计算在15℃时,pH值为6时,次氯酸HCIO所占比例?(其中平衡常数Kl=3.0x10-8mol/L)
正确答案:
由平衡常数
Kl=[H+][ClO-]/[HClO]([HClO]X100)/([HCIO]+[ClO-])=100÷(1+[ClO-]/[HCIO])
=100÷(1+K1/[H+])
=100÷(1+3.0x10-8/10-6)
=100÷(1+3X10-2)
=100÷1.03=97.1%
次氯酸占97.1%
Kl=[H+][ClO-]/[HClO]([HClO]X100)/([HCIO]+[ClO-])=100÷(1+[ClO-]/[HCIO])
=100÷(1+K1/[H+])
=100÷(1+3.0x10-8/10-6)
=100÷(1+3X10-2)
=100÷1.03=97.1%
次氯酸占97.1%
答案解析:有
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