Final answer:
To find the pH and pOH of the solution, we need to calculate the concentration of H+ and OH-. Using the dissociation reaction of benzoic acid and the equilibrium constant expression, we can calculate the concentration of H+. Given the concentrations provided, we can find that the pH of the solution is 1.2 and the pOH is 12.8.
Step-by-step explanation:
To find the pH and pOH of the solution, we need to calculate the concentration of H+ and OH-. The benzoic acid will dissociate in water to produce H+ ions and benzoate ions, while the sodium benzoate will dissociate to produce sodium ions and benzoate ions. The equation for the dissociation of benzoic acid is:
Benzoic acid ⇌ H+ + Benzoate-
To calculate the concentration of H+ ions, we can use the equilibrium constant expression for benzoic acid:
Ka = [H+][Benzoate-] / [Benzoic acid]
We can rearrange the equation to solve for [H+]:
[H+] = (Ka * [Benzoic acid]) / [Benzoate-]
Given that the concentration of benzoic acid is 5 M and the concentration of benzoate is 0.005 M, we can substitute these values into the equation:
[H+] = (6.3 x 10⁻⁵ * 5) / 0.005
[H+] = 6.3 x 10⁻⁵ x 1000
[H+] = 6.3 x 10⁻² M
Therefore, the pH of the solution is 2 - log(6.3 x 10⁻²) = 1.2.
Since pH + pOH = 14, we can calculate the pOH:
pOH = 14 - pH = 14 - 1.2 = 12.8.