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vera_ewing

  • one year ago

chem

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  1. vera_ewing
    • one year ago
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    @Photon336

  2. vera_ewing
    • one year ago
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    @Photon336

  3. Photon336
    • one year ago
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    Ask yourself what happens at your equivalent point? You've added equimolar quantities of acid/base hence neutralization

  4. cuanchi
    • one year ago
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    what do you think? do you know where is the equivalence point in the graph?

  5. vera_ewing
    • one year ago
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    I'm pretty sure it's C. Is that right?

  6. cuanchi
    • one year ago
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    I think you are right, I would choose the same answer

  7. vera_ewing
    • one year ago
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    Thanks! :)

  8. Photon336
    • one year ago
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    That's still within the equivalence point it, says "above equivalence point" so I think it would be 12

  9. Photon336
    • one year ago
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    You can eliminate 2 and 8 right away because one is the equivalent point where the other is past it. If you see at the equivalencen point you add equimolar quantities of acid and base so the term A-/HA drops out and you are left with pH= pKA and at that flat line where 8 is you'll have the greatest resistance to change in pH.

  10. Photon336
    • one year ago
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    Sorry pH of 2 is not past but before the equivalence point.

  11. Photon336
    • one year ago
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    I agree with @Cuanchi

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