To find the percentage dissociation of benzoic acid (C6H5COOH) in a solution with a given pH, we need to use the acid dissociation constant (Ka) and the pH value. Benzoic acid is a weak acid, and its dissociation in water can be represented by the following equilibrium reaction:
C6H5COOH ⇌ C6H5COO- + H+
The acid dissociation constant (Ka) for this reaction is given as 6.5 x 10^-5.
First, we need to find the concentration of the benzoic acid and the concentration of the hydronium ions (H+) in the solution.
- Calculate the concentration of H+ ions using the pH: pH = -log[H+] 2.59 = -log[H+]
[H+] = 10^(-pH) [H+] = 10^(-2.59) ≈ 2.47 x 10^(-3) mol/L
- Since benzoic acid is a monoprotic acid, the initial concentration of benzoic acid (C6H5COOH) is equal to the concentration of hydronium ions (H+).
[C6H5COOH] = [H+] ≈ 2.47 x 10^(-3) mol/L
- Now, we can set up an expression for the dissociation of benzoic acid:
Ka = [C6H5COO-][H+] / [C6H5COOH]
We already know the values of Ka and [C6H5COOH], and we need to find [C6H5COO-].
Let x be the concentration of the dissociated benzoic acid (C6H5COO-).
Ka = x * x / (2.47 x 10^(-3) - x)
At this point, we can make an assumption that the dissociation of benzoic acid is small compared to its initial concentration. This means that x is much smaller than 2.47 x 10^(-3). Therefore, we can approximate 2.47 x 10^(-3) - x as approximately 2.47 x 10^(-3).
Now, we can solve for x:
6.5 x 10^(-5) = x * x / (2.47 x 10^(-3))
x^2 = 6.5 x 10^(-5) * 2.47 x 10^(-3)
x^2 ≈ 1.6055 x 10^(-7)
x ≈ √(1.6055 x 10^(-7))
x ≈ 1.267 x 10^(-4) mol/L
Finally, we can calculate the percentage dissociation:
Percentage Dissociation = (Dissociated Concentration / Initial Concentration) * 100
Percentage Dissociation ≈ (1.267 x 10^(-4) mol/L / 2.47 x 10^(-3) mol/L) * 100
Percentage Dissociation ≈ 5.12%
Therefore, the percentage dissociation of benzoic acid in the given solution with pH=2.59 is approximately 5.12%.