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3 September, 12:55

What is the ph of a solution that contains 1.0 l of 0.10 m ch3cooh and 0.080 m nach3coo after 0.03 moles of naoh added?

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  1. 3 September, 13:09
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    Answer: pH = 4.996

    Explanation:

    No of moles = molarity x volume

    :• no of moles of CH3COOH = 0.1M x 0.1L

    n (CH3COOH) = 0.1mol

    Since 0.03mole of NaOH is added, then 0.03 mole of CH3COOH will be converted to the conjugate.

    Therefore, Moles of CH3COOH becomes,

    0.1 - 0.03 = 0.07 mol

    Subsequently, the moles of CH3COONa increases and becomes,

    0.08 + 0.03 = 0.11 mol

    Using the Hendersom-Hasselbach equation,

    pH = pKa + log [Moles of conjugate: moles of Ch3COOH]

    From literature, pKa of Ch3COOH is 4.8

    Thus,

    pH = 4.8 + log [0.11/0.07]

    pH = 4.8 + 0.1963

    pH = 4.996
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