A 0.26 M solution of a weak acid HA is 0.35 % ionized. What are the H+

, A-
, and HA concentrations
at equilibrium? What is Ka?

To find the H+, A-, and HA concentrations at equilibrium, we need to use the ionization percentage and the initial concentration of the weak acid HA.

Let's assume the initial concentration of HA is 'C' M.

Given that the solution is 0.35% ionized, we can conclude that 0.35% of the initial concentration dissociates into H+ and A-. Therefore, the concentration of H+ and A- can be calculated as:

[H+] = 0.35% × C = (0.35/100) × C M
[A-] = 0.35% × C = (0.35/100) × C M

Now, since HA is a weak acid, we assume that it dissociates incompletely. Thus, the remaining concentration of HA at equilibrium can be calculated as:

[HA] = initial concentration - dissociated concentration
= C - (0.35/100) × C
= (100 - 0.35)/100 × C
= 0.9965C M

So, at equilibrium:
[H+] = (0.35/100) × C M
[A-] = (0.35/100) × C M
[HA] = 0.9965C M

To find Ka, we can use the equilibrium concentrations of HA, H+, and A-.

Ka (acid dissociation constant) is defined as the ratio of [H+] × [A-] to [HA].

Ka = [H+] × [A-] / [HA]
Ka = ((0.35/100) × C) × ((0.35/100) × C) / (0.9965C)

Simplifying the expression:
Ka = ((0.35/100)^2 × C^2) / (0.9965C)
Ka = 0.1225 / 0.9965
Ka ≈ 0.123 M

So, the concentrations at equilibrium are:
[H+] ≈ (0.35/100) × C M
[A-] ≈ (0.35/100) × C M
[HA] ≈ 0.9965C M

And the value of Ka is approximately 0.123 M.