chemistry

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Calculate the pH of a solution obtained by mixing 30 mL of 0.75 M CH3COOH with 15 mL of 1.5 M NaOH.

  • chemistry -

    millimoles NaOH = mL x M = 15 x 1.5 = 22.5
    mmoles CH3COOH = 30 x 0.75 = 22.5 where CH3COOH = HAc.
    .........HAc + NaOH ==> NaAc + H2O
    initial..22.5...22.5.....0......0
    change..-22.5..-22.5....+22.5..22.5
    equil.....0......0.......22.5....

    So you can see that the CH3COOH is exactly neutralized by the NaOH and neither free CH3COOH nor free NaOH come through the reaction. We have just a solution of CH3COONa (NaAc) and that is hydrolyzed. The pH is determined by that. (Ac^-) = mmoles/mL = 22.5/45 = 0.5M
    ..........Ac^- + HOH ==> HAc + OH^-
    initial..0.50....0.......0.....0
    change.....-x..............x......x
    equil......0.50-x..........x.......x

    Kb for Ac^- = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)
    Kw = 1E-14 for water.
    Ka = 1.8E-5 for acetic acid
    x = (HAc) = (OH^-)
    (Ac^-) = 0.50-x
    Substitute, solve for x (which = OH^-), convert to pOH, then to pH.

  • chemistry -

    gibbe that answer b0ss

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