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What would the [H3O+] be after addition of 5.15 mL of 0.1000 M HCl to 25.00 mL of 0.1000 B (a weak base, Kb = 6.31e-5)?

  • college chemistry -

    millimols BOH = 25 mL x 0.1M = 2.5
    millimols HCl = 5.25 mL x 0.1M = 0.515

    ..........BOH + HCl ==> BCl + H2O
    I.........2.5....0.......0......0
    added.........0.515...............
    C......-0.515..-0.515...+0.515
    E...... 1.985....0.......0.515

    I would substitute the E line into the Henderson-Hasselbalch equation and solve for pH, then H^+.
    Note: you have Kb and you need pKa. Convert Kb to pKb, the pKa + pKb = pKw = 14 and solve for pKa.
    You don't need to use the HH equation; it's just easier to do it that way.

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