[jalen]

Calculate the pH of a sample of vinegar that contains 0.83mol/L acetic acid. What is the percent dissociation of the vinegar?

Ka = [CH₃COO][H₃O] / [CH₃COOH]
    = x x / [0.83-x]
    = 1.8 x 10<sup>-5[/s]

[CH₃COOH] = 0.83
          Ka      1.8 x 10-5
                   = 4.6 x 104

Quadratic answer: x=3.86 x 10-3

pH = -log(3.86 x 10-3)
     = 2.41

Q: How do you get the percent dissociation of the vinegar? Please show steps. Thank you.</sup>

[Borek]

pH is OK.

For % dissociation just use definition - you know how much acid dissociated ([CH₃COO^-] = [H₃O⁺]), you know how much acid is present.

http://www.chembuddy.com/?left=pH-calculation&right=introduction-acid-base-equilibrium

Percent dissociation and dissociation fraction are the same thing.