To solve this question, we'll use the information provided and the concept of equilibrium concentrations.
Let's assume that the final equilibrium concentration of HA is x M. After dissociation, the concentration of H+ and A- will also be x M each.
We can set up an ICE (Initial, Change, Equilibrium) table to track the changes in concentration:
HA <-> H+ + A-
Initial 0.30 0 0
Change -x +x +x
Equilibrium 0.30 - x x x
Now, we can use the equilibrium constant expression (K) to set up an equation:
K = [H+][A-] / [HA]
K = 5.0 x 10^-9
Plugging in the equilibrium concentrations, we get:
5.0 x 10^-9 = x * x / (0.30 - x)
Simplifying the equation, we get:
5.0 x 10^-9 = x^2 / (0.30 - x)
Next, we can cross-multiply and rearrange the equation:
5.0 x 10^-9 * (0.30 - x) = x^2
1.5 x 10^-9 - 5.0 x 10^-9x = x^2
Rearranging and simplifying, we get a quadratic equation:
x^2 + 5.0 x 10^-9x - 1.5 x 10^-9 = 0
At this point, we can solve the quadratic equation using the quadratic formula. However, this equation has coefficients involving very small numbers ("10^-9"), which means that x will be much smaller than 0.30. Therefore, we can assume that "0.30 - x" is approximately equal to 0.30:
x^2 + 5.0 x 10^-9x - 1.5 x 10^-9 ≈ 0
Simplifying further:
x^2 + 5.0 x 10^-9x - 1.5 x 10^-9 ≈ 0
x(x + 5.0 x 10^-9) - 1.5 x 10^-9 ≈ 0
Approximating 5.0 x 10^-9 to 0 gives us:
x^2 - 1.5 x 10^-9 ≈ 0
Solving this simplified equation, we can take the square root of both sides:
x ≈ √(1.5 x 10^-9)
Calculating this value, we find:
x ≈ 3.87 x 10^-5
Therefore, the final equilibrium concentrations are approximately as follows:
HA ≈ 0.30 - x ≈ 0.30 - 3.87 x 10^-5 ≈ 0.2999613 M
H+ ≈ A- ≈ x ≈ 3.87 x 10^-5 M
So, the final equilibrium concentration of HA is approximately 0.29996 M, and the final equilibrium concentrations of H+ and A- are approximately 3.87 x 10^-5 M.