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15 January, 07:34

For the titration of 50.0 ml of 0.020 m aqueous salicylic acid with 0.020 m koh (aq), calculate the ph after the addition of 55.0 ml of the base. for salycylic acid, pka = 2.97.'

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  1. 15 January, 08:50
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    First by getting moles of salicylic acid:

    moles of salicylic acid = molarity * volume

    = 0.02 * 0.05 L

    = 0.001 mol

    then moles of base KOH = molarity * volume

    = 0.02 * 0.055L

    = 0.0011

    when total volume = 0.05 + 0.055 = 0.105 L

    [salisalic acid] = moles / total volume

    = 0.001 / 0.105

    = 0.0095

    [KOH] = moles / total volume

    = 0.0011 / 0.105

    = 0.01

    by using H-H equation, we can get the PH:

    PH = Pka + ㏒[salt/acid]

    by substitution:

    PH = 2.97 + ㏒[0.01 / 0.0095]

    = 2.99
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