为什么摩尔电导率与无限稀释电导率之比为解离度

2025-03-18 05:32:21
推荐回答(1个)
回答1:

1.离子迁移速率 :r(+-)=u(+-)*(dE/dL), 这里u(+-)表示正(负)离子在单位电位梯度时的迁移速率,dE表示电位差,dL表示迁移距离。
2.摩尔电导率:1mol某电解质在距离为1m的两个平行电极之间的电导。公式:A=k/c (k为改电解质在浓度为c时的电导率)
对于乙酸(用HA表示),可设计电解池使其总反应为乙酸解离反应,浓度为c的HA在溶液中解离设点解读为a,有浓度:
              HA  =  H+   +   A-
平衡时   c*(1-a)   c*a       c*a
若H+的迁移速率为r(+),则单位时间内向阴极移动通过任意截面的正离子所迁移的电荷量为(电流):
Q(+)/t = I(+) = [c*a*Ar(+)]*F , A为任意截面面积,F为法拉第常数
同理对A- , 也有
Q(-)/t = I(-) = [c*a*A*r(-)]*F
则,总电流:Q/t = Q(+) + Q(-)/t = I = c*a*A*[r(+) + r(-)]*F
将r(+-)=u(+-)*(dE/dL)带入上式,有电流:
I = c*a*[u(+) +u(-)]*F*A*E/L (a)
因为电导率k = L*I/(E*A)  (b) ,  (G=k*A/L, G= I/E)
将(a)和(b)合并,有:
k = c*a*[u(+) +u(-)]*F
则摩尔电导率A = k/c = a*[u(+) +u(-)]*F
用A‘表示无限稀释摩尔电导率,u(+-)'表示无限稀释单位电位梯度离子迁移率,
因为无限稀释溶液 a = 1 , 有A' = u(+)'F + u(-)'F
假定离子迁移率随浓度变化可以忽略,即u(+-) = u(+-)', 有:A/A' = a
证明完毕

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