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=1.701.70 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