用100ml0.1mol/l的naoh滴定0.1mol/l的盐酸，需要饱和氢氧化钠溶液多少毫升

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正0.1%：溶液中OH-的浓度C=0.1mol/L*(0.1%*100mL)/(100mL+100mL)=5*10^(-5)mol/LpOH=-lg5*10^(-5)=5-lg5=4.3 pH=14-4.3=9.7负0.1%：氢氧化钠量不足,盐酸有剩余,剩余量n=0.1mol/L*0.1L-0.1mol/L*[0.1L*(1-0.1%)=0.00001molC(H+)=N/(0.1L+0.1L)=5*10^(-5)mol/L所以pH=-lgC(H+)=4.3