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Calculate the pH at the equivalence point for the titration of 0.25 M CH3COOH with 0.25
-5
M NaOH. (For CH3COOH, Ka= 1.8 â‹… 10 )

  • chemistry -

    If we let HAc stand for acetic acid (CH3COOH), then the titration of acetic acid and NaOH is
    HAc + NaOH ==>NaAc + H2O
    At the equivalence point we have exactly neturalized HAc and NaOH and neither is in excess. All we have is the salt, NaAc, in water; therefore the pH is determined by the hydrolysis of the salt.
    The (Ac^-) at the equivalence point is 0.125M (that's half the starting M of acetic acid and NaOH).
    .......Ac^- + HOH ==> HAc + OH^-
    I.....0.125............0.....0
    C.......-x.............x.....x
    E....0.125-x...........x.....x

    Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.125-x)
    Solve for x = (OH^-) and convert to pH.

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