Chemistry

A 30 mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: 30 mL, 35 mL, 36 mL, 37 mL, and 40 mL.

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  1. millimols KOH = 30 mL x 0.150 = 4.50
    millimols HClO4 = mL x 0.125M = ??
    ..........HClO4 + KOH ==> KClO4 + H2O
    begin..... 0......4.50........0.......0
    add 30....3.75....0....................
    react....-3.75...-3.75....+3.75...+3.75
    final......0......0.75....+3.75...+3.75

    HClO4 is a strong acid; KOH is a strong base. So the equivalence point will be 7.0 and neither cation nor anion will be hydrolyzed. Thus, the pH for the 30 mL addition will be determined by the excess KOH which will be 0.75 mmoles/(30+30 mL) = 0.75/60 =0.0125 = (OH^-), a pOH of 1.90, and a pH of 12.
    Continue with the additions. It will be helpful if you determine where the equivalence point is.

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