Answer:
10.71%
Explanation:
The dissociation of acetic acid can be well expressed as follow:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:
Then:
The I.C.E Table is expressed as follows:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Initial 0.0014 0 0
Change - x +x +x
Equilibrium (0.0014 - x) x x
Recall that:
Ka for acetic acid CH₃COOH = 1.8×10⁻⁵
∴
![K_a = \dfrac{[x][x]]}{[0.0014-x]}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cdfrac%7B%5Bx%5D%5Bx%5D%5D%7D%7B%5B0.0014-x%5D%7D)
![1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}](https://tex.z-dn.net/?f=1.8%2A10%5E%7B-5%7D%20%3D%20%5Cdfrac%7B%5Bx%5D%5Bx%5D%5D%7D%7B%5B0.0014-x%5D%7D)
![1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}](https://tex.z-dn.net/?f=1.8%2A10%5E%7B-5%7D%20%3D%20%5Cdfrac%7B%5Bx%5D%5E2%7D%7B%5B0.0014-x%5D%7D)
![1.8*10^{-5}(0.0014-x) = x^2](https://tex.z-dn.net/?f=1.8%2A10%5E%7B-5%7D%280.0014-x%29%20%3D%20x%5E2)
![2.52*10^{-8} -1.8*10^{-5}x = x^2](https://tex.z-dn.net/?f=2.52%2A10%5E%7B-8%7D%20-1.8%2A10%5E%7B-5%7Dx%20%3D%20x%5E2)
![2.52*10^{-8} -1.8*10^{-5}x - x^2 =0](https://tex.z-dn.net/?f=2.52%2A10%5E%7B-8%7D%20-1.8%2A10%5E%7B-5%7Dx%20-%20x%5E2%20%3D0)
By rearrangement:
![- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0](https://tex.z-dn.net/?f=-%20x%5E2%20-1.8%2A10%5E%7B-5%7Dx%20%2B2.52%2A10%5E%7B-8%7D%3D%200)
Multiplying through by (-) and solving the quadratic equation:
![x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0](https://tex.z-dn.net/?f=x%5E2%20%2B1.8%2A10%5E%7B-5%7Dx-2.52%2A10%5E%7B-8%7D%3D%200)
![(-0.00015 + x) (0.000168 + x) =0](https://tex.z-dn.net/?f=%28-0.00015%20%2B%20x%29%20%280.000168%20%2B%20x%29%20%3D0)
x = 0.00015 or x = -0.000168
We will only consider the positive value;
so x=[CH₃COO⁻] = [H⁺] = 0.00015
CH₃COOH = (0.0014 - 0.00015) = 0.00125
However, the percentage fraction of the dissociated acetic acid is:
![= \dfrac{ 0.00015}{0.0014}\times 100](https://tex.z-dn.net/?f=%3D%20%5Cdfrac%7B%200.00015%7D%7B0.0014%7D%5Ctimes%20100)
= 10.71%