问题标题:
25度时0.1mol/lL醋酸的电离度为1.32%,求该醋酸溶液中c(H+)及Ka
问题描述:

25度时0.1mol/lL醋酸的电离度为1.32%,求该醋酸溶液中c(H+)及Ka

糜宏斌回答:
  CH3COOHH++CH3COO-0.100(未电离时)0.1(1-1.32%)0.1*1.32%0.1*1.32%(电离后)=0.098680.001320.00132c(H+)=0.00132mol/LK...
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