2.nKF(c1os-·Voy-1/6·c2o32·Ws2032)×5=0.7242mmo cK=0.7242/25.00=0.02897mo/L 3.A=-gT=-lg0.300=0.523 0.523 E =1.5×04 Lmol.cn bc2.0×1.7×10 x=②x)m=2054 =20.54% R=20.56-2048=0.12% d ∑以1|0.06+0.01+0.040.06+001+0.06 =0.046% CV=÷×100=_5 x100=0.22% 20.54 E Ea=20.54-2045=0.09% 72045×100=044% 分析化学试卷答案第2页共2页分析化学试卷答案 第2页 共 2 页 2. n KI= (c IO3-·VIO3-－1/6·cS2O32-·VS2O32- )×5＝0.7242 mmol cKI= 0.7242/25.00=0.02897 mol/L 3. A = −lg T = −lg 0.300 = 0.523 4 -1 -1 5 0.523 1.5 10 L mol cm 2.0 1.7 10 A bc  − = = =    4． (2