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A weak acid HA (pKa = 5.00) was titrated with 1.00 M KOH. The acid solution had a volume of 100.0 mL and a molarity of 0.100 M. Find the pH at the following volumes of base added and make a graph of pH versus Vb: Vb = 0, 1, 5, 9, 9.9, 10, 10.1, and 12 mL.

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    1. First calculate the volume KOH needed to reach the equivalcnce point. That will tell you if the volume(s) in the problem are BEFORE or AFTER the equivalence point.
    HA + KOH ==> KA + H2O

    2. For the beginning, zero mL.
    .........HA ==> H^+ + A^-
    I........0.1....0......0
    C.........-x....x......x
    E.......0.1-x....x......x

    Ka = (H^+)(A^-)/(HA)
    Substitute from the E line and solve for x = (H^+) then convert to pH.

    b. All points before the eq. pt. I will use 5 mL KOH as an example.
    100 mL HA x 0.1M = 10 millimoles
    5 mL KOH x 1M = 5 mmoles KOH.
    10-5 = 5 mmols HA remains.
    5 mmols KA formed. (HA) = 5 mmols/105 mL = ?
    (A^-) = 5 mmols/105 mL = ?
    Substitute into the Ka expression and solve for (H^+), then convert to pH.

    c. At the equivalence point the solution is KA and the A^- is a base.
    (KA) = (100 x 0.1)/110 mL = about 0.09M
    ..........A^- + HOH ==> HA^- + OH^-
    I.......0.09............0......0
    C.........-x............x......x
    E.......0.09-x..........x.......x

    Kb for A^- = (Kw/Ka for HA) = (x)(x)/(0.09-x)
    Substitute and solve for x = (OH^-), then convert to pH.

    d. All points after the eq pt.
    mmoles HA initially.
    mmols KOH added.
    KOH will be in excess; therefore,
    mmoles KOH-mmoles HA = mmols KOH in excess. M KOH = mmol/mL and that gives you (OH^-). Then convert to pH.

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