Question

A 0.21-mol sample of a diprotic acid, H2A, is dissolved in 250 mL of water. The Ka1 of this acid is 1.0 × 10–5, and Ka2 is 1.0 × 10–10. Calculate the concentration of A2– in this solution.

a) 0.84 M
b) 1.0 × 10–5 M
c) 1.4 × 10–3 M
d) 1.0 × 10–10 M
e) 2.9 × 10–3 M

Answer 1

To calculate the concentration of A2− for a diprotic acid H2A in solution, we need to consider both dissociation steps and their equilibrium expressions. However, the concentration of A2− cannot be determined from the given information without knowing the pH of the solution or the concentration of H+.

Step-by-step explanation:

To calculate the concentration of A2− in a solution of a diprotic acid H2A, we need to consider the two dissociation steps of the acid and their respective dissociation constants (Ka1 and Ka2). Since the Ka1 is significantly larger than Ka2, we can assume that the first dissociation reaches equilibrium much faster and to a greater extent than the second. Therefore, the concentration of HA− produced from the first dissociation will be close to the initial concentration of H2A. For the second dissociation, we will have a very small concentration of A2− relative to HA−. Thus, we can write the equations for the dissociation and equilibrium expressions:

H2A ⇌ HA− + H+

HA− ⇌ A2− + H+

The equilibrium expression for the second dissociation is:

Ka2 = [A2−][H+]/[HA−]

Since we know Ka2 and we assume [HA−] is approximately equal to the initial concentration of H2A, we can solve for [A2−]. However, without the pH of the solution or the concentration of H+, we cannot determine [A2−] with the information given. Additional data is needed to complete the calculation.