Ac-+H2OHAc+OH-Kh=Kw/Ka(HAc)[OH-]=[Kh·C(Ac-)]^(1/2)查表找Ka(HAc)=1.76*10^(-5),然后计算若醋酸钠溶液浓度为0.10mol/L[OH-]=7.5*10^(-6),pOH=5.1则pH=8.9
