Examine the PH and the fraction ( interms of percentage) of CH3COOH deprotonated In 0.2M CH3COOH ( aq ). (Ka=1.8×10^_8 ).
To determine the pH and fraction of CH3COOH deprotonated in 0.2M CH3COOH (aq), we need to calculate the concentration of each species in the solution and then use the dissociation constant (Ka) to find the percentage of CH3COOH that has been deprotonated.
First, let's set up the dissociation reaction of CH3COOH:
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
Let x be the concentration of CH3COO- and H+ produced from the dissociation of CH3COOH. Therefore, the concentrations at equilibrium are:
[CH3COOH] = 0.2 M - x
[CH3COO-] = x
[H+] = x
The Ka expression for the dissociation of CH3COOH is:
Ka = [CH3COO-][H+] / [CH3COOH]
Substitute in the equilibrium concentrations:
1.8 x 10^-5 = (x)(x) / (0.2 - x)
Since Ka is small compared to 0.2, we can assume that x is small compared to 0.2. Therefore, we can neglect x in the denominator:
1.8 x 10^-5 = (x)(x) / 0.2
Solving for x gives x = 0.001897 M
Since x represents the concentrations of both CH3COO- and H+ ions, the percentage of CH3COOH that has been deprotonated is given by:
Fraction of CH3COOH deprotonated = [CH3COO-] / [CH3COOH] * 100
Fraction of CH3COOH deprotonated = 0.001897 M / 0.2 M * 100
Fraction of CH3COOH deprotonated = 0.9485%
Next, we can calculate the pH of the solution. Since the concentration of H+ ions is equal to the concentration of CH3COO-, we can use the equation:
pH = -log[H+]
pH = -log(0.001897)
pH = 2.72
Therefore, the pH of the solution is 2.72 and approximately 0.95% of CH3COOH has been deprotonated.