反应前:0.05Lx0.15mol/L=0.0075molMH3;0.2mol/Lx0.05L=0.01molNH4Cl;0.1mol/Lx0.001L=0.0001molHClPKa(NH4Cl)=10PH=PKa+log[NH3]/[NH4+]=10+log(0.0075/0.01)=10-0.12=9.875NH3+HCl---NH4Cl反应后:0.0074molNH3;0.0101molNH4ClPH=PKa+log[NH3]/[NH4+]=10+log(0.0074/0.0101)=10-0.135=9.865