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  1. #1
    Registered User neilc777's Avatar
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    Chemistry help needed! (calculating the Ka or Kb value of salt in solution) *help?*

    heres the question
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  2. #2
    MiscReporter Pirate4lyfe's Avatar
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    Ka = [10^-9.25]^2/1
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    Originally Posted by Pirate4lyfe View Post
    Ka = [10^-9.25]^2/1
    smart dude
    It's not the size of the dog in the fight. It's the size of the fight in the dog

    i make love back to your cp's (;

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  4. #4
    Registered User neilc777's Avatar
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    sorry how do u do it?
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  5. #5
    Registered User mikay456's Avatar
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    Acetate breaks into equal concentrations of CH3COO- ion and H+ ion. Ka is: [H+ ion][Acetate anion]/[Starting concentration minus the concentrations of the ions]
    You calculate the [H+] like this: 10^(-9.25). This value is basically insignifcant so you can leave one as a denomiator. So Ka= [10^(-9.25)]^2
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    bum[p
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    Originally Posted by mikay456 View Post
    Acetate breaks into equal concentrations of CH3COO- ion and H+ ion. Ka is: [H+ ion][Acetate anion]/[Starting concentration minus the concentrations of the ions]
    You calculate the [H+] like this: 10^(-9.25). This value is basically insignifcant so you can leave one as a denomiator. So Ka= [10^(-9.25)]^2
    Nope, sorry. What are your sources?
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  8. #8
    Registered User mikay456's Avatar
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    Originally Posted by liftw8s View Post
    Nope, sorry. What are your sources?
    What I learned in AP chem.
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    Registered User TricepsNGirls's Avatar
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    pH = -log Ka
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  10. #10
    Unconquered Rockwilder's Avatar
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    You have to make an ICE chart and find the concentrations at equilibrium I think, so to find the concentration of H+ ions u need to do 10^(-9.25)=5.6234x10^-10. and then Ka=(concentration of products]/{concentration of reactants]. I cant make the ICE chart right now cuz I cant find my formula sheet.
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  11. #11
    Resident physician OhioUstud's Avatar
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    Originally Posted by mikay456 View Post
    Acetate breaks into equal concentrations of CH3COO- ion and H+ ion. Ka is: [H+ ion][Acetate anion]/[Starting concentration minus the concentrations of the ions]
    You calculate the [H+] like this: 10^(-9.25). This value is basically insignifcant so you can leave one as a denomiator. So Ka= [10^(-9.25)]^2
    That's wrong. Acetic acid falls apart into acetate and a proton.

    So overall I believe this would be the Kb because it is acetate (C2 H3 O2) not acetic acid. So pOH = 14 - pH. Then calculate from there.
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  12. #12
    Unconquered Rockwilder's Avatar
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    god this is pissing me off its easy if I knew how to set up the reaction.
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  13. #13
    Registered User mikay456's Avatar
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    Originally Posted by OhioUstud View Post
    That's wrong. Acetic acid falls apart into acetate and a proton.

    So overall I believe this would be the Kb because it is acetate (C2 H3 O2) not acetic acid. So pOH = 14 - pH. Then calculate from there.
    Okay, so it is acetic acid that, my bad. But why is Kb involved?

    Edit: forget I asked, aceate is the conjugate weak base, forgot that.
    Last edited by mikay456; 07-17-2007 at 12:33 AM.
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  14. #14
    Unconquered Rockwilder's Avatar
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    cuz Kb is used for the concentration of OH ions because weak bases in water react incompletely to produce OH- ions
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  15. #15
    Resident physician OhioUstud's Avatar
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    K final answer... You were given sodium acetate so you must calculate the Kb of acetate since it is a weak base.

    pOH = 4.75

    Kb = [10^(-4.75)]^2.
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  16. #16
    Unconquered Rockwilder's Avatar
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    k if u find out how to right the equation this is what u do.
    do an ICE chart, then using the equilibrium values that u got go Kb= [porducts]/ [reactants].
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  17. #17
    Resident physician OhioUstud's Avatar
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    Originally Posted by Rockwilder View Post
    k if u find out how to right the equation this is what u do.
    do an ICE chart, then using the equilibrium values that u got go Kb= [porducts]/ [reactants].
    ICE chart isn't needed in this case because the solution is already at equilibrium. You only do ICE charts when you're mixing solutions or changing the pressure or molarities.
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  18. #18
    Unconquered Rockwilder's Avatar
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    okay lol, this problem was pissing me off.
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