concentration initially remains constant due to persulfate ions oxidising iodide ions to iodine but then sodium thiosulfate reducing iodine immediately back down to iodide ions again.

    Hope that makes da sense
    Hey, there's something I don't get, how the Sodium thiosulphate is used to find the concentration of Iodine of it reduces it back continuously, like when it is used up, there would still be Iodide ions that would oxidise to Iodine (hope you understand what I mean )
 
 
 
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