a pAg a 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
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~B作图 注：0=0.999 [C1]= 8-1-0.999 +[Ag] 1+0.999 0=1.001 (Ag'7=101- +[C1j 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,c&。=c=0.02mo/L时Y滴定Ca2, 计算0=0.5,1.0,1.5时的pCa。 解：无副反应，Ca+Y=CaY Ig B=10.7 0=05 Ca”1=80-0-002X0-05=667a0 1+0 1+0.5 pCa=2.18 0=1.0 [Ca2*]=[Y],[CaY]=Cca.eo =0.01 [Ca2] [CaY] 0.01 V100.7 =447107 pCa=6.35 或 ca=gB+0)=07+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
0=1.5 Y]= 0.021.5-) 1+1.5 ICaYl=co0 CCa a]上 CaY] 1+0 BY] Bs0- P0-1）10715-1) =10-104 1+0 pCa=10.40 12 10 pCa~o作图 8 pCa 6 2 0 0.0 0.5 1.0 1.5 2.0 0
[ ] [ ] [ ] 10.40 10 10 (1.5 1) 1 ( 1) 1 1 ( 1) 1 1 , [ ] 1 1.5 0.02 (1.5 1) [ ] 1.5 10.4 0 10.7 0 0 = = − = − = + − + = = + = = + × − = = − pCa c c Y CaY Ca c Y CaY c Y Ca Ca Ca β θ θ θ β θ β θ θ pCa ~ θ 作图 0.0 0.5 1.0 1.5 2.0 0 2 4 6 8 10 12 pCa θ