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9 May, 14:45

A 100.0 ml sample of 0.18 m hclo4 is titrated with 0.27 m lioh. determine the ph of the solution after the addition of 100.0 ml of lioh.

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  1. 9 May, 14:55
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    (I think you have a mistake in your question as the addition is 30mL, not 100mL)

    when PH = - ㏒[H+]

    and here we have HClO4 is the strong acid

    So PH = - ㏒[HClO4]

    moles of HClO4 = 0.1 L * 0.18 m = 0.018 M

    moles of LiOH = 0.03 L * 0.27 m = 0.0081 M

    when the total volume = 0.1L + 0.03L = 0.13 L

    ∴ [HClO4] = (0.018-0.0081) / 0.13 L

    = 0.076 M

    PH = - ㏒ 0.076

    = 1.12
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