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Xxx 發問於 科學及數學化學 · 6 年前

show chem step plz

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  • 6 年前
    最愛解答

    Ka=[H+][ethanoate]/[ethanoic acid]

    assume sodium ethanoate had 100% dissociation

    in buffer, concentration of ethanoate and ethanoic acid is equal and is cancelled out

    Ka=[H+]

    [H+]= 10^-4.74=Ka

    number of mole of NaOH added= 4x5/1000=0.02mol

    CH3COOH + OH- --> CH3COO- +H2O

    number of mole of substance after reaction:

    ethanoate: 0.32

    ethanoic acid: 0.28

    (0.32)/(0.28)= 1.142857

    [H+]=10^-4.74/1.142857 = 1.592238x10^-5M

    pH = 4.797992=4.80

    NaOH adding into water, negate water's self dissociation

    [OH-]= 0.02/1.005=0.0199

    10^-14 = [H+][OH-]

    [H+]=10^-14/0.0199

    [H+]=5.025x10^-13

    pH= 12.3

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