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. 

Round your answer 2 significant digits.

3 Answers

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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

    Refer to: https://depts.washington.edu/eooptic/links/acidstr...

    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.

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