Calculate the pH of a solution that is 0.060 M in potassium propionate(C2H5COOK or KC3H5O2) and 0.085 M in propionic acid (C2H5COOH or HC3H5O2)
Use the Henderson-Hasselbalch equation.
To calculate the pH of a solution that contains both a weak acid (propionic acid) and its corresponding conjugate base (potassium propionate), we need to consider the dissociation of the weak acid and the reverse reaction of the conjugate base in water.
1. Write the balanced chemical equation for the dissociation of propionic acid (HC3H5O2):
HC3H5O2 ⇌ H+ + C3H5O2-
2. Identify the equilibrium expression for this dissociation reaction, known as the acid dissociation constant (Ka), which represents the ratio of the concentrations of the products (H+ and C3H5O2-) to the reactant (HC3H5O2) at equilibrium.
Ka = [H+][C3H5O2-] / [HC3H5O2]
3. Use the known values for the initial concentrations of the weak acid (0.085 M) and its conjugate base (0 M) to set up an ICE (Initial-Change-Equilibrium) table.
Initial: [HC3H5O2] = 0.085 M, [H+] = 0 M, [C3H5O2-] = 0 M
Change: -x, +x, +x (assuming the weak acid dissociates and forms x moles of H+ and C3H5O2- ions)
Equilibrium: [HC3H5O2] - x, x, x
4. Substituting these equilibrium concentrations into the acid dissociation constant expression:
Ka = [H+][C3H5O2-] / [HC3H5O2]
Ka = x * x / (0.085 - x)
5. Solve the quadratic equation for x (which represents the concentration of H+) using the given Ka value for propionic acid, which is 1.4 × 10^-5.
1.4 × 10^-5 = x^2 / (0.085 - x)
Since the value of x, representing the concentration of H+, is expected to be small compared to 0.085, we can assume that the equilibrium concentration of the weak acid (0.085 M) approximately equals the initial concentration.
1.4 × 10^-5 = x^2 / 0.085
Rearranging the equation, we get:
x^2 = 1.4 × 10^-5 * 0.085
x^2 = 1.19 × 10^-6
x ≈ 0.00109 M
6. Calculate the concentration of the conjugate base, C3H5O2-. Since propionic acid and potassium propionate are in a 1:1 ratio, the concentration of C3H5O2- will be the same as the concentration of x. Therefore, [C3H5O2-] = 0.00109 M.
7. Calculate the pH of the solution using the equation:
pH = -log[H+]
pH = -log(0.00109)
pH ≈ 2.96
So, the pH of the solution that is 0.060 M in potassium propionate and 0.085 M in propionic acid is approximately 2.96.