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?
A pH of 9.27 is between the 1st and 2nd equivalence points; therefore, the solution is a mixture of H2A^- and HA^2-.
I would use the Henderson-Hasselbalch equation and solve for the Base/Acid ratio. That should tell you which species predominates.
1)H3O+
2)H2A-
3)HA-2
4)A-3
THAT'S the answer choices I think is HA-2, IT IS RIGHT
I agree.
To determine which species is present in the highest concentration in the pH 9.27 solution of the weak triprotic acid, we need to consider the acid dissociation reactions and the pH range where each dissociation occurs.
Here are the dissociation reactions and their respective equilibrium constants (K values):
1st dissociation: H3A ⇌ H+ + HA-
K1 = 1.0 x 10^(-3)
2nd dissociation: HA- ⇌ H+ + A2-
K2 = 1.0 x 10^(-8)
3rd dissociation: A2- ⇌ H+ + A3-
K3 = 1.0 x 10^(-12)
At pH 9.27, the concentration of H+ ions is low and will primarily be consumed by reacting with the bases, resulting in a highly basic solution. Therefore, the H+ concentration is negligible and can be ignored.
To identify the species present in the highest concentration, we need to determine which dissociation reactions are significant at this pH.
At very basic conditions (pH above the pKa of the acid's conjugate base), only the first dissociation is significant. At pH 9.27, the first dissociation has already occurred:
H3A ⇌ H+ + HA-
Therefore, the major species present in the solution at pH 9.27 is the conjugate base HA- (since the H+ concentration is negligible).
Thus, the correct answer is choice A: HA-.