Question

16.在 0.5 mol⋅dm －3 MgCl2溶液中加入等体积的 0.10 mol⋅dm －3

Answer 1

混合后[Mg2+] = 0.25； [NH3] = 0.05 ；[NH4+] = 0.01 mol/L

由[OH-] = Kb（NH3）[NH3]/[NH4+] = 1.8*10^-5*0.05 / 0.01 = 9*10^-5

[Mg2+] [OH-] ^2 > Ksp[Mg(OH)2] ,有沉淀析出.

加入固体NH4Cl,让Mg(OH)2恰好沉淀,即求[NH4+]则：

[OH-] = { Ksp[Mg(OH)2] / [Mg2+] }^ 1/2 = Kb（NH3）[NH3]/[NH4+]

求出加了NH4Cl以后的[NH4+] 的浓度.

[1.2×10^-11/0.25]^1/2 = 1.8*10^10-5*0.05/[NH4+][NH4+]=0.13(mol/L)

等体积稀释后,原来[NH4+] 的浓度是0.01mol/L,实际还需[NH4+]=(0.13-0.01) = 0.12(mol/L)换算成稀释前的NH4Cl,浓度需乘以2.0.12*2*53.5 = 12.84(g)