呈中性,是氨水过量设盐酸xml。先反应生成0.1x/1000 mol的NH4Cl,剩余氨水0.005 -0.1x/1000 mol的氨水NH3.H2O和NH4Cl的缓冲溶液POH=PKb+lgc盐/c碱氨水Kb=1.8×10^-5,Pkb=4.747=4.74+lgcNH4Cl/cNH3.H2O182=cNH4Cl/cNH3.H2O0.1x/1000=182 x(0.005 -0.1x/1000 )0.1x=182(5-0.1x)x=5ml