A、点①反应后溶液是CH 3 COONa与CH 3 COOH物质的量之比为1:1的混合物,CH 3 COOH电离程度大于CH 3 COO - 的水解程度,故c(Na + )>c(CH 3 COOH),由电荷守恒可知: c(CH 3 COO - )+c(OH - )=c(Na + )+c(H + ),所以c(CH 3 COO - )+c(OH - )>c(CH 3 COOH)+c(H + ),故A错; B、点②pH=7,即c(H + )=c(OH - ),由电荷守恒知:c(Na + )+c(H + )=c(CH 3 COO - )+c(OH - ),故c(Na + )=c(CH 3 COO - ),故B错; C、点③说明两溶液恰好完全反应生成CH 3 COONa,因CH 3 COO - 水解,且程度较小,c(Na + )>c(CH 3 COO - )>c(OH - )>c(H + ),故C错; D、当CH 3 COOH较多,滴入的碱较少时,则生成CH 3 COONa少量,可能出现c(CH 3 COOH)>c(CH 3 COO - )>c(H + )>c(Na + )>c(OH - ),故D正确; 故选:D. |