For a weak diprotic acid H2X, the ionization steps are:
H2X⇌H++HX− with Ka1=2.5×10−8
HX−⇌H++X2− with Ka2=1.0×10−13
Since Ka1≫Ka2, the concentration of H+ is primarily determined by the first ionization.
[H+]≈Ka1⋅C=2.5×10−8×0.1=2.5×10−9=5×10−5M
Also, from the first ionization, [HX−]≈[H+]=5×10−5M
For the second ionization:
Ka2=[HX−][H+][X2−]
Substituting the values: 1.0×10−13=5×10−5(5×10−5)[X2−]
Thus, [X2−]=Ka2=1.0×10−13M
To express this in the form x×10−15M:
1.0×10−13=100×10−15M
The value is 100.