College Chemistry (DrBob222)

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A solution of an unknown weak acid, HA, is titrated with 0.100 M NaOH solution. The
equivalence point is achieved when 36.12 mL of NaOH have been added. After the
equivalence point is reached, 18.06 mL of 0.100 M HCl are added to the solution and the pH
at that point is found to be 5.20. What is the pKa of this unknown acid?

  • College Chemistry (DrBob222) -

    If I didn't mess up somewhere the pKa = 5.20. Interesting problem.
    mmols NaOH added = 36.12 x 0.1 = 3.612
    ...........HA + NaOH ==> NaA + H2O
    So at the equivalence point you have 3.612 mmols NaA in solution.

    Now you add 18.06 mL x 0.1M = 1.806 mmols
    ..........A^- + H^+ ==> HA
    I........3.612...0......0
    add............1.806.......
    C......-1.806.-1.806....+1.806
    E........1.806....0.....+1.806

    Use the Henderson-Hasselbalch equation and solve for pKa
    5.20 = pKa + log (1.806)/(1.806)
    pKa = ?

  • College Chemistry (DrBob222) -

    Thank you very much!!! Now it makes sense

  • College Chemistry (DrBob222) -

    Excellent job! Thanks a lot

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