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Calculate the ph for ch3co2h,3% dissociated at a concentration of 0.02 mol dm-3.

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    Use caps where needed. ch3co2h means nothing. I assume you meant to write CH3COOH. Let me call that HAc (H for the last H on the right, the acid H, and Ac for the rest of the molecule).
    ............3% = 0.03 dissociated
    .............HAc ==> H^+ + Ac^-
    initial.....0.02M....0.....0
    change.....-6E-4....6E-4...6E-4
    equil...0.0194.....6E-4....6E-4
    (Note:6E-4 = 0.02*0.03
    Then plug H^+ into pH = -log(H^+)

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