2、一元弱酸弱碱的电离平衡 例:求0.10 mol/dm3HAc水溶液的酸度,已知 K=1.76×105 解:HAc+H2OH3O++Ac 0.10-X 设H3O+的平衡浓度为x 则Ka=[H3O+]AcT/[HAcl x2/(0.10-x)≈x2/0.10 1.76×105 X=[H3O=(0.10×1.76×105)05 =13×103( mold3) 电离度a=x/C0=13×103/0.10 0.013(1.3%)2、一元弱酸弱碱的电离平衡 例：求0.10 mol/dm3 HAc水溶液的酸度，已知 Ka = 1.76 × 10–5 解：HAc + H2O H3O+ + Ac– 0.10-x x x 设 H3O+ 的平衡浓度为 x 则 Ka = [H3O+ ] [Ac– ] / [HAc] = x2 / (0.10-x)  x 2 / 0.10 = 1.76 × 10–5 x = [H3O+ ] = (0.10 × 1.76 × 10–5 ) 0.5 = 1.3× 10–3 (mol/dm3 ) 电离度 = x / C0 = 1.3× 10–3 / 0.10 = 0.013 (1.3%)