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A 30 mL sample of .165 M propanoic acid is titrated with .300M KOH. Calculate the pH at each volume of added base: 0 mL, 5 mL, 10, equivalence point, one-half equivalence point, 20 mL, 25 mL. Use calculations to make a sketch of the titration curve.

  • chemistry -

    eq pt.
    mL acid x M acid = mL KOH x M KOH
    Solve for mL KOH.

    0 mL:
    .........HPr ==> H^+ + Pr^-
    I......0.165.....0.....0
    C.........-x......x....x
    E.......0.165-x...x....x

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

    For additions up to but not including eq pt use the Henderson-Hasselbalch equation.

    For the eq pt the pH is determined by the hydrolysis of the salt. (salt) = (Pr^-) = (0.165 x 30 mL)/(total mL) where total mL = 30 + mL to eq pt. I will call this 0.1 M but you need to do it more accurately
    ........Pr^- + HOH ==> HPr + OH^-
    I......0.1..............0.....0
    C........-x.............x.....x
    E......0.1-x............x.....x

    Kb for Pr^- = (Kw/Ka for acid) = (x)(x)/(0.1-x). Solve for x = (OH^-) and convert to pH.

    For all points past the eq pt it will be excess OH^-.
    mols base = M x L
    mols acid to eq pt = ?
    excess OH^- = difference.
    (OH^-) = mols OH excess/total L.
    Then convert to pH.

  • chemistry -

    2.8?

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