Examine the PH and the fraction ( interms of percentage) of CH3COOH molecule deprotonated in 2.0M CH3COOH ( aq ). ( Ka=1.8×10^-5).

Answer 1

To find the pH and fraction of CH3COOH molecules deprotonated, we can use the Ka expression for the dissociation of acetic acid:

CH3COOH ⇌ CH3COO- + H+

First, let's calculate the initial concentration of CH3COOH assuming it completely dissociates in solution:

[CH3COOH] = 2.0 M

Next, let's assume that x is the concentration of CH3COO- and H+ ions formed due to dissociation. Hence, the equilibrium concentration of CH3COOH will be 2.0 - x.

The Ka expression for this reaction is:

Ka = [CH3COO-][H+] / [CH3COOH]

Substitute the values into the equation:

1.8×10^-5 = x*x / (2.0 - x)

Let's solve this equation to find the value of x:

1.8×10^-5 = x^2 / (2.0 - x)

x^2 = 1.8 x 10^-5(2.0 - x)

x^2 = 3.6 x 10^-5 - 1.8 x 10^-5X

x^2 + 1.8 x 10^-5X - 3.6 x 10^-5 = 0

x ≈ 0.001898 M

Now, the fraction of CH3COOH molecules that deprotonated can be calculated as:

Fraction deprotonated = [CH3COO-] / [CH3COOH] = 0.001898 / 2.0 = 0.000949

Finally, we can find the pH of the solution using the equation:

pH = -log[H+]

Since the concentration of H+ ions is equal to the concentration of CH3COO- ions at equilibrium, the pH is:

pH = -log(0.001898) = 2.72

Therefore, in a 2.0M CH3COOH solution, approximately 0.0949% of CH3COOH molecules are deprotonated, and the pH of the solution is 2.72.