A.NaHC 2 O 4 溶液呈酸性说的HC 2 O 4 - 的电离程度大于水解程度,所以c(C 2 O 4 2- )>c(H 2 C 2 O 4 ),故A错误; B.等物质的量的碳酸钠和碳酸氢钠溶液中存在电荷守恒,即c(H + )+c(Na + )=c(OH - )+2c(CO 3 2- )+c(HCO 3 - ),故B错误; C.溶液呈中性,即c(OH - )=c(H + ),溶液中存在电荷守恒c(Cl - )+c(OH - )=c(NH 4 + )+c(H + ),所以c(Cl - )=c(NH 4 + ),故C错误; D.根据电荷守恒得c(H + )+c(Na + )=c(CH 3 COO - )+c(OH - ),根据物料守恒得c(CH 3 COO - )+c(CH 3 COOH)=2c(Na + ),所以得c(CH 3 COO - )-c(CH 3 COOH)=2[c(H + )-c(OH - )],故D正确; 故选D. |