n(Al3+)=0.5*0.075=0.0375mol
n(Al(OH)3)=1.56/78=0.02mol
c1=0.02*3/0.1=0.6mol/L
c2=(0.02*3+0.0175*4)/0.1=1.3mol/L
向75ml 0.5mol/l明矾溶液中,滴入100mlNaOH溶液后,得1.56g白色沉淀,求原NaOH溶液的物质的量浓度
我这个题目记得有2种情况但是忘了
n(Al3+)=0.5*0.075=0.0375mol
n(Al(OH)3)=1.56/78=0.02mol
c1=0.02*3/0.1=0.6mol/L
c2=(0.02*3+0.0175*4)/0.1=1.3mol/L