HCl + NaOH = NaCl + H2O
1MoL 1MoL
9×0.001×0.1MoL 10×0.001×0.1MoL
→NaOH过量 其剩余量为1×0.001×0.1MoL
故其浓度为 1×0.001×0.1/((10+9)×0.001)=5.26×10*（-3）
PH=2.28