To calculate the pH of the solution, we need to consider the dissociation of acetic acid.
CH3COOH + H2O ⇌ CH3COO- + H3O+
The equilibrium constant expression for the dissociation of acetic acid is:
Ka = [CH3COO-][H3O+]/[CH3COOH]
At equilibrium, the concentration of CH3COOH will be decreased by x and the concentrations of CH3COO- and H3O+ will be increased by x. Therefore, we can assume that the initial concentration of CH3COOH is equal to the concentration of CH3COO- and H3O+.
Let x be the concentration of H3O+ and CH3COO- at equilibrium.
Ka = x^2 / (0.35 - x)
Since the value of x is small compared to 0.35, we can simplify the equation by assuming that 0.35 - x ≈ 0.35.
Ka = x^2 / 0.35
Rearranging the equation, we can solve for x:
x = √(Ka * 0.35) = √(1.75*10^-5 * 0.35) = 1.49*10^-3 M
The pH of the solution can be calculated using the concentration of H3O+:
pH = -log[H3O+] = -log(1.49*10^-3) = 2.83
Therefore, the pH of a 0.35 M CH3COOLi solution is 2.83.