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biochemistry

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The pH of 0.02M solution of a weak acid was measured at 7.6. What is the [OH-] in this
solution?
B) The Ka for benzoic acid is 6.4 x 10-5M. 150 ml of 0.1 M NaOH is added to 200 ml of 0.1 M
benzoic acid, and water is added to give a final volume of 1 L. What is the pH of the final solution?

  • biochemistry -

    A). pH = -log(H^+)
    7.6 = -log(H^+). Solve for (H^+). You should get approximately 2.5E-8.

    B)
    150 mL x 0.1M NaOH = 15 millimiles.
    200 mL x 0.1M benzoic acid = 20 mmols.

    Let's call benzoic acid HB, then
    ............HB + NaOH ==> NaB + H2O
    initial....20......0........0.....0
    add................15..............
    change.....-15...-15........15...15
    equil.......5......0.........15
    pH = pKa + log (base)/(acid)
    pH = 4.19 + log (15/5); solve for pH.

    Note and this is important:
    I used millimoles (the equivalent of mol) INSTEAD of concentration. The HH equation calls for concentration. I used millimoles as a shortcut. The answer will be the same; however, some profs will count off if you use mols (or millimoles) instead of concentration. So you need to do one of three things if your prof is picky (as I am).
    (base) = mmols/mL = 15/1000 = ?
    (acid) = mmols/mL = 5/1000 = ?
    #1.
    pH = 4.19 + log(0.015/0.005) = ?
    #2.
    pH = 4.19 + log(15/1000/5/1000) =
    = cancel the denominator since both are the same (by the way they will ALWAYS be the same) so
    = 4.19 + log(15/5) = ?
    or third,
    pH = 4.19 + log(15/volume/(5/volume) =
    cancel v and
    pH = 4.19 + log (15/5) = ?

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