If we look at this as H2T, then
H2T ==> H^+ + HT^-
HT^-==> H^+ + T^-2
Ignoring k2 for the moment,
H^+ from k1 is
(H^+)(HT^-)/(H2T).
Substituting and solving for (H^+) gives us (H^+)=0.00214 M.
(HT^-) = 0.00214 EXCEPT that k2 it is so close that it will ionize almost as easily as k1 will; therefore, we will get more (H^+) from k2 which probably makes (H^+) the most concd of the ions @ about 0.0021+. The unionized acid, H2T, is about 0.098 which should make it the highest of the species. That leaves HT^- and T^-2 to be placed. The HT^- is about 0.0021- ionization products and T^-2 is about k2. So I'm making an educated guess that the next to the highest is H3O^+. I'm ranking them
H2T, H3O^+, HT^-, T2^-, OH^- which is B, A, C, D, E. Check my thinking.
In a 0.10 M solution of glutaric acid, HO2C(CH2)3CO2H (K1 = 4.6 ´ 10-5, K2 = 3.9 ´ 10-6), the species present in the next to highest concentration is
A. H3O+(aq).
B. HO2C(CH2)3CO2H(aq).
C. HO2C(CH2)3COO-(aq).
D. -O2C(CH2)3COO-(aq).
E. OH-(aq).
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