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Equilibria, Energetics and Elements (F325) - June 2011 Exam. watch

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    (Original post by amyyy24)
    Can anyone help with this? I kind of understand it, but don't at the same time (I'm no good with dm/dilutions and all that)

    500cm3 of 2moldm-3 of CH3COOH was added to 4g of Li.
    Ka = 1.7x10-5
    What is the pH of the buffer?

    I got as far as knowing there is 1mol of CH3COOH and 0.58mol of Li.

    But then my teacher went on to x2 the moles of 0.84 and 1.16 and it all got a bit confusing.

    Any help?
    Ok, I understand the moles. What's the base supposed to be? Is it Li? Did you get an equation?
    I didn't think we need to know that :eek: I can calculate the pH of buffer's using the conjugate base.
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    (Original post by INeedToRevise)
    Ok, I understand the moles. What's the base supposed to be? Is it Li? Did you get an equation?
    I didn't think we need to know that :eek: I can calculate the pH of buffer's using the conjugate base.

    Oh the equation!

    CH3COOH + Li ---> CH3COO-Li+ + 1/2 H2
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    (Original post by amyyy24)
    Oh the equation!

    CH3COOH + Li ---> CH3COO-Li+ + 1/2 H2
    Haha Do you have an answer for the problem? I think I have an answer but its probably wrong so I dont want to post it
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    (Original post by INeedToRevise)
    Haha Do you have an answer for the problem? I think I have an answer but its probably wrong so I dont want to post it
    well, I think it's either 4.91 or 5.91. I really don't know
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    (Original post by amyyy24)
    well, I think it's either 4.91 or 5.91. I really don't know
    Ohh I got 5.00. Its probably wrong though because I don't understand why you multiply the moles of Li by 2.
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    (Original post by amyyy24)
    Can anyone help with this? I kind of understand it, but don't at the same time (I'm no good with dm/dilutions and all that)

    500cm3 of 2moldm-3 of CH3COOH was added to 4g of Li.
    Ka = 1.7x10-5
    What is the pH of the buffer?

    I got as far as knowing there is 1mol of CH3COOH and 0.58mol of Li.

    But then my teacher went on to x2 the moles of 0.84 and 1.16 and it all got a bit confusing.

    Any help?
    0.58mol of Li reacts with 0.58 mol of CH3COOH, so there will be 0.58 mol of CH3COO-Li+ and 0.42mol of CH3COOH
    H+ = Kax[ACID]/[CONJ. BASE]
    [acid]=0.42/0.5=0.84 mol dm3
    base = 0.58/0.5 = 1.16mol dm3
    so H+ = 0.000017x 0.84/1.16 = 0.00001231

    -log[H+] = pH = 4.9
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    (Original post by wilsea05)
    0.58mol of Li reacts with 0.58 mol of CH3COOH, so there will be 0.58 mol of CH3COO-Li+ and 0.42mol of CH3COOH
    H+ = Kax[ACID]/[CONJ. BASE]
    [acid]=0.42/0.5=0.84 mol dm3
    base = 0.58/0.5 = 1.16mol dm3
    so H+ = 0.000017x 0.84/1.16 = 0.00001231

    -log[H+] = pH = 4.9
    Where does 0.42 come from? I thought CH3COOH was 1 mole :confused:
    Why do you divide by 0.5?
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    (Original post by INeedToRevise)
    Where does 0.42 come from? I thought CH3COOH was 1 mole :confused:
    Why do you divide by 0.5?
    because you know that there is 0.58 mol of Li, the acid is in excess so all of the Li will react with the acid forming 0.58 mols of the salt (CH3COO-Li+)
    we started with 1 mol of acid, 0.58 of it has reacted, so 1-0.58=0.42 mols of acid remaining in the buffer solution

