Estimate ph and the percentage of CH3COOH deprotonated in 0.2 M Ch300H aq,given ka =1.8x10^-5.with Ice table.
To estimate the pH and percentage of CH3COOH deprotonated in a 0.2 M CH3COOH solution, we can use an ICE table to track the changes in concentration of CH3COOH and its conjugate base CH3COO- as the acid dissociates in water.
The dissociation of CH3COOH in water can be represented as:
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
Given that the initial concentration of CH3COOH is 0.2 M and the Ka value for CH3COOH is 1.8x10^-5, we can set up an ICE table:
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
I: 0.2 M 0 M 0 M
C: -x +x +x
E: 0.2-x x x
Using the equilibrium expression for Ka:
Ka = [CH3COO-][H+] / [CH3COOH]
1.8x10^-5 = x^2 / (0.2-x)
Since x is likely to be small compared to 0.2, we can assume that 0.2 - x is approximately 0.2. Solving for x:
1.8x10^-5 = x^2 / 0.2
x^2 = 0.2 * 1.8x10^-5
x = √(0.2 * 1.8x10^-5)
x = 0.00212
This gives us the concentration of [H+] ions in the solution. To find the pH, we take the negative logarithm of the [H+] concentration:
pH = -log[H+]
pH = -log(0.00212)
pH ≈ 2.67
To find the percentage of CH3COOH deprotonated, we can calculate the degree of dissociation (α) using x:
α = [CH3COO-] / [CH3COOH]
α = 0.00212 / 0.2
α ≈ 0.0106
Therefore, approximately 1.06% of CH3COOH is deprotonated in the 0.2 M CH3COOH solution.