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grimsnightmare

  • one year ago

What is the OH^- if the pH is 3.2, H3O^+ is 0.0006 M and the pOH is 10.8? Please explain how you found.

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  1. Photon336
    • one year ago
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    \[[H30+][OH-] = 1x10^-14 at 25C \] \[pH = -Log[H+] = -\log[0.0006] = 3.22 \] \[1x10^-14/[H3O+] = [OH-] 1x10^-14/(6x10^-4) = (1.7x10^-11) = pOH (1.7x10^-11)(6x10^-4) = 10^-14

  2. Photon336
    • one year ago
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    The pH+pOH = 14 we have 3.2+10.8 = 14 also the [H3O+][OH-] = 10^-14 at 25 degrees celsius we know that the pH = -log[H] so the -log[0.0006] = 3.2 so we take 0.0006 = [H3O+] and we divide this by 1.0x10^-14 to get the concentration of OH- ions. oh in the last line it's 1.0x10^-14

  3. Photon336
    • one year ago
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    pOH = -Log[OH-]

  4. grimsnightmare
    • one year ago
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    Ok so the H3O and OH together equal 1*10^-14 so, we do what we do with pH and pOH only instead of subtracting the one you have by what they equal together it's dividing?

  5. Photon336
    • one year ago
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    Yeah so if you find one you know the other one b/c the product of both = 1.0x10^-14 at 25C, of course if you change the temperature this value changes.

  6. Photon336
    • one year ago
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    FYI it's called Kw

  7. grimsnightmare
    • one year ago
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    Ok that totally makes sense (as if it wouldn't) Thank you for explaining that!! Feel kinda silly for even asking

  8. Photon336
    • one year ago
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    nah its' totally fine.. feel free to ask me anything

  9. Photon336
    • one year ago
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    Another thing is that your Kw will change depending on the temp, but they usually give you the one at 25*C.

  10. grimsnightmare
    • one year ago
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    Good to know will make a note of that! Thanks again

  11. Photon336
    • one year ago
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    NP

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