(例)计算pH=4.0,pH=8.0时0.1mol/LHAc溶液的[HAc], [Ac]. 解：pH=4.0 104 104 6= 10-4+10474 10-35-1007=0.85 6。=0.15 [HAc]=0.085(mol/L),[Ac]=0.015(mol/L) pH=8.0 10-8 10-8+10-479=10-326 60= [HAc]=10426(mol/L),[Ac]=0.1(mol/L) 〈例〉计算 pH = 4.0, pH = 8.0 时 0.1 mol/L HAc 溶液的[HAc], [Ac-]。 解：pH = 4.0 0.15 10 0.85 10 10 10 10 10 0 0.07 3.93 4 4 4.74 4 1 = = = = + = − − − − − − δ δ [HAc] = 0.085 (mol/L), [Ac-] = 0.015 (mol/L) pH = 8.0 1 10 10 10 10 0 3.26 8 4.74 8 1 = = + = − − − − δ δ [HAc] = 10-4.26 (mol/L), [Ac-] = 0.1 (mol/L)