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17 November, 20:27

What is the ph of a 0.135 m aqueous solution of potassium acetate, kch3coo? (ka for ch3cooh = 1.8*10-5) ?

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  1. 17 November, 20:55
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    [OH-] = √ (Cs*Kw) / Ka

    [OH-] = √ (0,135*10^-14) / 1,8*10^-5

    [OH-] = √0,075*10^-9 = √75*10^-12 = 8,66*10^-6

    pOH = - log[OH-]

    pOH = - log[8,66*10^-6]

    pOH ≈ 5,0625

    pH + pOH = 14

    pH = 14-5,0625

    pH = 8,9375

    :•)
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