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Buffers equation help!

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    Hey guys,

    a buffer solution is formed from the addition of propanoic acid to sodium hydroxide.

    The reason why i need to form two equations is because the following question says :

    When a further small amount of sodium hydroxide is added in the titration, the pH changes very little.

    The answer talks about the OH- reacting with the H+ which is confusing.

    Basically i need two equations which will help me understand this process. Cheers
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    Write the ionic equation first, that might make it easier.
    You don't really need two.
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    hiya,
    a base is a mixture of weak acid (propanoic acid) and its conjugate base and H+ (as its dissociated).

    ok sodium hydroxide is a base, so upon the addition of a base (OH-) it would react with the H+ from the dissociation of acid (in the buffer) forming water. this would mean that more acid has to dissociate to make up for the deficit in H+ ions,

    do that make sense?
    Hope it helps
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    (Original post by James A)
    Hey guys,

    a buffer solution is formed from the addition of propanoic acid to sodium hydroxide.

    The reason why i need to form two equations is because the following question says :

    When a further small amount of sodium hydroxide is added in the titration, the pH changes very little.

    The answer talks about the OH- reacting with the H+ which is confusing.

    Basically i need two equations which will help me understand this process. Cheers
    CH3CH2COOH + NaOH = CH3CH2COO-Na+ + H2O

    THAT IS A WEAK ACID AND A SALT OF A WEAK ACID THUS A BUFFER SOLUTION

    WHEN H+ ADDED
    CH3CH2COO- + H+ = CH3CH2COOH

    WHEN OH- ADDED
    CH3CH2COOH + OH- = CH3CH2COO- + H2O

    SO POTENTIAL pH CHANGE IS PREVENTED
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    You only need 1 equation. The second one is simply hydrogen ions reacting with OH ions to form water. Propanoic acid dissociating is your buffer solution. Don't consider NaOH. it's a reversible reaction since propanoic acid dissociates partially to give hydrogen ions and CH3CH2COO- ions. adding MORE hydroxide ions remove hydrogen ions from the solution, so the equilibrium shifts from left to right to produce more of the H+ ions and restoring the pH.
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    (Original post by hayjes02)
    hiya,
    a base is a mixture of weak acid (propanoic acid) and its conjugate base and H+ (as its dissociated).

    ok sodium hydroxide is a base, so upon the addition of a base (OH-) it would react with the H+ from the dissociation of acid (in the buffer) forming water. this would mean that more acid has to dissociate to make up for the deficit in H+ ions,

    do that make sense?
    Hope it helps
    Thanks, it does, but I'm struggling to figure out the equation of the dissociation of propanoic acid.

    The correct way is:

    CH3CH2COOH ⇌ CH3CH2COO- + H+

    however i also stumble across books saying:

    CH3CH2COOH + H20 ⇌ H30+ + CH3CH2COO-


    why have they included the (h20) in this equation? I thought that when a weak acid dissociates, it needs H20 in the reaction to do so?
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    (Original post by James A)
    Thanks, it does, but I'm struggling to figure out the equation of the dissociation of propanoic acid.

    The correct way is:

    CH3CH2COOH ⇌ CH3CH2COO- + H+

    however i also stumble across books saying:

    CH3CH2COOH + H20 ⇌ H30+ + CH3CH2COO-


    why have they included the (h20) in this equation? I thought that when a weak acid dissociates, it needs H20 in the reaction to do so?
    You know in reality H+ ions don't exist, it's H3O ions. Something dissociates because it reacts with water. Whether you consider the role of water or not should make no difference though
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    Oh and if that makes no sense, then think of HCl. In gaseous form it's just a gas. You test it with MOIST blue litmus paper. it has acidic properties ONLY when dissolved in water. That's because H+ ions are H3O+ ions really, but for simplicity we ignore the role of water
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    (Original post by James A)
    Thanks, it does, but I'm struggling to figure out the equation of the dissociation of propanoic acid.

    The correct way is:

    CH3CH2COOH ⇌ CH3CH2COO- + H+

    however i also stumble across books saying:

    CH3CH2COOH + H20 ⇌ H30+ + CH3CH2COO-


    why have they included the (h20) in this equation? I thought that when a weak acid dissociates, it needs H20 in the reaction to do so?

    CH3CH2COOH + H20 ⇌ H30+ + CH3CH2COO-

    Kc = [ CH3CH2COO-][ H3O+]/ [CH3CH2COOH][H2O]


    Kc*[H2O] = Ka

    therefore Ka = [ CH3CH2COO-][ H3O+]/ [CH3CH2COOH]

    H+ exist as H3O+
    but in the equation always use H+

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Updated: June 11, 2012
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