Estimate the PH and the fraction in terms of percentage of CH3COOH molecules deprotonated in 0.2M CH3COOH.ka=1.8x10^-5

Answer 1

To estimate the pH and fraction of CH3COOH molecules deprotonated, we can use the equation for the dissociation of acetic acid (CH3COOH) in water:

CH3COOH + H2O ↔ CH3COO- + H3O+

Since Ka = [CH3COO-][H3O+]/[CH3COOH], we can calculate the concentration of CH3COO- and H3O+ at equilibrium.

Let x = concentration of CH3COO- and H3O+ at equilibrium

The initial concentration of CH3COOH is given as 0.2M, and since CH3COOH is a weak acid, we can assume that x << 0.2M. Therefore, the concentration of CH3COO- and H3O+ at equilibrium is approximately x.

Using the equilibrium constant (Ka = 1.8x10^-5), we can set up the equilibrium expression as:

1.8x10^-5 = x^2 / (0.2-x)

Since x << 0.2M, simplifying the equation gives:

1.8x10^-5 ≈ x^2 / 0.2

x ≈ sqrt(1.8x10^-5 * 0.2) ≈ 0.0085M

Now, to determine the fraction of CH3COOH molecules deprotonated, we can calculate the ratio of [CH3COO-] to the initial concentration of CH3COOH:

Fraction deprotonated = [CH3COO-]/[CH3COOH] ≈ x/0.2 ≈ 0.0085/0.2 ≈ 0.0425 or 4.25%

Therefore, approximately 4.25% of CH3COOH molecules are deprotonated in a 0.2M CH3COOH solution. To calculate the pH, we can use the equation:

pH = -log10([H3O+])

Since [H3O+] ≈ x = 0.0085M:

pH = -log10(0.0085) ≈ 2.07

Therefore, the estimated pH of a 0.2M CH3COOH solution is approximately 2.07.