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    (Original post by thad33)
    Does anyone have a nice way of explaining cathodes and anodes? I keep mixing them up. I know the direction that the electrons flow in but everywhere I look to help explain the terminals, they contradict each other.
    CATHODES are positive terminals that attract ANIONS and ANODES are negative terminals that attract CATIONS. They only was I can remember it is by thinking that cats make me feel positive! (I don't know if this is what you were looking for?)

    In cell potentials it's the half cell with most negative electrode potential that loses its electrons, therefore it acts as the ANODE/negative terminal!
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    (Original post by EllieO)
    CATHODES are positive terminals that attract ANIONS and ANODES are negative terminals that attract CATIONS. They only was I can remember it is by thinking that cats make me feel positive! (I don't know if this is what you were looking for?)

    In cell potentials it's the half cell with most negative electrode potential that loses its electrons, therefore it acts as the ANODE/negative terminal!
    so oxidation happens at negative terminal?


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    (Original post by Dinasaurus)
    For writing ionic equations of say Ca(OH)2 + 2Pyruvic Acid ----> (Pyruvate)2Ca2+ + 2H2O

    The ionic equation is apparently just H+ and OH- ----> H2O

    Can someone explain how, I don't get why the Pyruvic Acid to Pyruvate isn't also in the ionic equation if it lost a Hydrogen?
    On the left hand side 2 Pyruvic acid breaks down into 2 Pyruvate and 2 H+.

    On the right, (Pyruvate)2Ca2 breaks down to 2 Pyruvate and Ca2 2+.

    The 2 Pyruvates on the left and right cancel each other out. So they are not present in the ionic equation.
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    CO + 2H2 > CH3OH

    How could CH3OH be removed? Is it by decreasing pressure shifting eq to the left?
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    Guys how do I calculate the pH of a buffer solution of a week acid, is it not Ka=[H]^2 / (HA)

    So you do -log of the sqrt[HA]Ka
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    Could anyone explain how you would describe an experiment for enthaky change of solution?
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    When we get those crazy equations to balance (usually near the end) how do you know whether to use H+ ions, OH-, H20, or electrons? EG last Q on 2014 paper
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    (Original post by mechanism)
    Can anyone please explain Jan 2013 Q8)C)ii ? I've worked it out before but I really don't remember how it's done.
    I posted it earlier:
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    (Original post by Dinasaurus)
    Guys how do I calculate the pH of a buffer solution of a week acid, is it not Ka=[H]^2 / (HA)

    So you do -log of the sqrt[HA]Ka
    It's [H+] = Ka* [HA]/[A-] (I remember it by saying Kacid-over-salt because it sounds like a holiday destination lol)
    Then pH=-log[H+]
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    (Original post by lai812matthew)
    0.64-0.4 quick
    how do I do this question please, I get that 0.24 represents the HCOOH.

    But why do we need the 0.4? I Tthought the formula was Ka= Hsquared/ [HCOOH]
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    Time to rest. Best of luck tomorrow guys
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    (Original post by Dinasaurus)
    Guys how do I calculate the pH of a buffer solution of a week acid, is it not Ka=[H]^2 / (HA)

    So you do -log of the sqrt[HA]Ka
    No..... Buffer is the normal equation of Ka=[H+][A-]/[HA]

    [H+] doesn't equal [A-] so you can't make that assumption, thats only for weak acids.
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    (Original post by HJS14)
    When we get those crazy equations to balance (usually near the end) how do you know whether to use H+ ions, OH-, H20, or electrons? EG last Q on 2014 paper
    If you have oxygen on one side of the equation you need to add it to the other side with water.
    Then since you've added water you need to balance the hydrogens by adding hydrogen ions to the other side.
    You use OH- if it specifies an alkaline conditions or if it says something like "oxygen reacts with water" so obviously you can't balance the oxygens by adding water to the other side so you use OH- instead
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    (Original post by HJS14)
    When we get those crazy equations to balance (usually near the end) how do you know whether to use H+ ions, OH-, H20, or electrons? EG last Q on 2014 paper
    ****ing hate these questions, I dont even know electrolytes are:
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    (Original post by Dinasaurus)
    how do I do this question please, I get that 0.24 represents the HCOOH.

    But why do we need the 0.4? I Tthought the formula was Ka= Hsquared/ [HCOOH]
    That's the equation for calculating pH of a weak acid But for a buffer solution the equation is different becuase you cant assume that [A-]=[H+] So instead of [H+]sqrd you use:
    Ka=([A-][H+])/[HA]
    So 0.4 represents the salt (A-)
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    (Original post by BioStudentx)
    No..... Buffer is the normal equation of Ka=[H+][A-]/[HA]

    [H+] doesn't equal [A-] so you can't make that assumption, thats only for weak acids.
    So Ka is actually on the fraction?
    So Ka/[HA] = [H][A]

    So Ka/HA x [A] = H?

    I am so lost
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    (Original post by HJS14)
    When we get those crazy equations to balance (usually near the end) how do you know whether to use H+ ions, OH-, H20, or electrons? EG last Q on 2014 paper
    I have no idea from the 2014 paper how someone would have known, but they accepted it without OH- but usually there is an indication ie last year talking about adding potassium hydroxide so we knew it was adding OH- ions
    It would indicate if there is some sort of acid added, it's usually in the block of text that can easily be missed
    If you know for sure it's not acidic or alkali then you must only add water
    Electrons are always included in half equations
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    (Original post by Dinasaurus)
    So Ka is actually on the fraction?
    So Ka/[HA] = [H][A]

    So Ka/HA x [A] = H?

    I am so lost
    Ka=([A][H])/[HA]
    Which rearranges to [H]= Ka*[HA]/[A]
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    (Original post by Dinasaurus)
    how do I do this question please, I get that 0.24 represents the HCOOH.

    But why do we need the 0.4? I Tthought the formula was Ka= Hsquared/ [HCOOH]
    hcooh+naoh-->hcoona+h2o, 0.4 moles of naoh getting 0.4 moles of hcoona.
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    (Original post by Rust Cohle)
    ****ing hate these questions, I dont even know electrolytes are:
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    memorise the hydrogen fuel cell in acidic and alkali conditions - the two half equations then it's simply just recall
 
 
 
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