Estimate the pH and the fraction (in terms of percentage) of CH3COOH molecules deprotonated in 0.2M CH3COOH. (Ka= 1.8 × 10^-5)
To estimate the pH and fraction of CH3COOH molecules deprotonated, we first need to calculate the concentration of acetic acid (CH3COOH) that has deprotonated in a 0.2M solution.
Let x be the concentration of CH3COOH molecules that have deprotonated. Since acetic acid is a weak acid, we can assume that x is very small compared to 0.2M.
The equilibrium reaction for the dissociation of acetic acid is:
CH3COOH + H2O ⇌ CH3COO- + H3O+
The expression for the equilibrium constant (Ka) is:
Ka = [CH3COO-][H3O+] / [CH3COOH]
Substitute the given values:
1.8 × 10^-5 = x * x / (0.2 - x)
Since x is very small compared to 0.2M, we can assume that 0.2 - x ≈ 0.2. Therefore, the equation simplifies to:
1.8 × 10^-5 = x^2 / 0.2
Rearranging the equation:
x^2 = 1.8 × 10^-5 * 0.2
x^2 = 3.6 × 10^-6
x = √(3.6 × 10^-6)
x ≈ 1.9 × 10^-3
Now we can calculate the fraction of CH3COOH molecules deprotonated:
Fraction deprotonated = x / 0.2
Fraction deprotonated = 1.9 × 10^-3 / 0.2
Fraction deprotonated ≈ 0.0095
Finally, to calculate the pH, we need to find the concentration of H3O+ ions produced from the dissociation of acetic acid. Since acetic acid is a weak acid, we can assume that the concentration of H3O+ ions is equal to the concentration of x:
[H3O+] = 1.9 × 10^-3 M
Now, calculate the pH:
pH = -log[H3O+]
pH = -log(1.9 × 10^-3)
pH ≈ 2.72
Therefore, the estimated pH of the 0.2M CH3COOH solution is approximately 2.72, and the fraction of CH3COOH molecules deprotonated is about 0.95%.