Question

.计算0.1mol/L的HAc溶液的[H+]和PH值.Ka＝1.76×10-5

Answer 1

c(H+)=1.33*10^-3 mol/L,pH=2.88

Ka=c(H+)*c(CH3COO-)/c(CH3COOH)

c(CH3COOH)=0.1mol/L

c(H+)=√【Ka*c(CH3COOH)】=√(1.76*10^-5*0.1)=1.33*10^-3 mol/L

pH=-lgc(H+)=2.88