Final answer:
By calculating the molar concentrations of hydronium ions, [H+], and conjugate base, [A-], from the degree of dissociation and initial concentration of HA, and then using the Ka equation, we find that the Ka for the weak acid is 8.9 x 10^-5.
Step-by-step explanation:
To calculate the acid ionization constant (Ka) for a weak acid that undergoes dissociation, we can use the formula:
Ka = ([H+][A-]) / [HA], where [H+] is the concentration of hydronium ions, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the un-dissociated acid.
The question states that a 0.13 M solution of the weak acid is 2.6% dissociated. We interpret this percentage as the ratio of the acid that has dissociated compared to the initial concentration. Therefore, we can find the concentrations of [H+] and [A-] by the following calculation:
[H+] = [A-] = 0.13 M * 2.6% = 0.13 M * 0.026 = 0.00338 M
The concentration of [HA] remaining is the initial concentration minus the concentration that has dissociated:
[HA] = 0.13 M - 0.00338 M = 0.12662 M
Now, we plug the values back into the Ka equation:
Ka = (0.00338 M2) / 0.12662 M = 8.9 x 10-5
Therefore, the Ka for the weak acid (HA) is 8.9 x 10-5.