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29 January, 15:53

Calculate the ph of a buffer solution that contains 0.25 m benzoic acid (c6h5co2h) and 0.15m sodium benzoate (c6h5coona). [ka = 6.5 x 10-5 for benzoic acid]

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  1. 29 January, 16:05
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    When C6H5COO - is the conjugate base & the C6H5CO2H is the weak acid

    -and we have [C6H5COO-] = 0.15 M

    and [C6H5CO2H] = 0.25 M

    - and we have Ka = 6.5 x 10^-5 so, we can use it and get Pka

    ∴ Pka = - ㏒Ka

    = - ㏒ (6.5 x 10^-5)

    = 4.2

    So, by using H-H equation:

    PH = Pka + ㏒[C6H5COO-]/[C6H5CO2H]

    = 4.2 + (0.15 / 0.25)

    = 4.8
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