A 0.15M solution of weak triprotic acid is adjusted (base added) so that the pH is 9.27(you can assume minimal volume change). The triprotic acid(H3A) has the following acid ionization constants: Kal=1.0*e-3, ka2=1*e-8,ka3=1*e-12. which species given below is present in the highest concentration in this pH 9.27 solution?

1)H3O+
2)H2A-
3)HA-2
4)A-3
THAT'S the answer choices I think is HA-2, IT IS RIGHT

See your post below.

To determine which species is present in the highest concentration in a pH 9.27 solution of the weak triprotic acid, we need to consider the acid ionization constants and the given pH value.

The triprotic acid (H3A) can donate three protons (H+) in successive steps, each with its own acid ionization constant (Ka). As the pH of the solution is 9.27, we can assume that the concentration of H3O+ (hydronium ion) is higher than 10^-9.27 M.

To determine which species is present in the highest concentration, we must consider the dissociation reactions and equilibrium concentrations at pH 9.27.

The dissociation reactions for the triprotic acid can be written as follows:

H3A ⇌ H2A- + H+
H2A- ⇌ HA-2 + H+
HA-2 ⇌ A-3 + H+

At each step, the acid ionization constant (Ka) comes into play. Given that H3A is a weak acid, the concentrations of its conjugate base (HA-2 and A-3) will be higher than the concentration of H3A itself.

To determine which species is present in the highest concentration, note the following:

The pH of 9.27 indicates the presence of a higher concentration of hydronium ions (H3O+). As we move from H3A to H2A-, the concentration of H3A decreases, and the concentration of H2A- increases. Similarly, as we move from H2A- to HA-2, the concentration of H2A- decreases, and the concentration of HA-2 increases. Finally, from HA-2 to A-3, the concentration of HA-2 decreases, and the concentration of A-3 increases.

Therefore, it can be concluded that the species present in the highest concentration in this pH 9.27 solution is HA-2 (choice 3).

You are correct in selecting HA-2 as the answer.