N: EDTA= nCa= mcao, 2nco= 2ndEDTA, VEDTA=25. 26mL, V'EDTA=2650mL Tco·×10000.001000×2500 (I)CEDT 5608×22.29 ×1000=0.02000(mol/L) 20 EDTA EDTA 2×0.02000×24.50×39.10 (2) =5.600% 0.6842×1000 解:V1=2567mL;V2=2476mL:V3=676mL 依题意nB= CEDTA V1,npb+n:= CEDTA V2,ncd= CEDTA VEDTA= CUbaNO3pV3 Cro,1Ma10.02479×2567×2090 =2748% m样×1000 0.4840×1000 PNo23c002174×676×1124 341% 样×1000 04840×1000 形(C11(四3)Mm=0247×2476-002174×676)×2072=1999% m×1000 0.4840×10004. 解：nEDTA= nCa= nCaO，nK= 2nCo= 2n′EDTA，VEDTA=25.26mL，V′EDTA=26.50mL (1) 1000 56.08 22.29 0.001000 25.00 M V T 1000 CaO EDTA CaO × × × = ⋅ × = V CEDTA =0.02000 (mol/L) (2) m 1000 2CEDTA VEDTA MK × ⋅ ′ ⋅ = 样 wK = 0.6842 1000 2 0.02000 24.50 39.10 × × × × =5.600% 5. 解：V1=25.67mL；V2=24.76mL；V3=6.76mL 依题意 nBi= CEDTAV1，nPb+nCd= CEDTAV2，nCd= CEDTAVEDTA=CPb(NO3)2V3 ∴ 0.4840 1000 0.02479 25.67 209.0 m 1000 EDTA 1 Bi × × × = × = 样 C V M wBi =27.48% 0.4840 1000 0.02174 6.76 112.4 m 1000 Pb(NO3 )2 3 Cd × × × = × = 样 C V M wCd =3.41% m 1000 ( ) EDTA 2 Pb(NO3 )2 3 Pb × − = 样 C V C V M wPb = 0.4840 1000 (0.02479 24.76 0.02174 6.76) 207.2 × × − × × =19.99%