What is the concentration of x2? in a 0.150 m solution of the diprotic acid h2x? for h2x, ka1=4.5?10?6 and ka2=1.2?10?11?

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