pa,=can+lg)=2+1=65 APZn pZn pZngp 5.7-6.5=-0.8 E=10-104 ×1009%=100-100 VCin.p·knr V103x10=0.019% 3解:Tm,1 5 X CKMmO,×Mr×10- =5×0.0200×55.84×103 =5.584×10-gmL1 5xco,10 TKMO IF30 1 =5x020x15969x10- =7.984×10-3gmL 4解:}KK+K2 Vc+KI 因为 c》20Ka1,K2>20K 所以[H]=√Ka1·K2=V4.2×10×5.6×101=4.8×10°mol.L pl=8.325.7 6.5 0.8 (2 11) 6.5 2 1 ( lg ) 2 1 . ep sp sp Zn sp ZnY PZn pZn pZn pZn pc k 0.019% 10 10 10 10 100% 10 10 2.00 11 0.8 0.8 , Zn sp ZnY pM pM t c k E 3 1 3 3 n e / 5.584 10 g mL 5 0.0200 55.84 10 1 5 c 10 3. 4 4 - - - = 解: = = KM O F KMno Fe M T 3 1 3 3 n e / 7.984 10 g mL 159.69 10 2 1 5 0.0200 1 10 2 1 5 c 4 2 3 4 2 3 - - - = = = KM O F O KMnO Fe O M T a1 a1 a2 c c 4. : K K K K H W + ( + ) 解 因为 c 》20Ka1,Ka2>20KW 所以 [H+]= 7 11 9 1 1 2 4.2 10 5.6 10 4.8 10 . Ka Ka mol L pH=8.32