The first dissociation for H2X: H2X +H2O ↔ HX + H3O initial 0.15 0 0 change -X +X +X at equlibrium 0.15-X X X because Ka1 is small we can assume neglect x in H2X concentration Ka1 = [HX][H3O]/[H2X] 4.5x10^-6 =( X )(X) / (0.15) X = √(4.5x10^-6*0.15) ∴X = 8.2 x 10-4 m ∴[HX] & [H3O] = 8.2x10^-4 the second dissociation of H2X HX + H2O↔ X^2 + H3O 8.2x10^-4 Y 8.2x10^-4 Ka2 for Hx = 1.2x10^-11 Ka2 = [X2][H3O]/[HX] 1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4) ∴y = 1.78x10^-5 ∴[X^2] = 1.78x10^-5 m
