Question

醋酸钠的PH值是多少?

Answer 1

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