(1)CH3COONH4溶液中,CH3COO-和NH4+都发生水解,且水解程度相等,CH3COONH4溶液呈中性,将CH3COONH4加入到0.1 mol?L-1醋酸中使pH增大,说明CH3COO-抑制了醋酸的电离,其它物质的水溶液都呈碱性,不能用于证明;CH3COONH4 溶液为中性,所以加入固体CH3COONH4 对原溶液的pH无影响,
故答案为:B;不变;
(2)①将0.010molCH3COONa和0.004molHCl溶于水,由于发生CH3COO-+H+?CH3COOH,溶液中存在CH3COO-、Cl-、OH-、H+、Na+等离子,共5种,故答案为:5;
②根据物料守恒可知,0.010mol CH3COONa在溶液中以CH3COOH和CH3COO-存在,n(CH3COOH)+n(CH3COO-)=0.010mol,
故答案为:CH3COOH;CH3COO-;
③溶液遵循电荷守恒,存在:n(H+)+n(Na+)=n(Cl-)+n(CH3COO-)+n(OH-),
则n(CH3COO-)+n(OH-)-n(H+)=n(Na+)-n(Cl-)=0.010mol-0.004mol=0.006mol,
故答案为:0.006mol.