Final answer:
The initial pH of a 0.065M CH3COOH solution is found using its Ka value and the dissociation expression for weak acids. The pH is calculated to be approximately 2.77.
Step-by-step explanation:
The pH of a 0.065M CH₃COOH solution can be found by using its acid dissociation constant (Ka). Acetic acid, CH₃COOH, is a weak acid and partially dissociates in water according to the equilibrium:
CH₃COOH <=> H+ + CH₃COO-
Using the formula for the equilibrium constant, Ka = [H+][CH₃COO-] / [CH₃COOH], we can set up an expression to solve for the concentration of hydrogen ions [H+], which are present when the acid dissociates. Given that Ka for acetic acid is 1.8 x 10^-5, we have:
Ka = [H+]^2 / [CH₃COOH_initial - H+]
Assuming that the concentration of H+ will be much smaller than the initial concentration of CH₃COOH, the expression simplifies to:
Ka ≈ [H+]^2 / [CH₃COOH_initial]
Solving for [H+] gives us [H+] = √(Ka × [CH₃COOH_initial]). When we plug in the numbers:
[H+] = √(1.8 x 10^-5 × 0.065) ≈ 1.71 x 10^-3 M
The pH is then calculated using the formula pH = -log([H+]). Therefore:
pH = -log(1.71 x 10^-3) ≈ 2.77
The initial pH of the 0.065M CH₃COOH solution is approximately 2.77.