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