例〉0.1mo/ LAgNO:滴定01 moll Nac,计算0=0.,0.5, 1,1,5时pAg 解: 0<1[Cj 1+0 0=01(1]=01×(-01=082 1+0.1 K。32×0 [Ag I 3.9×10° Cl]0.082 pAg=8:41 pAg=8:01 6=0.5 pAS 0=1[Ag]=(C]=K DAs pAg- p K=-×949=474 6=1.5[Ag]= (0-1)01X1 002 1+6 1+1.5 pAg=1.70
1.70 0.02 1 1.5 0.1 (1.5 1) 1 ( 1) 1.5 [ ] 0 = = + × − = + − = = + + pAg c Ag Ag θ θ θ 〈例〉0.1 mol/L AgNO3滴定0.1 mol/L NaCl，计算θ = 0.1, 0.5, 1, 1.5时pAg。 9.49 4.74 2 1 2 1 1 [ ] [ ] = = × = = = = + − sp sp pAg pK θ Ag Cl K 0.5 8.01 8.41 3.9 10 0.082 3.2 10 [ ] [ ] 0.082 1 0.1 0.1 (1 0.1) 0.1 [ ] 1 (1 ) 1 [ ] 9 10 0 = = = = × × = = = + × − = = + − < = − − − + − − − pAg pAg Cl K Ag Cl c Cl sp Cl θ θ θ θ 解： θ
DAs pAg Ag 0.050 8.45 0.999 524 0.100 8.41 0.200 8.31 1.001 0.400 8.12 1.010 3.30 0.500 8.01 2.32 0.600 7.89 1.20 2.04 0800 7.54 1.88 0.900 721 0.990 6.19 1.48
θ pAg θ pAg 0.050 8.45 0.999 5.24 0.100 8.41 1.000 4.74 0.200 8.31 1.001 4.25 0.400 8.12 1.010 3.30 0.500 8.01 1.100 2.32 0.600 7.89 1.20 2.04 0.800 7.54 1.30 1.88 0.900 7.21 1.50 1.70 0.990 6.19 2.00 1.48
pAg~作图 注:O=0.99 C1=(1-099 -+lAg I 1+0.999 0=1001 LAg I +Cl I 1+1.001
pAg ~ θ 作图 注：θ = 0.999 1+1.001 [ ] (1.001 1) [ ] 1.001 [ ] 1 0.999 (1 0.999) [ ] 0 0 + − − + + − = = + + − = + − Cl c Ag Ag c Cl Ag Cl θ 0.0 0.5 1.0 1.5 2.0 1 2 3 4 5 6 7 8 9 pAg θ
例)pH=12,ca=c=0.02moL时Y滴定Ca2 计算6=0.5,1.0,15时的pCa 解:无副反应,Ca+Y= CaY lg B=10.7 0=05Ca21= C1(1-6)0.02×1-0 667×0 1+ 1+0.5 0=10[Ca2]=],ICaY]=ca=0.01 2+ 4.47Ⅺ0 107 pCa=6.35 或pCas=(gB+pC)=(107+2)=635
〈例〉pH = 12， 0 0 Ca Y c = c = 0.02 mol/L 时 Y 滴定 Ca2+, 计算θ = 0.5, 1.0, 1.5 时的 pCa。 解：无副反应，Ca + Y CaY lg β = 10.7 (10.7 2) 6.35 21 (lg ) 21 6.35 4.47 10 10 [ ] 0.01 [ ] 1.0 [ ] [ ], [ ] 0.01 2.18 6.67 10 1 0.5 0.02 (1 0.5) 1 (1 ) 0.5 [ ] 7 10.7 2 . 2 3 0 2 = + = + = = = = = × = = = = = = × + × − = + − = = + − + + − pCa pC pCa CaY Ca Ca Y CaY c pCa c Ca eq CaY Ca eq Ca β β θ θ θ θ 或