Answer:
percent ionization = 50.01%; pH = 4.75
Explanation:
To solve this question we must write the acetic acid equilibrium (Where HX will be acetic acid and X⁻ the sodium acetate):
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
Where equilibrium constant, Ka, is defined as:
Ka = 1.76x10⁻⁵ = [H⁺] [X⁻] / [HX]
<em>Where the concentration of each ion is:</em>
[H⁺] = X
[X⁻] = 0.10M + X
[HX] = 0.10M - X
Replacing in Ka expression:
1.76x10⁻⁵ = [X] [0.10-X] / [0.10+X]
1.76x10⁻⁶ + 1.76x10⁻⁵X = 0.10X - X²
X² - 0.0999824 X + 1.76×10⁻⁶ = 0
X ≈ 0.1M → False solution. Decreases a lot the concentration of HX
X = 0.0000176M → Right solution.
The concentration of each ion is:
[H⁺] = 0.0000176062M
[X⁻] = 0.10M + 0.0000176M = 0.1000176M
[HX] = 0.10M - 0.0000176M = 0.0999824M
Percent ionization:
[X-] / [X-] + [HX] * 100 =
0.1000176M / 0.2M =
<h3>50.01%</h3><h3 />
And pH = -log [H+]
<h3>pH = 4.75</h3><h3 />
As you can see, [H+]≈ Ka
