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Find the pH during the titration of a 20.00mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka=1.54 x10^-5), with the addition of 10 mL of 0.1000 M NaOH solution??

  • chemistry -

    Butanoic acid we'll call HB and you watch the significant figures.
    millimols HB = 20 mL x 0.1 M = 2
    mmols NaOH added = 10 mL x 0.1M = 1

    ..........HB + OH^-==> H2O + B^-
    initial...2......0......0
    add.............1...........
    change....-1....-1......1.....1
    equil.....1......0.......1....1.

    So you have formed 1 mmol base and have left 1 mmols HB (this is a buffer) so use the Henderson-Hasselbalch equation to solve for pH.
    pH = pKa + log(base)/(acid)
    base = 1
    acid = 1
    pKa = -log Ka
    Solve for pH. You should get 4.81

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