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Chemistry

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Ok, moving on from my last question, you have 100mL of 0.10M acetic acid with ka = 1.8x10(-5) and add 50mL (0.10M) NaOH.

  • Chemistry - ,

    100 mL HAc x 0.1M = 10 mmoles.
    50 mL NaOH x 0.1M = 5 mmoles.

    .........HAc + OH^- => Ac^- + H2O
    begin....10....0........0.......0
    add............5.0.................
    change..-5.0...-5.0.....+5.0....+5.0
    equil....5.0.....0.......5.0
    Then I would use the Henderson-Hasselbalch equation to solve for pH.
    Since (base) = (acid), then log base/acid = log 1 = 0 so pH = pKa or about 4.74

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