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March 25, 2017

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calculate the ph 25.00 mL solution of 0.125 M acetic acid (Ka= 1.75 x 10^-5) after the addition of each of the following volumes of 0.100 M KOH: 10.00 mL, Vep and 35.00 mL

  • Chemistry - ,

    KOH + HAc ==> H2O + KAc
    Divide the problem up into
    a. beginning (I've worked this but the problem doesn't ask for it.)
    b. equivalence point
    c. everything between a and b
    d. everything after b.

    What volume is b?
    mols HAc = 0.025 x 0.125 = 0.003125
    mols KOH = 0.003125(The coefficients are 1 mol KOH to 1 mol HAc.)
    M KOH = mols KOH/L KOH solve for L; I obtained 31.25 mL. The pH is determined by the hydrolysis of the salt. Salt concn is M = mol/L = 0.003125/(25.00+ 31.25) = about 0.055M (that's an estimate)
    .......Ac^- + HOH ==> HAc + OH^-
    I....0.055.............0.....0
    C.......-x.............x......x
    E.....0.055-x .........x......x

    Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.055-x)
    Solve for x = OH^- and convert to pH.

    a........ HAc ==> H^+ + Ac^-
    I.......0.125.....0......0
    C..........-x.....x......x
    E......0.125-x....x......x

    Substitute into Ka expression and solve for x = (H^+) and convert pH.

    c. Use the Henderson Hasselbalch equation for these.

    d. mols HAc = ?
    mols KOH = ? and this is greater than mols HAc.
    Excess KOH = subtract mols KOH-mols HAc.
    Excess = KOH = OH^- ? and convert to pH.

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