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A student prepared a 0.10 M solutino of acetic acid, CH3COOH. Acetic acid has a Ka of 1.75 x 10^-5. What are the hydronium ion concentration and the pH of the solution?

  • Chemistry -

    Let's let CH3COOH be represented by HAc (H is the H of the COOH and Ac stands for the remaining part of the molecule.).
    HAc ==> H^+ + Ac^-

    Beginning concentrations:
    (HAc) = 0.1
    (H^+) = 0
    (Ac^-) = 0

    After ionization:
    (H^+) = x
    (Ac^-) = x
    (HAc) = 0.1 - x

    Write the Ka expression and plug the after ionization into it in the appropriate places, then solve for x. Following that, use pH = -log(H^+).

    Post your work if you get stuck.

  • Chemistry -

    1.75 x 10^-5 = [x][x]/[0.10-x]?

    I'm completely lost...

  • Chemistry -

    See my response to your later post above.
    Repost if you still don't get it.

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