Final answer:
The pH of a 0.23 M solution of potassium generate (KR-COO) can be calculated using the equilibrium constant Kₐ for the acid R-COOH. By substituting the concentration of KR-COO for R-COOH in the Kₐ equation, we can solve for the concentration of H+ ions and then calculate the pH. The pH of the solution is approximately 7.57.
Step-by-step explanation:
To determine the pH of a solution containing potassium generate (KR-COO), we need to consider the dissociation of the acid R-COOH. The acid dissociates into its conjugate base R-COO- and a hydrogen ion (H+).
The equilibrium constant for this dissociation is given as Kₐ = 2.7 x 10⁻⁸. To calculate the pH, we need to calculate the concentration of H+ ions. Since 0.23 M of KR-COO is present, the concentration of R-COOH is also 0.23 M.
Using the equation for Kₐ:
Kₐ = [H+][R-COO-] / [R-COOH]
Since the initial concentration of R-COOH is equal to the concentration of KR-COO, we can substitute 0.23 M for both [R-COO-] and [R-COOH] in the equation:
Kₐ = [H+] (0.23) / (0.23)
Simplifying the equation gives:
[H+] = Kₐ
Therefore, the pH of the solution is equal to the negative logarithm (base 10) of Kₐ:
pH = -log(Kₐ)
Plugging in the given value for Kₐ:
pH = -log(2.7 x 10⁻⁸) ≈ 7.57
The pH of the 0.23 M solution of potassium generate (KR-COO) is approximately 7.57.