NH3+H2O<=>NH4++OH-Kb=[OH-][NH4+]/[NH3]=约=[OH-]n(NH4+)/n(NH3)则n(NH4+)/n(NH3)=Kb/[OH-]=1.8*10^-5/10^-5=1.8因此氯化铵的物质的量=0.18mol质量=0.18*53.5=9.63g
