To calculate the pKa of the acid, we first need to determine the concentration of the acid that is ionized. In an acidic solution with a pH of 3.478, we can use the formula pH = -log[H+], where [H+] represents the concentration of hydrogen ions in the solution.
Therefore, 10^(-3.478) = [H+], which is approximately 3.60 x 10^(-4) M. Since the acid is 3.478% ionized, the concentration of ionized acid is 0.03478 * 3.60 x 10^(-4) = 1.25 x 10^(-5) M.
Finally, we can calculate the pKa of the acid using the formula pKa = -log(Ka), where Ka is the acid dissociation constant. Assuming that most of the acid dissociates into H+ and A-, the equilibrium equation for the dissociation is as follows:
HA = H+ + A-
Ka = [H+][A-]/[HA]
Since only 3.478% of the acid is ionized, we know that [H+] = 1.25 x 10^(-5) M and [A-] = 1.25 x 10^(-5) M as well. We can assume that the initial concentration of HA is equal to the initial concentration of the acid, so [HA] = 1.25 x 10^(-5)/0.03478 = 3.59 x 10^(-4) M.
Inserting these values into the equation for Ka and taking the negative logarithm gives:
Ka = (1.25 x 10^(-5))^2 / 3.59 x 10^(-4) = 4.35 x 10^(-10)
pKa = -log(4.35 x 10^(-10)) = 9.362
Therefore, the pKa of the acid is approximately 9.362.
In an acidic solution with a pH of 3.478% of the acid is ionized ,what is the pka of the acid?
1 answer