Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

# Chemistry help?

Calculate the percent dissociation of acetic acid in a 0.35M aqueous solution of the stuff.

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• 2 months ago

Use the Ka equation to solve for [H+] . The Ka is small enough to avoid the use of the quadratic route

Ka = [H+]² / 0.35

[H+]² = (1.8*10^-5) * 0.35

[H+]² = 6.30*10^-6

[H+] = 2.51*10^-3M

% dissociation = (2.51*10^-3) / 0.35 * 100% = 0.72%

• 2 months ago

Ka for acetic acid (CH₃COOH) = 1.8 × 10⁻⁵

Let y% be the percent dissociation of acetic acid.

CH₃COOH(aq) + H₂O(ℓ)  ⇌ CH₃COO⁻(aq) + H₃O⁺(aq)   Ka = 1.8 × 10⁻⁵

Initial:         0.35 M                               0 M                 0 M

Change:  -0.35y% M                      +0.35y% M       +0.35y% M

Eqm:   0.35(1 - y%)  M                    0.35y% M         0.35y% M

≈ 0.35 M                          = 0.0035y M     = 0.0035y M

At equilibrium:

Ka = [CH₃COO⁻] [H₃O⁺] / [CH₃COOH]

1.8 × 10⁻⁵ = (0.0035y)² / 0.35

1.8 × 10⁻⁵ = 0.0035²y² / 0.35

y = √[0.35 × (1.8 × 10⁻⁵) / 0.0035²]

y = 0.72

Percent dissociation of acetic acid = 0.72%

• Anonymous
2 months ago

Sorry, my answer is "Anonymous" as your post is Anonymous. There is no need for an Anonymous question in this category, Most 'Anonymous' posts are trolls, maybe yours is not, but the high likelihood is there. If you post as yourself, I would be glad to help you with any problem you have.

I spend a lot of time on each problem, I just want to make sure it's for a worthwhile cause.