例2.计算0.050moL1A1CL3溶液中C和A3+的活度。 解: 1=2∑cZ-0030x3+3x050x)=030ml-L 查附录表3,C的a=300,A13+的8=900;B=0.00328 =0.18 Ycr=0.66ar=Ycr[C1]=0.66×3×0.050=0.099(mo1.L1) -lgy4=0.512×32 √0.30 1+00328×900x030 =0.96 Y4=0.11aA.=y[A1+]=0.11x0.050=0.0055m01E6 例2.计算0.050 mol·L-1AlCl3溶液中Cl-和Al3+的活度。 解: 查附录表3,Cl- 的å =300,Al3+的å =900;B =0.00328 1 2 2 i i i I c Z = 1 [ ] 0.66 3 0.050 0.099( ) Cl Cl a Cl mol L − − g − − = = = 1 2 2 1 (0.050 3 3 0.050 1 ) 0.30( ) 2 mol L− = + = 2 0.30 lg 0.512 1 0.18 1 0.00328 300 0.30 Cl g − − = = + 0.66 Cl g − = 3 2 0.30 lg 0.512 3 0.96 1 0.00328 900 0.30 Al g + − = = + 3 0.11 Al g + = 3 3 3 1 [ ] 0.11 0.050 0.0055( ) Al Al a Al mol L + + g + − = = =