51问答网 > 25℃时，有0.01mol／L的醋酸溶液，1.写出醋酸的电离平衡常数表达式, 2。达平衡时，溶液中氢离子浓度是多少

25℃时，有0.01mol／L的醋酸溶液，1.写出醋酸的电离平衡常数表达式, 2。达平衡时，溶液中氢离子浓度是多少

2025-03-21 15:41:55
推荐回答（1个）
回答1：

HAc=可逆=Ac- + H+
Ka=[H+][Ac-]/[HAc]
cKa/Kw=0.01*10^(-4.757)/10^(-14)=10^7.243>20,忽略水解离
c/Ka=0.01/10^(-4.757)=571.48>400,忽略酸解离
[H+]=(cKa)^0.5=4.18*10^(-4)mol/L

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