Question

现有200mlMgCl2.AlCl3的混合液,其中c(Mg2+)=0.2ml,c (cl-)=1.3mol/L

Answer 1

1.求Al3+的物质的量浓度；

电荷守恒

2c(Mg2+) + 3c(Al3+) = c(Cl-)

2*0.2+3c(Al3+) = 1.3

c(Al3+)=0.3mol/L

2、若要使Mg2+转换为Mg（OH）2,并使Mg2+,Al3+分离开来,至少需要4mol/L NaOH溶液多少ml

NaOH涉及到的反应是

Mg2+ + 2OH- = Mg(OH)2

Al3+ + 3OH- = Al(OH)3

Al(OH)3 + OH- = AlO2^- + 2H2O

即,Mg2+ ----2NaOH,Al3+ ------4NaOH

至少,n(NaOH)=0.2*0.2*2 + 0.2*0.3*4=0.32mol

体积,0.08L,即80mL