Estimate the Ph and the fraction (interms of percentage) of CH3COOH molecule deprotonated in 0.2M CH3COOH. (Ka =1.8 ×10^-5)

Answer 1

To estimate the pH and fraction of CH3COOH molecule deprotonated in a 0.2M CH3COOH solution, we can use the Ka expression for the dissociation of acetic acid:

CH3COOH ⇌ CH3COO- + H+

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

Given that the initial concentration of CH3COOH is 0.2M and assuming that x is the concentration of CH3COO- formed and H+ ions, this gives:

0.2-x = [CH3COOH]
x = [CH3COO-] = [H+]

Now, we can substitute these values into the Ka expression:

1.8 x 10^-5 = x^2/(0.2-x)

Solving this equation gives x = 0.0042 M

This concentration corresponds to the concentration of both CH3COO- and H+. The fraction deprotonated can be calculated as the ratio of x to the initial concentration of CH3COOH:

Fraction deprotonated = x / [CH3COOH] = 0.0042 / 0.2 = 0.021 = 2.1%

To determine the pH of the solution, we can use the equation for pH:

pH = -log[H+]

pH = -log(0.0042) = 2.376

Therefore, the estimated pH of the 0.2M CH3COOH solution is 2.376, and the fraction of CH3COOH molecules deprotonated is 2.1%.