Question

In the solution containing both 0.10 M acetic acid and 0.10 M sodium acetate, the acetic acid undergoes ionization. The chemical equation for this ionization reaction is the same as for a solution containing acetic acid alone. The difference is that the initial concentration of acetate ion (before any ionization reaction occurs) for the solution containing acetic acid alone is zero, whereas the initial concentration of acetate ion is 0.10 M in your solution containing both acetic acid and sodium acetate. Calculate the percent ionization and the expected initial pH for the solution that contained both 0.10 M acetic acid and 0.10 M sodium acetate. (Hint: Again, you will need to use Ka for acetic acid.)

Answer 1

percent ionization = 50.01%; pH = 4.75

Step-by-step 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]

Where the concentration of each ion is:

[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 =

50.01%

And pH = -log [H+]

pH = 4.75

As you can see, [H+]≈ Ka