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10 February, 07:12

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 16.0 mL of KOH.

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  1. 10 February, 07:39
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    moles HBr = 0.0500 L x 0.15 M=0.0075

    moles KOH = 0.0160 L x 0.25 M=0.0040

    moles H + in excess = 0.0075 - 0.0040 = 0.0035

    total volume = 0.070 L

    [H+] = 0.0035/0.070=0.050 M

    pH = 1.55
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