Ky=Kpc2a草酸=1.42×103 S=Kp/C草酸=1.42×10mol/L 2. 1 Sb2S3 92 Sbe3S96Fe*g6/5 KMnO nsb=2×5/6×0.002000×30.20=0.1007mmol Sb%=(0.1007/1000×121.70.2357×100%=520% 3.5.00=4.75+lg([ Ac-(HAc]) g(Ac]/HAc])=0.25 [ Ac/HAc]=1.78[Ac]=1.78×020=0.36 nAaC3H2o=0.36×10×1361=49(g VHAc=1.0×020/20=0.10(L)=100(mL) 2500=01635m01 0.02002×10.15 c(Cd) =0.00813mol 25.00 0.02015×30.16-0.02002×10.15 c(Pb) =0.016l8mol 25.00 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 Ksp ’ =KspCa草酸=1.42×10-8 S=Ksp ’ /C 草酸＝1.42×10-7 mol/L 2. 1 Sb2S3 ≌2 Sb≌3S≌6Fe2+≌6/5 KMnO4 nSb= 2×5/6×0.002000×30.20=0.1007 mmol Sb%= (0.1007/1000)×121.7/0.2357×100％＝5.20% 3. 5.00＝4.75＋lg([Ac-]/[HAc]) lg([Ac- ]/[HAc])＝0.25 [Ac- ]/[HAc]＝1.78 [Ac- ]＝1.78×0.20＝0.36 ∴ mNaAc·3H2O＝0.36×1.0×136.1＝49(g) V HAc＝1.0×0.20/2.0＝0.10(L)＝100(mL) 4. 0.02015 20.28 -1 (Bi) 0.01635mol L 25.00 c  = = 0.02002 10.15 1 (Cd) 0.00813mol L 25.00 c  − = = 0.02015 30.16 0.02002 10.15 1 (Pb) 0.01618mol L 25.00 c  −  − = =