北京化工大学 仪器分析习题解答 董慧茹 编 2 0 1 0 年 6 月 第二章 电化学分析法习题解答 2 5 . 解: pHs = 4.00 , Es = 0.209V pHx = pHs +059.0EsEx (1) pHx1 = 4.00 + 059.0209.0312.0 = 5.75 (2) pHx2 = 4.00 +059.0209.0088.0 = 1.95 (3) pHx3 = 4.00 +059.0209.0017.0 = 0.17 2 6 . 解: [HA] = 0.01mol/L , E = 0.518V [A-] = 0.01mol/L , Φ SCE = 0.2438V E = Φ SCE - Φ 2H+/H2 0.518 = 0.2438 - 0.059 lg[H+] [H+] = ka][][AHA = 01.001.0ka 0.518 = 0.2438 - 0.059 lg01.001.0ka lgka = - 4.647 ka = 2.25×10-5 2 7 . 解: 2Ag+ + CrO24 = Ag2CrO4 [Ag+]2 = ][24CrOKsp AgCrOAgSCEE/42 - 0.285 = 0.2438 - [0.799 + 224)][lg(2059.0CrOKsp] ][lg24CrOKsp = - 9.16 , ][24CrOKsp = 6.93× 10-10 [CrO24 ] = 10121093.6101.1 = 1.59×10-3 (mol/L) 2 8 . 解:pBr = 3 , aBr- = 10-3mol/L pCl = 1 , aCl- = 10-1mol/L 百分误差 = BrClClBraaK,× 100 = 3131010106 × 100 = 60 因为干扰离子Cl-的存在,使测定的aBr- 变为: aBr = aBr +K ClBr .× aCl = 10-3+6×10-3×10-1=1.6×10-3 即 aBr 由 10-3mol/L 变为1.6×10-3mol/L 相差3.0 - 2.8 = 0.2 pBr 单位 2 9 . 解: (1) 由pH―V 图查得 V终点 = 9.10ml 由Δ pH/Δ V―V 图查得 V 终点 = 9.10ml (2) 计算[HA] [HA] =00.1010.91000.0 = 0.09100 (mol/L) (3) 化学计量点的 pH 值 pH Δ 2pH/Δ V2 6.80 +42.5 9.10 -40.0 pH终点 = 6.80 + 0.405.428.61.9×42.5 = 7.98 3 0 . 解: R = G1 = k R1 = 11G = 1k = 0630.00.20 ≈ 317(Ω ) R2 = 21G = 2k = 750.00.20 = 26.7(Ω ) 所对应的电阻值范围是26.7-317Ω 。 3 1 . 解:根据所给电阻值,求出电导值,具体如下: 从右上图可得化学计量点时HCl 的体积为3.25ml 3.25 × 1.00 = Cx × 100 Cx = 0.0325(mol/L) 即溶液中 NaOH 的浓度为0.0325 mol/L。 3 2 . 解: (1) ΦCdCd/2 = Φ θCdCd/2 + ]lg[2059.02Cd = - ...