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两种金属离子同时被 滴定时的终点误差计算

两种金属离子同时被 滴定时的终点误差计算. 丁雷. 单一离子滴定的终点误差. pM’ sp =(pc sp +lgK’) Δ pM’=pM’ ep -pM’ sp E t =([Y]’ ep -[M]’ ep )/c M ep =(10 - Δ pM’ [M]’ sp -10 - Δ pY’ [Y]’ sp =(10 Δ pM -10 - Δ pM )/(K’c M sp ) 1/2. 两种离子滴定的终点误差. [Y’] sp =[M’] sp +[N’] sp Δ pY’= Δ p([M’]+[N’])

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两种金属离子同时被 滴定时的终点误差计算

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  1. 两种金属离子同时被滴定时的终点误差计算 丁雷

  2. 单一离子滴定的终点误差 • pM’sp=(pcsp+lgK’) • ΔpM’=pM’ep-pM’sp • Et=([Y]’ep-[M]’ep)/cMep =(10-ΔpM’[M]’sp-10-ΔpY’[Y]’sp =(10ΔpM-10-ΔpM)/(K’cMsp)1/2

  3. 两种离子滴定的终点误差 • [Y’]sp=[M’] sp +[N’] sp • ΔpY’=Δp([M’]+[N’]) • Y+MIn=MY+In;Y+NIn=NY+In • KM=lgK’MY-lgK’M-In • KN=lgK’NY-lgK’N-In

  4. pM’ep=lgK’M-In • [N’]ep=(k’MYcNep)/([M’]epk’NYcMep) • pN’ep=lgk’NY-lgk’MY+pM’ep • pY’ep=lgk’MY+pcMep-pM’ep • Et=([Y’]ep-[M’]ep-[N’]ep)/ cM+Nep

  5. 例题 • 一种仅由Zn,Fe,Cd组成的合金经消解后配成的试液中cZn,cFe,cCd均约为0.02mol/L.今平行移取25.00mL试液,分别作以下操作: • A.加入酒石酸令[Tart]ep=0.02mol/L,以六次 甲基四胺为缓冲溶液在pH=6.0处用 0.04025mol/LEDTA滴定,平均消耗24.15mL

  6. B.加入磺基水杨酸[L]ep=1x10-6mol/L,以六次甲基四胺为缓冲溶液在pH=6.0处用同浓度EDTA滴定,平均消耗26.84mLB.加入磺基水杨酸[L]ep=1x10-6mol/L,以六次甲基四胺为缓冲溶液在pH=6.0处用同浓度EDTA滴定,平均消耗26.84mL • C.调pH=8,加入KCN，加入一定量 甲醛，此时有[CN-]=2x10-5mol/L,调pH=6.0,以六次甲基四胺为缓冲溶液用同浓度EDTA滴定,平均消耗25.33mL

  7. (1)求出试液中cZn,cFe,cCd • (2)若系统误差仅由终点误差引起，求出(1)中各结果的系统误差

  8. (lgKZnY=16.5,lgKFeY=14.32, lgKCdY=16.46,α(H)6.0=4.65,

  9. 解答 • (1)cZn=cEDTAx(VB+VC-VA)/2 • =0.02256mol/L • cCd=cEDTAx(VA+VC-VB)/2 • =0.01823mol/L • cFe=cEDTAx(VA+VB-VC)/2 • =0.02066mol/L

  10. (2)A:α(Zn) =β1[Tart]ep+ β2[Tart]ep2+1 =104.92 α(Cd) = β1[Tart]ep+1 =101.13 αY(Zn) =KZnYcZnep/ α(Zn)+1=109.58 α(Y) = αY(Zn)+αY(H)6.0-1=109.58 lgK’CdY=lgKCdY-lgα(Y)-lg α(Cd) = 5.75

  11. lgK’FeY=lgKFeY-lgα(Y)= 4.74 • pCd’ep=lgK’Cd-In-lgα(Cd) =4.37 • pFe’ep=lgk’FeY-lgk’CdY+pCd’ep=3.36 • pY’ep=lgk’CdY+pcCdep-pCd’ep=3.38 • EtA=([Y’]ep-[Cd’]ep-[Fe’]ep)/ cCd+Feep • =-0.0031

  12. B:α(Zn) =β1[L]ep+β2[L]ep2+1=100.34 α(Cd)=β1[L]ep+β2[L]ep2+1=1017.08 α(Fe) =β1[L]ep+ β2[L]ep2+1=100.26 αY(Cd)=KCdYcCdep/ α(Cd)+1=1 α(Y) = αY(Cd)+αY(H)6.0-1=104.65 lgK’ZnY=lgKZnY-lgα(Y)-lg α(Zn) = 11.51

  13. lgK’FeY=lgKFeY-lgα(Y)-lgα(Fe) • = 9.41 • pZn’ep=lgK’Zn-In-lgα(Zn) =6.16 • pFe’ep=lgk’FeY-lgk’ZnY+pZn’ep=4.06 • pY’ep=lgk’ZnY+pcZnep-pZn’ep=7.35 • EtB=([Y’]ep-[Zn’]ep-[Fe’]ep)/ cZn+Feep • =-0.0044

  14. C:α(Zn) =β4[CN]ep+1=1α (Cd)=β1[CN]ep+β2[CN]ep2+β3 [CN]3ep+β4[CN]ep4+1=100.99 α(Fe) =β4[CN]4ep+ 1=1015 αY(Fe)=KFeYcFeep/ α(Fe)+1=1 α(Y) = αY(Fe)+αY(H)6.0-1=104.65 lgK’ZnY=lgKZnY-lgα(Y)-lg α(Zn) = 11.85

  15. lgK’CdY=lgKCdY-lgα(Y)-lgα(Cd) • = 10.82 • KZn=lgK’ZnY-lgK’Zn-In=5.35 • KCd=lgK’CdY-lgK’Cd-In+lgα(Cd) =6.31 • pZn’ep=lgK’Zn-In=6.50 • pCd’ep=lgk’CdY-lgk’ZnY+pZn’ep=5.47 • pY’ep=lgk’ZnY+pcZnep-pZn’ep=7.35 • EtC=([Y’]ep-[Cd’]ep-[Zn’]ep)/ cCd+Znep • =-0.00018

  16. ER(Zn)/R=(EtBVB+EtCVC-EtAVA) / (VB+VC-VA) =-0.0017 ER(Cd)/R=(EtAVA+EtCVC-EtBVB) / (VA+VC-VB) =0.0017 ER(Fe)/R=(EtBVB+EtAVA-EtCVC) / (VB+VA-VC) =0.0073

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