Chem: pH of acids and bases

Calculate the concentrations of all species in a 0.320-M solution of H2X if K1 = 6.31E-06 and K2 = 3.71E-09

[H2X] =

[H3O1+] =

[HX1-] =

[X2-] =

Work:
H2X + H2O --> HX- + H3O+
.320 M

This is literally the only thing I know. How can there be two different Ks??

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  1. You started very well. Just go from there. There can be two Ka values because there are two H ions to dissociate.

    .......H2X + H2O ==> H3O^+ + HX^-
    I....0.320............0.......0
    C.....-x..............x.......x
    E...0.320-x...........x.......x

    Now Ka1 = (H3O^+)(HX^-)/(H2X)
    Substitute the E line into the Ka1 expression and solve for x = (H3O^+) = (HX^-). You get (H2X) from 0.320-x = ?

    Then for ka2, it is
    .....HX^- + H2O ==> H3O^+ + X^2-
    You don't need to fill this in.
    Ka2 = (H3O^+)(X^2-)/(HX^-)
    You just solved from Ka1 that (H3O^+) = (HX^-). One of those is in the numerator of Ka2 and the other one in the denominator of Ka2 so they cancel which leaves Ka2 = (X^2-).
    Case closed.

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