Question

大学化学题求解

10mL 0.2mol/L MgCl2 和 10mL 0.2mol/L 氨水混合，并向溶液中投入固体NH4Cl，至少要加入多少克NH4Cl方不至产生Mg(OH)2沉淀？
Ksp=1.2x10^-11，Kb=1.77x10^-5

Answer 1

[Mg2+][OH-]^ 2= 1.2X10^-11 => (0.1)[OH-]^2= 1.2X10^-11 => [OH-] = 1.1x10-7
不产生Mg(OH)2沉淀 [OH-] =< 1.1x10-7

[NH4+][OH-] [NH4+] (1.1x10-7) [NH4+]
__________ = _______________ = Kb =1.77x10^-5 => _________ = 161
[NH3] [NH3] [NH3]

当投入大量固体NH4Cl [NH3] =0.1 => [NH4+] =16.1mol/L
1mol NH4Cl 重 14+1x4+35.5 = 53.5 g , 20 ml 16.1mol/L溶液要投入固体NH4Cl 0.322mol = 17.2g

Answer 2

[Mg2+]=0.1mol/L
[OH-]=1.1*10^-5mol/L
[NH4+]=Kb[NH3]/[OH-]=0.16mol/L
m(NH4Cl)=0.02*0.16*53.5=0.17g

Answer 3

既然是临界点，把K带进去算得了
根据Ksp=[Mg2+][OH-]^2
得到[OH-]=
然后根据Kb=[NH4+][OH-]
得到[NH4+]=
n=cV得到总的n(NH4+)
减去溶液里原有的NH4+

最后得到 n=@#%$%&%^*#