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March 29, 2017

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I did an experiment using 20ml of 0.1M acetic acid and added 8ml of 0.1M NaOH. pH obtained after adding NaOH was 5.2. How do i calculate the mols of NaOH that reacted? mols of acetate formed, acetic acid initially present and acetic acid unreacted. Also how do i determine the [H+], [CH3CO2-] and [CH3CO2H] and Ka

  • Chemistry - ,

    To save some typing, I will call acetic acid, HAc and acetate is Ac^-
    millimoles HAc = mL x M = 20 x 0.1 = 2.0
    mmoles NaOH = 8 x 0.1 = 0.8

    ...............HAc + NaOH ==> NaAc + H2O
    initially......2.0....0........0.......0
    added.................0.8...............
    change.........-0.8...-0.8.....+0.8..+0.8
    equil.......... 1.2.....0......0.8...0.8

    (HAc) = mmoles/mL = 1.2/28 = ??
    NaAc = same process = mmoles/mL.

    pH = pKa + log [(NaAc)/(HAc)] = ??

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