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Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.230 M HClO (aq) with 0.230 M KOH (aq). The ionization constant (Ka) for HClO is 4.00 x 10^-8.

(b) after addition of 25.0 mL of KOH
(d) after addition of 50.0 mL of KOH

  • Chemistry -

    You must learn to recognize where you are on the titration curve.
    1. Start by calculating mL KOH to reach the equivalence point.
    50.0 x 0.230 = mL x 0.230
    Obviously it will be 50.0 mL KOH and that will answer part. That is determined by the hydrolysis of the salt, KClO.
    .........ClO^- + HOH ==> HClO + OH^-
    initial..0.150M............0.....0
    change....-x...............x.....x
    equil....0-.150-x...........x.....x

    Kb for ClO^- = (Kw/Ka for HClO) = (x)(x)/(0.150-x)
    Solve for x = (OH^-) and convert to pH.

    For part a.
    ........HClO + KOH ==> KClO + H2O
    ....50*0.230 + 25.0*0.230 .....
    mmol....11.5.. 5.75......0.....0
    change..-5.75..-5.75....5.75...5.75
    equil..4.25......0.......5.75..5.75
    Ka = (H^+)(ClO^-)/(HClO)
    Plug in x for (H^+), get HClO and ClO^- from the ICE chart, solve for x and convert to pH.
    Alternatively, you can use the Henderson-Hasselbalch equation.

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