[CO]=K2[HCOV[H'] 62,00X103×10103107=935X10 2解：pE=pE+1 nIg Pco2a125[Hr]VPc012s =2.87+lg(0.35p)0125×1070.65p0125 =.416 Eh=0.059pE=-4.16X0.059 -0.25 3解：K=-K+a.(KHr+KfOH]) =1.6+0.1(0+4.9×107×1056y ≈1.6d1[CO3 - ] = K2[HCO3 - ]/[H+ ] = 2.00×10-3×10-10.33/10-7= 9.35×10-7 mol/l [OH- ] =10-14/ 10-7 =10-7mol/l 2 解：pE=pE0+1/n lg PCO2 0.125[H+ ]/PCH40.125 =2.87+lg（0.35p）0.125×10-7 /(0.65p)0.125 =-4.16 Eh=0.059pE=-4.16×0.059 =-0.25 3 解：Kh =[Kn+  w(Ka[H+ ]+ Kb[OH- ])] =1.6+0.1（0+4.9×10-7×10-5.6）  1.6d-1