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vera_ewing

  • one year ago

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  1. Photon336
    • one year ago
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    it's definitely not D. let me see. BTW do you have some kind of time limit for this?

  2. Photon336
    • one year ago
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    @Rushwr what do you think?

  3. Rushwr
    • one year ago
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    thinking and reading now !

  4. Rushwr
    • one year ago
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    what if we consider hydrolysis !

  5. Photon336
    • one year ago
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    I think i have it \[1.76x10^-5 = \frac{ [x][x] }{ [0.28-x] }\]

  6. Photon336
    • one year ago
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    Vera i got 2.65 ill show you how i got it

  7. Photon336
    • one year ago
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    \[ \sqrt{ }(1.76x10^-5)*0.28 = 0.00221M = [H^+]\]

  8. Photon336
    • one year ago
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    pH = -Log[H+] = -Log[0.00221] = 2.65

  9. Rushwr
    • one year ago
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    @Photon336 I agree I got the same answer

  10. Photon336
    • one year ago
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    you have to set up an ice table.. you know that CH3COOH --> CH3COO- + H+ \[\frac{ [CHCOO-][H^+] }{ [CH3COOH]}\] \[\frac{ [+x][+x] }{ [CH3COOH-x }\] concentration of CH3COOH will go down by x and the concentration of your acid and the conjugate base CH3COO- will both be x. you can assume that because this is a weak acid we know we will have more of CH3COO- so we can ignore the -x here for CH3COO-

  11. Photon336
    • one year ago
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    re-arranges to give \[\frac{ x ^{2} }{ 0.28 } = 1.76x10^-5\]

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