c(H+)=1.33*10^-3 mol/L，pH=2.88Ka=c(H+)*c(CH3COO-)/c(CH3COOH)c(CH3COOH)=0.1mol/Lc(H+)=√【Ka*c(CH3COOH)】=√(1.76*10^-5*0.1)=1.33*10^-3 mol/LpH=-lgc(H+)=2.88
