HAc Ac- + H+因为Ka＝1.8X10-5Ka=[H+][Ac-]/(0.1-[H+])=[H+]^2/(0.1-[H+])=1.8x10^-5[H+]=0.00133 mol/LpH=2.88
分析化学课后习题及答案酸碱滴定法计算0.1摩尔每升醋酸钠水溶液的PH值
HAc Ac- + H+因为Ka＝1.8X10-5Ka=[H+][Ac-]/(0.1-[H+])=[H+]^2/(0.1-[H+])=1.8x10^-5[H+]=0.00133 mol/LpH=2.88