文档分类
=0.34 - lg[H+]2
-lg[H+]2 = 2.36
∴pH = 2.0
12.解 依题意 正极-负极得
I2(aq)→ I2(s)
E=Eθ+(0.0592/n)lg[I2]=0
Eθ=-(0.0592/2)lg[I2]=0.621-0.535
lg[I2]=-(0.086/0.0296) [I2]=1.24×10-3(mol/L)
相关推荐: