举报
iensteven
PH=2 c1(H+)=c1(CH3COO-)+c1(OH-)=0.01mol/L pH=3 c2(H+)=c2(CH3COO-)+c2(OH-)=0.001mol/L 所以 c1(H+) = 10 c2(H+) 又 c1(H+) ·c1(OH-) =c2(H+)·c2(OH-)=1*10^-14 c1(H+) ·c1(OH-) =c2(H+)·c2(OH-) 把 c1(H+) = 10 c2(H+)代入上式 得 10 c2(H+)·c1(OH-) =c2(H+)·c2(OH-) 两边同除以 c2(H+)得 10 c1(OH-) =c2(OH-) 所以 c1(OH-) = 0.1 c2(OH-) c1(OH-) 1 ———— = —— c2(OH-) 10 酸性溶液中也有OH-,为了精确比较醋酸根,只能考虑OH-