【用戶】Ivan Lia
【年級】高三下
【評論內容】[CH3COOH] = 6/60/1 = 0.1M[CH3COO-] = 8.2/82/1 = 0.1M加NaOH:[OH-]:10-3M,[H+]:10-11M CH3COOH→ H+ +CH3COO-初: 0.1 10-11 0.1平: (0.1+x) (x+10-11) (0.1+x) Ka=10-5=[CH3COO-][H+]/[CH3COOH]=0.1x/0.1 x=10-5pH ≅ pKa = -log(10-5) = 5(等濃度共軛酸鹼對,可知形成緩衝溶液,加入少量強酸或強鹼不會影響溶液的pH值...