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Answer the following two questions about the equivalence point for the titration of 27.8 mL of 0.235 M hypobromous acid (pKa = 8.64) with 0.214 M KOH.

1)Predict if the solution will be acidic, basic or neutral at the equivalence point.

2)Calculate the pH of the solution at the equivalence point.

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    HBrO + KOH ==> KBrO + H2O
    1)What do you have at the equivalence point? That is a soln of KBrO. It is the salt of a strong base and a weak acid; therefore, the solution MUST be basic.

    2) The pH at the equivalence point is determined by the hydrolysis of the salt.
    ............BrO^- + HOH ==> HBrO + OH^-
    Set up ICE chart.

    You have M x L = 27.8 x 0.235 = about .0065 moles HBrO(you need to be more exact). The equivalence point will be about 0.0065/0.214 = about 30 mL KOH.
    The salt (KBrO) is then M = moles/L = 0.0065/(28+30) = about 0.11M
    Substitute into Keq as follows:
    Kb = (Kw/Ka) = (x)(x)/(0.11)
    Solve for x, convert to pOH, then convert to pH.

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