Answer:
From the equilibrium pH, we can find the concentration of H+ ions in solution using the relation:
[H+] = 10^(-pH)
[H+] = 10^(-2.546) = 2.177 × 10^(-3) M
Now we can use the fact that the acid is a weak acid and only partially dissociates to form H+ ions and its conjugate base. Therefore, we can assume that [HA] at equilibrium is equal to the initial concentration of the acid minus the concentration of H+ ions that were produced from the dissociation of the acid.
[HA] at equilibrium = initial concentration of acid - [H+]
[HA] at equilibrium = 1.497 M - 2.177 × 10^(-3) M
[HA] at equilibrium = 1.497 M (since the concentration of H+ ions is negligible compared to the initial concentration of the acid)
Now we can plug in the values we obtained into the Henderson-Hasselbalch equation:
2.546 = pKa + log([A-]/[HA])
2.546 = pKa + log(0/[HA])
2.546 = pKa - log([HA])
log([HA]) = pKa - 2.546
[HA] = 10^(pKa - 2.546)
Since we assumed that the concentration of the conjugate base at equilibrium is negligible, we can assume that [A-] ≈ 0.
Therefore, we have:
pKa = log([HA]/0) + 2.546
pKa = log([HA]) + 2.546
pKa = log(1.497) + 2.546
pKa = 0.174 + 2.546
pKa = 2.72
Therefore, the pKa of the acid is approximately 2.72.