Chemistry

I need help please. The question is vinegar, or a dilute of CH3COOH (acetic acid), should have a molarity of around 0.83 M. Using the Ka for acetic acid, what should the pH of vinegar be??

Thank you!!

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asked by Lilly
  1. CH3COOH>> H+ + CH3COO-

    k=(h+)(CH3COOH-)/.83

    k=2H/.83
    solve for H
    I get about 1.7E-5*.83/2=.70E-5 about
    the pH=-logH=about 5.1
    Use your table for Ka, and you calculator to check, this is estimated.

  2. Given 0.84M HOAc and Ka =1.8x10^5
    For a weak monoprotic acid the [H] can be calculated from [H] = (Ka[Acid)^1/2
    = [1.75x10^-5(.83)]^1/2
    = 3.8x10^-3M(H)
    pH(vinegar) = -log[H] = -log(3.8E-3)
    pH = 2.42

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