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14.10: Equações-chave

  • Page ID
    198454
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    K w = [H 3 O +] [OH ] = 1,0××10 −14 (a 25 °C)
    pH=−registro[H3O+]pH=−registro[H3O+]
    pOH = −log [OH ]
    [H 3 O +] = 10 −pH
    [OH ] = 10 −pOH
    pH + pOH = p K w = 14,00 a 25 °C
    Kuma=[H3O+][UMA][HA]Kuma=[H3O+][UMA][HA]
    Kb=[HB+][OH][B]Kb=[HB+][OH][B]
    Eu sou××K b = 1,0××10 −14 = K w
    Porcentagem de ionização=[H3O+]eq[HA]0×100Porcentagem de ionização=[H3O+]eq[HA]0×100
    p K a = −log K a
    p K b = −log K b
    pH=pKuma+tora[UMA][HA]pH=pKuma+tora[UMA][HA]