    the equation to find out H+ you need concentration of the acid and conjugate base, not moles.
    there is 0.5dm of sulution (500/1000)
    and concentration = moles/volume
    so you do the moles of acid (0.42) and moles of conj. base(0.58) divided by 0.5, to get the concentration of them. then you get the concentration of H+, so you can work out pH
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    (Original post by wilsea05)
    because you know that there is 0.58 mol of Li, the acid is in excess so all of the Li will react with the acid forming 0.58 mols of the salt (CH3COO-Li+)
    we started with 1 mol of acid, 0.58 of it has reacted, so 1-0.58=0.42 mols of acid remaining in the buffer solution

    the equation to find out H+ you need concentration of the acid and conjugate base, not moles.
    there is 0.5dm of sulution (500/1000)
    and concentration = moles/volume
    so you do the moles of acid (0.42) and moles of conj. base(0.58) divided by 0.5, to get the concentration of them. then you get the concentration of H+, so you can work out pH
    Oh ok. Thanks
    I hope something like that doesn't come up haha. Its hard!
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    (Original post by INeedToRevise)
    Oh ok. Thanks
    I hope something like that doesn't come up haha. Its hard!
    nah my teacher made us derive like every formula there is in the book D: after the first 1 or 2 you get it
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    Does anybody know the reason why you can't calculate lattice enthalpy directly?
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    I sat this paper in January but got an E, :rolleyes: so I'm resitting in June! The Joys of F325
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    Could someone help me on this question?

    The student diluted 0.015 mol dm-3 nitric acid with an equal volume of water and measured the pH of the diluted acid at 25 degrees.

    Calculate pH of the diluted acid. In the markscheme it says the [H+] = 0.0075. So you just -log0.0075 to give 2.12. I don't understand where 0.0075 comes from...
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    (Original post by INeedToRevise)
    Could someone help me on this question?

    The student diluted 0.015 mol dm-3 nitric acid with an equal volume of water and measured the pH of the diluted acid at 25 degrees.

    Calculate pH of the diluted acid. In the markscheme it says the [H+] = 0.0075. So you just -log0.0075 to give 2.12. I don't understand where 0.0075 comes from...
    you've got x amount of nitric acid, lets say 0.1dm
    and because you're adding an equal volume of water, x, that's gonna be 0.1dm too
    so you're doubling the volume, but the moles of nitric acid remain the same
    so the concentration of acid (which is equal to the concentration of H+) will be halved.
    you know its [H+]=[HNO3] because its a strong acid.
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    (Original post by P4TRiiCE)
    I was running out of things to do with the lack of past papers because of it being a new spec and all that.
    Thank you very much for this!!!
    easy badboy thanks for adding them don
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    (Original post by wilsea05)
    you've got x amount of nitric acid, lets say 0.1dm
    and because you're adding an equal volume of water, x, that's gonna be 0.1dm too
    so you're doubling the volume, but the moles of nitric acid remain the same
    so the concentration of acid (which is equal to the concentration of H+) will be halved.
    you know its [H+]=[HNO3] because its a strong acid.
    Omd thanks I get it now You are so good at all this!
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    I really hope there's a born-haber cycle on this one, because there wasn't in January I love born-haber cycles, they're easy-peasy!
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    After much pain and deliberation I have found/collected all the F325 past papers + all the legacy 2814, 2815 and 2816 papers from 2001. Do people want me to upload these online? I can send them via email If people PM me. Hopefully this will help others as well. Rep me if you want
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    (Original post by ghostkillah)
    After much pain and deliberation I have found/collected all the F325 past papers + all the legacy 2814, 2815 and 2816 papers from 2001. Do people want me to upload these online? I can send them via email If people PM me. Hopefully this will help others as well. Rep me if you want
    Yes Please! thank you!! can you upload it on here plz?
    Rep given!
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    (Original post by thefirstnotlastsamurai)
    Does anybody know the reason why you can't calculate lattice enthalpy directly?
    Correct me if I'm wrong but I think it's because lattice enthalpy is the enthalpy change that accompanies the formation of one mole of an ionic compound from its gasous ions. It would be impossible to react gasous ions in such small quantities as to produce one mole of the ionic compound experimentally.
 
 
 
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