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7.10: Key Equations

  • Page ID
    456088
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    Kw = [H3O+][OH] = 1.0 \times 10^{−14} (at 25 °C)
    pOH = −log[OH]
    [H3O+] = \times 10^{−pH}
    [OH] = \times 10^{−pOH}
    pH + pOH = pKw = 14.00 at 25 °C
    Ka Kb = 1.0 \times 10^{−14} = Kw
    pKa = −log Ka
    pKb = −log Kb

     


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